SUMMERZETT/gemma-2-2b-proofnet-finetuned
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Rudin|exercise_1_1a | If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational. | \begin{proof}
If $r$ and $r+x$ were both rational, then $x=r+x-r$ would also be rational.
\end{proof} | theorem exercise_1_1a
(x : β) (y : β) :
( irrational x ) -> irrational ( x + y ) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_2 | Prove that there is no rational number whose square is $12$. | \begin{proof}
Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor. It follows that $m$ must be even, and therefore $n$ must be odd. Let $m=2 r$. Then we have $r^2=3 n^2$, so that $r$ is also odd. Let $r=2 s+1$ and $n=2 t+1$. Then
$$
4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3,
$$
so that
$$
4\left(s^2+s-3 t^2-3 t\right)=2 .
$$
But this is absurd, since 2 cannot be a multiple of 4 .
\end{proof} | theorem exercise_1_2 : Β¬ β (x : β), ( x ^ 2 = 12 ) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_5 | Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$. | \begin{proof}
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$. For brevity, let $\alpha=-\sup (-A)$. We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$. Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$. It follows that $x \geq-\sup (-A)$, i.e., $\alpha \leq x$. Thus $\alpha$ is a lower bound of $A$.
Now let $\beta$ be any lower bound of $A$. This means $\beta \leq x$ for all $x$ in $A$. Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$. This means $-\beta$ is an upper bound of $-A$. Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$, and so $-\sup (-A)$ is the greatest lower bound of $A$.
\end{proof} | theorem exercise_1_5 (A minus_A : set β) (hA : A.nonempty)
(hA_bdd_below : bdd_below A) (hminus_A : minus_A = {x | -x β A}) :
Inf A = Sup minus_A := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_11a | If $z$ is a complex number, prove that there exists an $r\geq 0$ and a complex number $w$ with $| w | = 1$ such that $z = rw$. | \begin{proof}
If $z=0$, we take $r=0, w=1$. (In this case $w$ is not unique.) Otherwise we take $r=|z|$ and $w=z /|z|$, and these choices are unique, since if $z=r w$, we must have $r=r|w|=|r w|=|z|, z / r$
\end{proof} | theorem exercise_1_11a (z : β) :
β (r : β) (w : β), abs w = 1 β§ z = r * w := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_13 | If $x, y$ are complex, prove that $||x|-|y|| \leq |x-y|$. | \begin{proof}
Since $x=x-y+y$, the triangle inequality gives
$$
|x| \leq|x-y|+|y|
$$
so that $|x|-|y| \leq|x-y|$. Similarly $|y|-|x| \leq|x-y|$. Since $|x|-|y|$ is a real number we have either ||$x|-| y||=|x|-|y|$ or ||$x|-| y||=|y|-|x|$. In either case, we have shown that ||$x|-| y|| \leq|x-y|$.
\end{proof} | theorem exercise_1_13 (x y : β) :
|(abs x) - (abs y)| β€ abs (x - y) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_16a | Suppose $k \geq 3, x, y \in \mathbb{R}^k, |x - y| = d > 0$, and $r > 0$. Prove that if $2r > d$, there are infinitely many $z \in \mathbb{R}^k$ such that $|z-x|=|z-y|=r$. | \begin{proof}
(a) Let w be any vector satisfying the following two equations:
$$
\begin{aligned}
\mathbf{w} \cdot(\mathbf{x}-\mathbf{y}) &=0, \\
|\mathbf{w}|^2 &=r^2-\frac{d^2}{4} .
\end{aligned}
$$
From linear algebra it is known that all but one of the components of a solution $\mathbf{w}$ of the first equation can be arbitrary. The remaining component is then uniquely determined. Also, if $w$ is any non-zero solution of the first equation, there is a unique positive number $t$ such that $t$ w satisfies both equations. (For example, if $x_1 \neq y_1$, the first equation is satisfied whenever
$$
z_1=\frac{z_2\left(x_2-y_2\right)+\cdots+z_k\left(x_k-y_k\right)}{y_1-x_1} .
$$
If $\left(z_1, z_2, \ldots, z_k\right)$ satisfies this equation, so does $\left(t z_1, t z_2, \ldots, t z_k\right)$ for any real number $t$.) Since at least two of these components can vary independently, we can find a solution with these components having any prescribed ratio. This ratio does not change when we multiply by the positive number $t$ to obtain a solution of both equations. Since there are infinitely many ratios, there are infinitely many distinct solutions. For each such solution $\mathbf{w}$ the vector $\mathbf{z}=$ $\frac{1}{2} \mathrm{x}+\frac{1}{2} \mathrm{y}+\mathrm{w}$ is a solution of the required equation. For
$$
\begin{aligned}
|\mathrm{z}-\mathbf{x}|^2 &=\left|\frac{\mathbf{y}-\mathbf{x}}{2}+\mathbf{w}\right|^2 \\
&=\left|\frac{\mathbf{y}-\mathbf{x}}{2}\right|^2+2 \mathbf{w} \cdot \frac{\mathbf{x}-\mathbf{y}}{2}+|\mathbf{w}|^2 \\
&=\frac{d^2}{4}+0+r^2-\frac{d^2}{4} \\
&=r^2
\end{aligned}
$$
and a similar relation holds for $|z-y|^2$.
\end{proof} | theorem exercise_1_16a
(n : β)
(d r : β)
(x y z : euclidean_space β (fin n)) -- R^n
(hβ : n β₯ 3)
(hβ : βx - yβ = d)
(hβ : d > 0)
(hβ : r > 0)
(hβ
: 2 * r > d)
: set.infinite {z : euclidean_space β (fin n) | βz - xβ = r β§ βz - yβ = r} := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_18a | If $k \geq 2$ and $\mathbf{x} \in R^{k}$, prove that there exists $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$ | \begin{proof}
If $\mathbf{x}$ has any components equal to 0 , then $\mathbf{y}$ can be taken to have the corresponding components equal to 1 and all others equal to 0 . If all the components of $\mathbf{x}$ are nonzero, $\mathbf{y}$ can be taken as $\left(-x_2, x_1, 0, \ldots, 0\right)$. This is, of course, not true when $k=1$, since the product of two nonzero real numbers is nonzero.
\end{proof} | theorem exercise_1_18a
(n : β)
(h : n > 1)
(x : euclidean_space β (fin n)) -- R^n
: β (y : euclidean_space β (fin n)), y β 0 β§ (inner x y) = (0 : β) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_1_19 | Suppose $a, b \in R^k$. Find $c \in R^k$ and $r > 0$ such that $|x-a|=2|x-b|$ if and only if $| x - c | = r$. Prove that $3c = 4b - a$ and $3r = 2 |b - a|$. | \begin{proof}
Since the solution is given to us, all we have to do is verify it, i.e., we need to show that the equation
$$
|\mathrm{x}-\mathrm{a}|=2|\mathrm{x}-\mathrm{b}|
$$
is equivalent to $|\mathrm{x}-\mathbf{c}|=r$, which says
$$
\left|\mathbf{x}-\frac{4}{3} \mathbf{b}+\frac{1}{3} \mathbf{a}\right|=\frac{2}{3}|\mathbf{b}-\mathbf{a}| .
$$
If we square both sides of both equations, we an equivalent pair of equations, the first of which reduces to
$$
3|\mathbf{x}|^2+2 \mathbf{a} \cdot \mathbf{x}-8 \mathbf{b} \cdot \mathbf{x}-|\mathbf{a}|^2+4|\mathbf{b}|^2=0,
$$
and the second of which reduces to this equation divided by 3 . Hence these equations are indeed equivalent.
\end{proof} | theorem exercise_1_19
(n : β)
(a b c x : euclidean_space β (fin n))
(r : β)
(hβ : r > 0)
(hβ : 3 β’ c = 4 β’ b - a)
(hβ : 3 * r = 2 * βx - bβ)
: βx - aβ = 2 * βx - bβ β βx - cβ = r := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_2_24 | Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. | \begin{proof}
We observe that if the process of constructing $x_j$ did not terminate, the result would be an infinite set of points $x_j, j=1,2, \ldots$, such that $d\left(x_i, x_j\right) \geq \delta$ for $i \neq j$. It would then follow that for any $x \in X$, the open ball $B_{\frac{\delta}{2}}(x)$ contains at most one point of the infinite set, hence that no point could be a limit point of this set, contrary to hypothesis. Hence $X$ is totally bounded, i.e., for each $\delta>0$ there is a finite set $x_1, \ldots, x_{N\delta}$such that $X=\bigcup_{j / 1}^{N\delta} B_\delta\left(x_j\right)$
Let $x_{n_1}, \ldots, x_{n N_n}$ be such that $X=\bigcup_{j / 1}^{N_n} B_{\frac{1}{n}}\left(x_{n j}\right), n=1,2, \ldots$ We claim that $\left\{x_{n j}: 1 \leq j \leq N_n ; n=1,2, \ldots\right\}$ is a countable dense subset of $X$. Indeed
25
if $x \in X$ and $\delta>0$, then $x \in B_{\frac{1}{n}}\left(x_{n j}\right)$ for some $x_{n j}$ for some $n>\frac{1}{\delta}$, and hence $d\left(x, x_{n j}\right)<\delta$. By definition, this means that $\left\{x_{n j}\right\}$ is dense in $X$.
\end{proof} | theorem exercise_2_24 {X : Type*} [metric_space X]
(hX : β (A : set X), infinite A β β (x : X), x β closure A) :
separable_space X := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_2_27a | Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that $P$ is perfect. | \begin{proof}
We see that $E \cap W$ is at most countable, being a countable union of at-most-countable sets. It remains to show that $P=W^c$, and that $P$ is perfect.
\end{proof} | theorem exercise_2_27a (k : β) (E P : set (euclidean_space β (fin k)))
(hE : E.nonempty β§ Β¬ set.countable E)
(hP : P = {x | β U β π x, Β¬ set.countable (P β© E)}) :
is_closed P β§ P = {x | cluster_pt x (π P)} := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_2_28 | Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. | \begin{proof}
If $E$ is closed, it contains all its limit points, and hence certainly all its condensation points. Thus $E=P \cup(E \backslash P)$, where $P$ is perfect (the set of all condensation points of $E$ ), and $E \backslash P$ is at most countable.
Since a perfect set in a separable metric space has the same cardinality as the real numbers, the set $P$ must be empty if $E$ is countable. The at-mostcountable set $E \backslash P$ cannot be perfect, hence must have isolated points if it is nonempty.
\end{proof} | theorem exercise_2_28 (X : Type*) [metric_space X] [separable_space X]
(A : set X) (hA : is_closed A) :
β Pβ Pβ : set X, A = Pβ βͺ Pβ β§
is_closed Pβ β§ Pβ = {x | cluster_pt x (π Pβ)} β§
set.countable Pβ := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_1a | Prove that convergence of $\left\{s_{n}\right\}$ implies convergence of $\left\{\left|s_{n}\right|\right\}$. | \begin{proof}
Let $\varepsilon>0$. Since the sequence $\left\{s_n\right\}$ is a Cauchy sequence, there exists $N$ such that $\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. We then have $\left| |s_m| - |s_n| \right| \leq\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. Hence the sequence $\left\{\left|s_n\right|\right\}$ is also a Cauchy sequence, and therefore must converge.
\end{proof} | theorem exercise_3_1a
(f : β β β)
(h : β (a : β), tendsto (Ξ» (n : β), f n) at_top (π a))
: β (a : β), tendsto (Ξ» (n : β), |f n|) at_top (π a) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_3 | If $s_{1}=\sqrt{2}$, and $s_{n+1}=\sqrt{2+\sqrt{s_{n}}} \quad(n=1,2,3, \ldots),$ prove that $\left\{s_{n}\right\}$ converges, and that $s_{n}<2$ for $n=1,2,3, \ldots$. | \begin{proof}
Since $\sqrt{2}<2$, it is manifest that if $s_n<2$, then $s_{n+1}<\sqrt{2+2}=2$. Hence it follows by induction that $\sqrt{2}<s_n<2$ for all $n$. In view of this fact,it also follows that $\left(s_n-2\right)\left(s_n+1\right)<0$ for all $n>1$, i.e., $s_n>s_n^2-2=s_{n-1}$. Hence the sequence is an increasing sequence that is bounded above (by 2 ) and so converges. Since the limit $s$ satisfies $s^2-s-2=0$, it follows that the limit is 2.
\end{proof} | theorem exercise_3_3
: β (x : β), tendsto f at_top (π x) β§ β n, f n < 2 := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_6a | Prove that $\lim_{n \rightarrow \infty} \sum_{i<n} a_i = \infty$, where $a_i = \sqrt{i + 1} -\sqrt{i}$. | \begin{proof}
(a) Multiplying and dividing $a_n$ by $\sqrt{n+1}+\sqrt{n}$, we find that $a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}$, which is larger than $\frac{1}{2 \sqrt{n+1}}$. The series $\sum a_n$ therefore diverges by comparison with the $p$ series $\left(p=\frac{1}{2}\right)$.
\end{proof} | theorem exercise_3_6a
: tendsto (Ξ» (n : β), (β i in finset.range n, g i)) at_top at_top := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_8 | If $\Sigma a_{n}$ converges, and if $\left\{b_{n}\right\}$ is monotonic and bounded, prove that $\Sigma a_{n} b_{n}$ converges. | \begin{proof}
We shall show that the partial sums of this series form a Cauchy sequence, i.e., given $\varepsilon>0$ there exists $N$ such that $\left|\sum_{k=m+1}^n a_k b_k\right|\langle\varepsilon$ if $n\rangle$ $m \geq N$. To do this, let $S_n=\sum_{k=1}^n a_k\left(S_0=0\right)$, so that $a_k=S_k-S_{k-1}$ for $k=1,2, \ldots$ Let $M$ be an uppper bound for both $\left|b_n\right|$ and $\left|S_n\right|$, and let $S=\sum a_n$ and $b=\lim b_n$. Choose $N$ so large that the following three inequalities hold for all $m>N$ and $n>N$ :
$$
\left|b_n S_n-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m S_m-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m-b_n\right|<\frac{\varepsilon}{3 M} .
$$
Then if $n>m>N$, we have, from the formula for summation by parts,
$$
\sum_{k=m+1}^n a_n b_n=b_n S_n-b_m S_m+\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k
$$
Our assumptions yield immediately that $\left|b_n S_n-b_m S_m\right|<\frac{2 \varepsilon}{3}$, and
$$
\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k\right| \leq M \sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right| .
$$
Since the sequence $\left\{b_n\right\}$ is monotonic, we have
$$
\sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right|=\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right)\right|=\left|b_m-b_n\right|<\frac{\varepsilon}{3 M},
$$
from which the desired inequality follows.
\end{proof} | theorem exercise_3_8
(a b : β β β)
(h1 : β y, (tendsto (Ξ» n, (β i in (finset.range n), a i)) at_top (π y)))
(h2 : monotone b)
(h3 : metric.bounded (set.range b)) :
β y, tendsto (Ξ» n, (β i in (finset.range n), (a i) * (b i))) at_top (π y) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_20 | Suppose $\left\{p_{n}\right\}$ is a Cauchy sequence in a metric space $X$, and some sequence $\left\{p_{n l}\right\}$ converges to a point $p \in X$. Prove that the full sequence $\left\{p_{n}\right\}$ converges to $p$. | \begin{proof}
Let $\varepsilon>0$. Choose $N_1$ so large that $d\left(p_m, p_n\right)<\frac{\varepsilon}{2}$ if $m>N_1$ and $n>N_1$. Then choose $N \geq N_1$ so large that $d\left(p_{n_k}, p\right)<\frac{\varepsilon}{2}$ if $k>N$. Then if $n>N$, we have
$$
d\left(p_n, p\right) \leq d\left(p_n, p_{n_{N+1}}\right)+d\left(p_{n_{N+1}}, p\right)<\varepsilon
$$
For the first term on the right is less than $\frac{\varepsilon}{2}$ since $n>N_1$ and $n_{N+1}>N+1>$ $N_1$. The second term is less than $\frac{\varepsilon}{2}$ by the choice of $N$.
\end{proof} | theorem exercise_3_20 {X : Type*} [metric_space X]
(p : β β X) (l : β) (r : X)
(hp : cauchy_seq p)
(hpl : tendsto (Ξ» n, p (l * n)) at_top (π r)) :
tendsto p at_top (π r) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_3_22 | Suppose $X$ is a nonempty complete metric space, and $\left\{G_{n}\right\}$ is a sequence of dense open sets of $X$. Prove Baire's theorem, namely, that $\bigcap_{1}^{\infty} G_{n}$ is not empty. | \begin{proof}
Let $F_n$ be the complement of $G_n$, so that $F_n$ is closed and contains no open sets. We shall prove that any nonempty open set $U$ contains a point not in any $F_n$, hence in all $G_n$. To this end, we note that $U$ is not contained in $F_1$, so that there is a point $x_1 \in U \backslash F_1$. Since $U \backslash F_1$ is open, there exists $r_1>0$ such that $B_1$, defined as the open ball of radius $r_1$ about $x_1$, is contained in $U \backslash F_1$. Let $E_1$ be the open ball of radius $\frac{r_1}{2}$ about $x_1$, so that the closure of $E_1$ is contained in $B_1$. Now $F_2$ does not contain $E_1$, and so we can find a point $x_2 \in E_1 \backslash F_2$. Since $E_1 \backslash F_2$ is an open set, there exists a positive number $r_2$ such that $B_2$, the open ball of radius $R_2$ about $x_2$, is contained in $E_1 \backslash F_2$, which in turn is contained in $U \backslash\left(F_1 \cup F_2\right)$. We let $E_2$ be the open ball of radius $\frac{r_2}{2}$ about $x_2$, so that $\bar{E}_2 \subseteq B_2$. Proceeding in this way, we construct a sequence of open balls $E_j$, such that $E_j \supseteq \bar{E}_{j+1}$, and the diameter of $E_j$ tends to zero. By the previous exercise, there is a point $x$ belonging to all the sets $\bar{E}_j$, hence to all the sets $U \backslash\left(F_1 \cup F_2 \cup \cdots \cup F_n\right)$. Thus the point $x$ belongs to $U \cap\left(\cap_1^{\infty} G_n\right)$.
\end{proof} | theorem exercise_3_22 (X : Type*) [metric_space X] [complete_space X]
(G : β β set X) (hG : β n, is_open (G n) β§ dense (G n)) :
β x, β n, x β G n := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_2a | If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $f(\overline{E}) \subset \overline{f(E)}$ for every set $E \subset X$. ($\overline{E}$ denotes the closure of $E$). | \begin{proof}
Let $x \in \bar{E}$. We need to show that $f(x) \in \overline{f(E)}$. To this end, let $O$ be any neighborhood of $f(x)$. Since $f$ is continuous, $f^{-1}(O)$ contains (is) a neighborhood of $x$. Since $x \in \bar{E}$, there is a point $u$ of $E$ in $f^{-1}(O)$. Hence $\frac{f(u)}{f(E)} \in O \cap f(E)$. Since $O$ was any neighborhood of $f(x)$, it follows that $f(x) \in \overline{f(E)}$
\end{proof} | theorem exercise_4_2a
{Ξ± : Type} [metric_space Ξ±]
{Ξ² : Type} [metric_space Ξ²]
(f : Ξ± β Ξ²)
(hβ : continuous f)
: β (x : set Ξ±), f '' (closure x) β closure (f '' x) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_4a | Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. | \begin{proof}
To prove that $f(E)$ is dense in $f(X)$, simply use that $f(X)=f(\bar{E}) \subseteq \overline{f(E)}$.
\end{proof} | theorem exercise_4_4a
{Ξ± : Type} [metric_space Ξ±]
{Ξ² : Type} [metric_space Ξ²]
(f : Ξ± β Ξ²)
(s : set Ξ±)
(hβ : continuous f)
(hβ : dense s)
: f '' set.univ β closure (f '' s) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_5a | If $f$ is a real continuous function defined on a closed set $E \subset \mathbb{R}$, prove that there exist continuous real functions $g$ on $\mathbb{R}$ such that $g(x)=f(x)$ for all $x \in E$. | \begin{proof}
Following the hint, let the complement of $E$ consist of a countable collection of finite open intervals $\left(a_k, b_k\right)$ together with possibly one or both of the the semi-infinite intervals $(b,+\infty)$ and $(-\infty, a)$. The function $f(x)$ is already defined at $a_k$ and $b_k$, as well as at $a$ and $b$ (if these last two points exist). Define $g(x)$ to be $f(b)$ for $x>b$ and $f(a)$ for $x<a$ if $a$ and $b$ exist. On the interval $\left(a_k, b_k\right)$ let
$$
g(x)=f\left(a_k\right)+\frac{x-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right) .
$$
Of course we let $g(x)=f(x)$ for $x \in E$. It is now fairly clear that $g(x)$ is continuous. A rigorous proof proceeds as follows. Let $\varepsilon>0$. To choose $\delta>0$ such that $|x-u|<\delta$ implies $|g(x)-g(u)|<\varepsilon$, we consider three cases.
i. If $x>b$, let $\delta=x-b$. Then if $|x-u|<\delta$, it follows that $u>b$ also, so that $g(u)=f(b)=g(x)$, and $|g(u)-g(x)|=0<\varepsilon$. Similarly if $x<a$, let $\delta=a-x$
ii. If $a_k<x<b_k$ and $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x\right)$. Since $|x-u|<\delta$ implies $a_k<u<b_k$, so that $g(u)=f\left(a_k\right)=f\left(b_k\right)=g(x)$, we again have $|g(x)-g(u)|=0<\varepsilon$. If $a_k<x<b_k$ and $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then if $|x-u|<\delta$, we again have $a_k<u<b_k$ and so
$$
|g(x)-g(u)|=\frac{|x-u|}{b_k-a_k}\left|f\left(b_k\right)-f\left(a_k\right)\right|<\varepsilon .
$$
iii. If $x \in E$, let $\delta_1$ be such that $|f(u)-f(x)|<\varepsilon$ if $u \in E$ and $|x-u|<\delta_1$. (Subcase a). If there are points $x_1 \in E \cap\left(x-\delta_1, x\right)$ and $x_2 \in E \cap\left(x, x+\delta_1\right)$, let $\delta=\min \left(x-x_1, x_2-x\right)$. If $|u-x|<\delta$ and $u \in E$, then $|f(u)-f(x)|<\varepsilon$ by definition of $\delta_1$. if $u \notin E$, then, since $x_1, x$, and $x_2$ are all in $E$, it follows that $u \in\left(a_k, b_k\right)$, where $a_k \in E, b_k \in E$, and $\left|a_k-x\right|<\delta$ and $\left|b_k-x\right|<\delta$, so that $\left|f\left(a_k\right)-f(x)\right|<\varepsilon$ and $\left|f\left(b_k\right)-f(x)\right|<\varepsilon$. If $f\left(a_k\right)=f\left(b_k\right)$, then $f(u)=f\left(a_k\right)$ also, and we have $|f(u)-f(x)|<\varepsilon$. If $f\left(a_k\right) \neq f\left(b_k\right)$, then
$$
\begin{aligned}
|f(u)-f(x)| & =\left|f\left(a_k\right)-f(x)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right)\right| \\
& =\left|\frac{b_k-u}{b_k-a_k}\left(f\left(a_k\right)-f(x)\right)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f(x)\right)\right| \\
& <\frac{b_k-u}{b_k-a_k} \varepsilon+\frac{u-a_k}{b_k-a_k} \varepsilon \\
& =\varepsilon
\end{aligned}
$$
(Subcase b). Suppose $x_2$ does not exist, i.e., either $x=a_k$ or $x=a_k$ and $b_k>a_k+\delta_1$. Let us consider the second of these cases and show how to get $|f(u)-f(x)|<\varepsilon$ for $x<u<x+\delta$. If $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\delta_1$. If $u>x$ we have $a_k<u<b_k$ and $f(u)=f\left(a_k\right)=f(x)$. If $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=$ $\min \left(\delta_1, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then, just as in Subcase a, we have $|f(u)-f(x)|<\varepsilon$.
The case when $x=b_k$ for some $k$ and $a_k<x-\delta_1$ is handled in exactly the same way.
\end{proof} | theorem exercise_4_5a
(f : β β β)
(E : set β)
(hβ : is_closed E)
(hβ : continuous_on f E)
: β (g : β β β), continuous g β§ β x β E, f x = g x := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_6 | If $f$ is defined on $E$, the graph of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact. | \begin{proof}
Let $Y$ be the co-domain of the function $f$. We invent a new metric space $E \times Y$ as the set of pairs of points $(x, y), x \in E, y \in Y$, with the metric $\rho\left(\left(x_1, y_1\right),\left(x_2, y_2\right)\right)=d_E\left(x_1, x_2\right)+d_Y\left(y_1, y_2\right)$. The function $\varphi(x)=(x, f(x))$ is then a mapping of $E$ into $E \times Y$.
We claim that the mapping $\varphi$ is continuous if $f$ is continuous. Indeed, let $x \in X$ and $\varepsilon>0$ be given. Choose $\eta>0$ so that $d_Y(f(x), f(u))<\frac{\varepsilon}{2}$ if $d_E(x, y)<\eta$. Then let $\delta=\min \left(\eta, \frac{\varepsilon}{2}\right)$. It is easy to see that $\rho(\varphi(x), \varphi(u))<\varepsilon$ if $d_E(x, u)<\delta$. Conversely if $\varphi$ is continuous, it is obvious from the inequality $\rho(\varphi(x), \varphi(u)) \geq d_Y(f(x), f(u))$ that $f$ is continuous.
From these facts we deduce immediately that the graph of a continuous function $f$ on a compact set $E$ is compact, being the image of $E$ under the continuous mapping $\varphi$. Conversely, if $f$ is not continuous at some point $x$, there is a sequence of points $x_n$ converging to $x$ such that $f\left(x_n\right)$ does not converge to $f(x)$. If no subsequence of $f\left(x_n\right)$ converges, then the sequence $\left\{\left(x_n, f\left(x_n\right)\right\}_{n=1}^{\infty}\right.$ has no convergent subsequence, and so the graph is not compact. If some subsequence of $f\left(x_n\right)$ converges, say $f\left(x_{n_k}\right) \rightarrow z$, but $z \neq f(x)$, then the graph of $f$ fails to contain the limit point $(x, z)$, and hence is not closed. A fortiori it is not compact.
\end{proof} | theorem exercise_4_6
(f : β β β)
(E : set β)
(G : set (β Γ β))
(hβ : is_compact E)
(hβ : G = {(x, f x) | x β E})
: continuous_on f E β is_compact G := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_8b | Let $E$ be a bounded set in $R^{1}$. Prove that there exists a real function $f$ such that $f$ is uniformly continuous and is not bounded on $E$. | \begin{proof}
The function $f(x)=x$ is uniformly continuous on the entire line, but not bounded.
\end{proof} | theorem exercise_4_8b
(E : set β) :
β f : β β β, uniform_continuous_on f E β§ Β¬ metric.bounded (set.image f E) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_12 | A uniformly continuous function of a uniformly continuous function is uniformly continuous. | \begin{proof}
Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be uniformly continuous. Then $g \circ f: X \rightarrow Z$ is uniformly continuous, where $g \circ f(x)=g(f(x))$ for all $x \in X$.
To prove this fact, let $\varepsilon>0$ be given. Then, since $g$ is uniformly continuous, there exists $\eta>0$ such that $d_Z(g(u), g(v))<\varepsilon$ if $d_Y(u, v)<\eta$. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d_Y(f(x), f(y))<\eta$ if $d_X(x, y)<\delta$
It is then obvious that $d_Z(g(f(x)), g(f(y)))<\varepsilon$ if $d_X(x, y)<\delta$, so that $g \circ f$ is uniformly continuous.
\end{proof} | theorem exercise_4_12
{Ξ± Ξ² Ξ³ : Type*} [uniform_space Ξ±] [uniform_space Ξ²] [uniform_space Ξ³]
{f : Ξ± β Ξ²} {g : Ξ² β Ξ³}
(hf : uniform_continuous f) (hg : uniform_continuous g) :
uniform_continuous (g β f) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_19 | Suppose $f$ is a real function with domain $R^{1}$ which has the intermediate value property: if $f(a)<c<f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also, for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous. | \begin{proof}
The contradiction is evidently that $x_0$ is a limit point of the set of $t$ such that $f(t)=r$, yet, $x_0$ does not belong to this set. This contradicts the hypothesis that the set is closed.
\end{proof} | theorem exercise_4_19
{f : β β β} (hf : β a b c, a < b β f a < c β c < f b β β x, a < x β§ x < b β§ f x = c)
(hg : β r : β, is_closed {x | f x = r}) : continuous f := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_4_24 | Assume that $f$ is a continuous real function defined in $(a, b)$ such that $f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$ for all $x, y \in(a, b)$. Prove that $f$ is convex. | \begin{proof}
We shall prove that
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
for all "dyadic rational" numbers, i.e., all numbers of the form $\lambda=\frac{k}{2^n}$, where $k$ is a nonnegative integer not larger than $2^n$. We do this by induction on $n$. The case $n=0$ is trivial (since $\lambda=0$ or $\lambda=1$ ). In the case $n=1$ we have $\lambda=0$ or $\lambda=1$ or $\lambda=\frac{1}{2}$. The first two cases are again trivial, and the third is precisely the hypothesis of the theorem. Suppose the result is proved for $n \leq r$, and consider $\lambda=\frac{k}{2^{r+1}}$. If $k$ is even, say $k=2 l$, then $\frac{k}{2^{r+1}}=\frac{l}{2^r}$, and we can appeal to the induction hypothesis. Now suppose $k$ is odd. Then $1 \leq k \leq 2^{r+1}-1$, and so the numbers $l=\frac{k-1}{2}$ and $m=\frac{k+1}{2}$ are integers with $0 \leq l<m \leq 2^r$. We can now write
$$
\lambda=\frac{s+t}{2},
$$
where $s=\frac{k-1}{2^{r+1}}=\frac{l}{2^r}$ and $t=\frac{k+1}{2^{r+1}}=\frac{m}{2^r}$. We then have
$$
\lambda x+(1-\lambda) y=\frac{[s x+(1-s) y]+[t x+(1-t) y]}{2}
$$
Hence by the hypothesis of the theorem and the induction hypothesis we have
$$
\begin{aligned}
f(\lambda x+(1-\lambda) y) & \leq \frac{f(s x+(1-s) y)+f(t x+(1-t) y)}{2} \\
& \leq \frac{s f(x)+(1-s) f(y)+t f(x)+(1-t) f(y)}{2} \\
&=\left(\frac{s+t}{2}\right) f(x)+\left(1-\frac{s+t}{2}\right) f(y) \\
&=\lambda f(x)+(1-\lambda) f(y)
\end{aligned}
$$
This completes the induction.
Now for each fixed $x$ and $y$ both sides of the inequality
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
are continuous functions of $\lambda$. Hence the set on which this inequality holds (the inverse image of the closed set $[0, \infty)$ under the mapping $\lambda \mapsto \lambda f(x)+(1-$ $\lambda) f(y)-f(\lambda x+(1-\lambda) y))$ is a closed set. Since it contains all the points $\frac{k}{2^n}$, $0 \leq k \leq n, n=1,2, \ldots$, it must contain the closure of this set of points, i.e., it must contain all of $[0,1]$. Thus $f$ is convex.
\end{proof} | theorem exercise_4_24 {f : β β β}
(hf : continuous f) (a b : β) (hab : a < b)
(h : β x y : β, a < x β x < b β a < y β y < b β f ((x + y) / 2) β€ (f x + f y) / 2) :
convex_on β (set.Ioo a b) f := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_5_2 | Suppose $f^{\prime}(x)>0$ in $(a, b)$. Prove that $f$ is strictly increasing in $(a, b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a<x<b)$. | \begin{proof}
For any $c, d$ with $a<c<d<b$ there exists a point $p \in(c, d)$ such that $f(d)-f(c)=f^{\prime}(p)(d-c)>0$. Hence $f(c)<f(d)$
We know from Theorem $4.17$ that the inverse function $g$ is continuous. (Its restriction to each closed subinterval $[c, d]$ is continuous, and that is sufficient.) Now observe that if $f(x)=y$ and $f(x+h)=y+k$, we have
$$
\frac{g(y+k)-g(y)}{k}-\frac{1}{f^{\prime}(x)}=\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}
$$
Since we know $\lim \frac{1}{\varphi(t)}=\frac{1}{\lim \varphi(t)}$ provided $\lim \varphi(t) \neq 0$, it follows that for any $\varepsilon>0$ there exists $\eta>0$ such that
$$
\left|\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}\right|<\varepsilon
$$
if $0<|h|<\eta$. Since $h=g(y+k)-g(y)$, there exists $\delta>0$ such that $0<|h|<\eta$ if $0<|k|<\delta$. The proof is now complete.
\end{proof} | theorem exercise_5_2 {a b : β}
{f g : β β β} (hf : β x β set.Ioo a b, deriv f x > 0)
(hg : g = fβ»ΒΉ)
(hg_diff : differentiable_on β g (set.Ioo a b)) :
differentiable_on β g (set.Ioo a b) β§
β x β set.Ioo a b, deriv g x = 1 / deriv f x := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_5_4 | If $C_{0}+\frac{C_{1}}{2}+\cdots+\frac{C_{n-1}}{n}+\frac{C_{n}}{n+1}=0,$ where $C_{0}, \ldots, C_{n}$ are real constants, prove that the equation $C_{0}+C_{1} x+\cdots+C_{n-1} x^{n-1}+C_{n} x^{n}=0$ has at least one real root between 0 and 1. | \begin{proof}
Consider the polynomial
$$
p(x)=C_0 x+\frac{C_1}{2} x^2+\cdots+\frac{C_{n-1}}{n} x^n+\frac{C_n}{n+1} x^{n+1},
$$
whose derivative is
$$
p^{\prime}(x)=C_0+C_1 x+\cdots+C_{n-1} x^{n-1}+C_n x^n .
$$
It is obvious that $p(0)=0$, and the hypothesis of the problem is that $p(1)=0$. Hence Rolle's theorem implies that $p^{\prime}(x)=0$ for some $x$ between 0 and 1 .
\end{proof} | theorem exercise_5_4 {n : β}
(C : β β β)
(hC : β i in (finset.range (n + 1)), (C i) / (i + 1) = 0) :
β x, x β (set.Icc (0 : β) 1) β§ β i in finset.range (n + 1), (C i) * (x^i) = 0 := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_5_6 | Suppose (a) $f$ is continuous for $x \geq 0$, (b) $f^{\prime}(x)$ exists for $x>0$, (c) $f(0)=0$, (d) $f^{\prime}$ is monotonically increasing. Put $g(x)=\frac{f(x)}{x} \quad(x>0)$ and prove that $g$ is monotonically increasing. | \begin{proof}
Put
$$
g(x)=\frac{f(x)}{x} \quad(x>0)
$$
and prove that $g$ is monotonically increasing.
By the mean-value theorem
$$
f(x)=f(x)-f(0)=f^{\prime}(c) x
$$
for some $c \in(0, x)$. Since $f^{\prime}$ is monotonically increasing, this result implies that $f(x)<x f^{\prime}(x)$. It therefore follows that
$$
g^{\prime}(x)=\frac{x f^{\prime}(x)-f(x)}{x^2}>0,
$$
so that $g$ is also monotonically increasing.
\end{proof} | theorem exercise_5_6
{f : β β β}
(hf1 : continuous f)
(hf2 : β x, differentiable_at β f x)
(hf3 : f 0 = 0)
(hf4 : monotone (deriv f)) :
monotone_on (Ξ» x, f x / x) (set.Ioi 0) := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Rudin|exercise_5_15 | Suppose $a \in R^{1}, f$ is a twice-differentiable real function on $(a, \infty)$, and $M_{0}, M_{1}, M_{2}$ are the least upper bounds of $|f(x)|,\left|f^{\prime}(x)\right|,\left|f^{\prime \prime}(x)\right|$, respectively, on $(a, \infty)$. Prove that $M_{1}^{2} \leq 4 M_{0} M_{2} .$ | \begin{proof}
The inequality is obvious if $M_0=+\infty$ or $M_2=+\infty$, so we shall assume that $M_0$ and $M_2$ are both finite. We need to show that
$$
\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2}
$$
for all $x>a$. We note that this is obvious if $M_2=0$, since in that case $f^{\prime}(x)$ is constant, $f(x)$ is a linear function, and the only bounded linear function is a constant, whose derivative is zero. Hence we shall assume from now on that $0<M_2<+\infty$ and $0<M_0<+\infty$.
Following the hint, we need only choose $h=\sqrt{\frac{M_0}{M_2}}$, and we obtain
$$
\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2},
$$
which is precisely the desired inequality.
The case of equality follows, since the example proposed satisfies
$$
f(x)=1-\frac{2}{x^2+1}
$$
for $x \geq 0$. We see easily that $|f(x)| \leq 1$ for all $x>-1$. Now $f^{\prime}(x)=\frac{4 x}{\left(x^2+1\right)^2}$ for $x>0$ and $f^{\prime}(x)=4 x$ for $x<0$. It thus follows from Exercise 9 above that $f^{\prime}(0)=0$, and that $f^{\prime}(x)$ is continuous. Likewise $f^{\prime \prime}(x)=4$ for $x<0$ and $f^{\prime \prime}(x)=\frac{4-4 x^2}{\left(x^2+1\right)^3}=-4 \frac{x^2-1}{\left(x^2+1\right)^3}$. This shows that $\left|f^{\prime \prime}(x)\right|<4$ for $x>0$ and also that $\lim _{x \rightarrow 0} f^{\prime \prime}(x)=4$. Hence Exercise 9 again implies that $f^{\prime \prime}(x)$ is continuous and $f^{\prime \prime}(0)=4$.
On $n$-dimensional space let $\mathbf{f}(x)=\left(f_1(x), \ldots, f_n(x)\right), M_0=\sup |\mathbf{f}(x)|$, $M_1=\sup \left|\mathbf{f}^{\prime}(x)\right|$, and $M_2=\sup \left|\mathbf{f}^{\prime \prime}(x)\right|$. Just as in the numerical case, there is nothing to prove if $M_2=0$ or $M_0=+\infty$ or $M_2=+\infty$, and so we assume $0<M_0<+\infty$ and $0<M_2<\infty$. Let $a$ be any positive number less than $M_1$, let $x_0$ be such that $\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$, and let $\mathbf{u}=\frac{1}{\left|\mathbf{f}^{\prime}\left(x_0\right)\right|} \mathbf{f}^{\prime}\left(x_0\right)$. Consider the real-valued function $\varphi(x)=\mathrm{u} \cdot \mathrm{f}(x)$. Let $N_0, N_1$, and $N_2$ be the suprema of $|\varphi(x)|,\left|\varphi^{\prime}(x)\right|$, and $\left|\varphi^{\prime \prime}(x)\right|$ respectively. By the Schwarz inequality we have (since $|\mathbf{u}|=1) N_0 \leq M_0$ and $N_2 \leq M_2$, while $N_1 \geq \varphi\left(x_0\right)=\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$. We therefore have $a^2<4 N_0 N_2 \leq 4 M_0 M_2$. Since $a$ was any positive number less than $M_1$, we have $M_1^2 \leq 4 M_0 M_2$, i.e., the result holds also for vector-valued functions.
Equality can hold on any $R^n$, as we see by taking $\mathbf{f}(x)=(f(x), 0, \ldots, 0)$ or $\mathbf{f}(x)=(f(x), f(x), \ldots, f(x))$, where $f(x)$ is a real-valued function for which equality holds.
\end{proof} | theorem exercise_5_15 {f : β β β} (a M0 M1 M2 : β)
(hf' : differentiable_on β f (set.Ici a))
(hf'' : differentiable_on β (deriv f) (set.Ici a))
(hM0 : M0 = Sup {(| f x | )| x β (set.Ici a)})
(hM1 : M1 = Sup {(| deriv f x | )| x β (set.Ici a)})
(hM2 : M2 = Sup {(| deriv (deriv f) x | )| x β (set.Ici a)}) :
(M1 ^ 2) β€ 4 * M0 * M2 := | import .common
open real complex
open topological_space
open filter
open_locale real
open_locale topology
open_locale big_operators
open_locale complex_conjugate
open_locale filter
noncomputable theory
|
Munkres|exercise_13_1 | Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$. | \begin{proof}
Since, from the given hypothesis given any $x \in A$ there exists an open set containing $x$ say, $U_x$ such that $U_x \subset A$. Thus, we claim that
$$
A=\bigcup_{x \in A} U_x
$$
Observe that if we prove the above claim, then $A$ will be open, being a union of arbitrary open sets. Since, for each $x \in A, U_x \subset A \Longrightarrow \cup U_x \subset A$. For the converse, observe that given any $x \in A, x \in U_x$ and hence in the union. Thus we proved our claim, and hence $A$ is an open set.
\end{proof} | theorem exercise_13_1 (X : Type*) [topological_space X] (A : set X)
(h1 : β x β A, β U : set X, x β U β§ is_open U β§ U β A) :
is_open A := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_13_4a1 | If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcap \mathcal{T}_\alpha$ is a topology on $X$. | \begin{proof}
Since $\emptyset$ and $X$ belong to $\mathcal{T}_\alpha$ for each $\alpha$, they belong to $\bigcap_\alpha \mathcal{T}_\alpha$. Let $\left\{V_\beta\right\}_\beta$ be a collection of open sets in $\bigcap_\alpha \mathcal{T}_\alpha$. For any fixed $\alpha$ we have $\cup_\beta V_\beta \in \mathcal{T}_\alpha$ since $\mathcal{T}_\alpha$ is a topology on $X$, so $\bigcup_\beta V_\beta \in \bigcap_\alpha \mathcal{T}_\alpha$. Similarly, if $U_1, \ldots, U_n$ are elements of $\bigcap_\alpha \mathcal{T}_\alpha$, then for each $\alpha$ we have $\bigcup_{i=1}^n U_i \in \mathcal{T}_\alpha$ and therefore $\bigcup_{i=1}^n U_i \in \bigcap_\alpha \mathcal{T}_\alpha$. It follows that $\bigcap_\alpha \mathcal{T}_\alpha$ is a topology on $X$.
\end{proof} | theorem exercise_13_4a1 (X I : Type*) (T : I β set (set X)) (h : β i, is_topology X (T i)) :
is_topology X (β i : I, T i) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_13_4b1 | Let $\mathcal{T}_\alpha$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$. | \begin{proof}
(b) First we prove that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$. Uniqueness of such topology is clear. For each $\alpha$ let $\mathcal{B}_\alpha$ be a basis for $\mathcal{T}_\alpha$. Let $\mathcal{T}$ be the topology generated by the subbasis $\mathcal{S}=\bigcup_\alpha \mathcal{B}_\alpha$. Then the collection $\mathcal{B}$ of all finite intersections of elements of $\mathcal{S}$ is a basis for $\mathcal{T}$. Clearly $\mathcal{T}_\alpha \subset \mathcal{T}$ for all $\alpha$. We now prove that if $\mathcal{O}$ is a topology on $X$ such that $\mathcal{T}_\alpha \subset \mathcal{O}$ for all $\alpha$, then $\mathcal{T} \subset \mathcal{O}$. Given such $\mathcal{O}$, we have $\mathcal{B}_\alpha \subset \mathcal{O}$ for all $\alpha$, so $\mathcal{S} \subset \mathcal{O}$. Since $\mathcal{O}$ is a topology, it must contain all finite intersections of elements of $\mathcal{S}$, so $\mathcal{B} \subset \mathcal{O}$ and hence $\mathcal{T} \subset \mathcal{O}$. We conclude that the topology $\mathcal{T}$ generated by the subbasis $\mathcal{S}=\cup_\alpha \mathcal{B}_\alpha$ is the unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$.
\end{proof} | theorem exercise_13_4b1 (X I : Type*) (T : I β set (set X)) (h : β i, is_topology X (T i)) :
β! T', is_topology X T' β§ (β i, T i β T') β§
β T'', is_topology X T'' β (β i, T i β T'') β T'' β T' := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_13_5a | Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. | \begin{proof}
Let $\mathcal{T}$ be the topology generated by $\mathcal{A}$ and let $\mathcal{O}$ be the intersection of all topologies on $X$ that contains $\mathcal{A}$. Clearly $\mathcal{O} \subset \mathcal{T}$ since $\mathcal{T}$ is a topology on $X$ that contain $\mathcal{A}$. Conversely, let $U \in \mathcal{T}$, so that $U$ is a union of elements of $\mathcal{A}$. Since each of this elements is also an element of $\mathcal{O}$, their union $U$ belongs to $\mathcal{O}$. Thus $\mathcal{T} \subset \mathcal{O}$ and the equality holds.
\end{proof} | theorem exercise_13_5a {X : Type*}
[topological_space X] (A : set (set X)) (hA : is_topological_basis A) :
generate_from A = generate_from (sInter {T | is_topology X T β§ A β T}) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_13_6 | Show that the lower limit topology $\mathbb{R}_l$ and $K$-topology $\mathbb{R}_K$ are not comparable. | \begin{proof}
Let $\mathcal{T}_{\ell}$ and $\mathcal{T}_K$ denote the topologies of $\mathbb{R}_{\ell}$ and $\mathbb{R}_K$ respectively. Given the basis element $[0,1)$ for $\mathcal{T}_{\ell}$, there is no basis element for $\mathcal{T}_K$ containing 0 and contained in $[0,1)$, so $\mathcal{T}_{\ell} \not \subset \mathcal{T}_K$. Similarly, given the basis element $(-1,1) \backslash K$ for $\mathcal{T}_K$, there is no basis element for $\mathcal{T}_{\ell}$ containing 0 contained in $(-1,1) \backslash K$, so $\mathcal{T}_K \not \subset \mathcal{T}_{\ell}$.
\end{proof} | theorem exercise_13_6 :
Β¬ (β U, Rl.is_open U β K_topology.is_open U) β§ Β¬ (β U, K_topology.is_open U β Rl.is_open U) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_13_8b | Show that the collection $\{(a,b) \mid a < b, a \text{ and } b \text{ rational}\}$ is a basis that generates a topology different from the lower limit topology on $\mathbb{R}$. | \begin{proof}
(b) $\mathcal{C}$ is a basis for a topology on $\mathbb{R}$ since the union of its elements is $\mathbb{R}$ and the intersection of two elements of $\mathcal{C}$ is either empty or another element of $\mathcal{C}$. Now consider $[r, s)$ where $r$ is any irrational number and $s$ is any real number greater than $r$. Then $[r, s)$ is a basis element for the topology of $\mathbb{R}_{\ell}$, but $[r, s)$ is not a union of elements of $\mathcal{C}$. Indeed, suppose that $[r, s)=\cup_\alpha\left[a_\alpha, b_\alpha\right)$ for rationals $a_\alpha, b_\alpha$. Then $r \in\left[a_\alpha, b_\alpha\right)$ for some $\alpha$. Since $r$ is irrational we must have $a_\alpha<r$, but then $a_\alpha \notin[r, s)$, a contradiction. It follows that the topology generated by $\mathcal{C}$ is strictly coarser than the lower limit topology on $\mathbb{R}$.
\end{proof} | theorem exercise_13_8b :
(topological_space.generate_from {S : set β | β a b : β, a < b β§ S = Ico a b}).is_open β
(lower_limit_topology β).is_open := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_16_4 | A map $f: X \rightarrow Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_{1}: X \times Y \rightarrow X$ and $\pi_{2}: X \times Y \rightarrow Y$ are open maps. | \begin{proof}
Exercise 16.4. Let $U \times V$ be a (standard) basis element for $X \times Y$, so that $U$ is open in $X$ and $V$ is open in $Y$. Then $\pi_1(U \times V)=U$ is open in $X$ and $\pi_2(U \times V)=V$ is open in $Y$. Since arbitrary maps and unions satisfy $f\left(\bigcup_\alpha W_\alpha\right)=\bigcup_\alpha f\left(W_\alpha\right)$, it follows that $\pi_1$ and $\pi_2$ are open maps.
\end{proof} | theorem exercise_16_4 {X Y : Type*} [topological_space X] [topological_space Y]
(Οβ : X Γ Y β X)
(Οβ : X Γ Y β Y)
(hβ : Οβ = prod.fst)
(hβ : Οβ = prod.snd) :
is_open_map Οβ β§ is_open_map Οβ := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_17_4 | Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$, and $A-U$ is closed in $X$. | \begin{proof}
Since
$$
X \backslash(U \backslash A)=(X \backslash U) \cup A \text { and } \quad X \backslash(A \backslash U)=(X \backslash A) \cup U,
$$
it follows that $X \backslash(U \backslash A)$ is closed in $X$ and $X \backslash(A \backslash U)$ is open in $X$.
\end{proof} | theorem exercise_17_4 {X : Type*} [topological_space X]
(U A : set X) (hU : is_open U) (hA : is_closed A) :
is_open (U \ A) β§ is_closed (A \ U) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_18_8b | Let $Y$ be an ordered set in the order topology. Let $f, g: X \rightarrow Y$ be continuous. Let $h: X \rightarrow Y$ be the function $h(x)=\min \{f(x), g(x)\}.$ Show that $h$ is continuous. | \begin{proof}
Let $A=\{x \mid f(x) \leq g(x)\}$ and $B=\{x \mid g(x) \leq f(x)\}$. Then $A$ and $B$ are closed in $X$ by (a), $A \cap B=\{x \mid f(x)=g(x)\}$, and $X=A \cup B$. Since $f$ and $g$ are continuous, their restrictions $f^{\prime}: A \rightarrow Y$ and $g^{\prime}: B \rightarrow Y$ are continuous. It follows from the pasting lemma that
$$
h: X \rightarrow Y, \quad h(x)=\min \{f(x), g(x)\}= \begin{cases}f^{\prime}(x) & \text { if } x \in A \\ g^{\prime}(x) & \text { if } x \in B\end{cases}
$$
is continuous
\end{proof} | theorem exercise_18_8b {X Y : Type*} [topological_space X] [topological_space Y]
[linear_order Y] [order_topology Y] {f g : X β Y}
(hf : continuous f) (hg : continuous g) :
continuous (Ξ» x, min (f x) (g x)) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_19_6a | Let $\mathbf{x}_1, \mathbf{x}_2, \ldots$ be a sequence of the points of the product space $\prod X_\alpha$. Show that this sequence converges to the point $\mathbf{x}$ if and only if the sequence $\pi_\alpha(\mathbf{x}_i)$ converges to $\pi_\alpha(\mathbf{x})$ for each $\alpha$. | \begin{proof}
For each $n \in \mathbb{Z}_{+}$, we write $\mathbf{x}_n=\left(x_n^\alpha\right)_\alpha$, so that $\pi_\alpha\left(\mathbf{x}_n\right)=x_n^\alpha$ for each $\alpha$.
First assume that the sequence $\mathbf{x}_1, \mathbf{x}_2, \ldots$ converges to $\mathbf{x}=\left(x_\alpha\right)_\alpha$ in the product space $\prod_\alpha X_\alpha$. Fix an index $\beta$ and let $U$ be a neighbourhood of $\pi_\beta(\mathbf{x})=x_\beta$. Let $V=\prod_\alpha U_\alpha$, where $U_\alpha=X_\alpha$ for each $\alpha \neq \beta$ and $U_\beta=U$. Then $V$ is a neighbourhood of $\mathbf{x}$, so there exists $N \in \mathbb{Z}_{+}$such that $\mathbf{x}_n \in V$ for all $n \geq N$. Therefore $\pi_\beta\left(\mathbf{x}_n\right)=x_n^\beta \in U$ for all $n \geq N$. Since $U$ was arbitrary, it follows that $\pi_\beta\left(\mathbf{x}_1\right), \pi_\beta\left(\mathbf{x}_2\right), \ldots$ converges to $\pi_\beta(\mathbf{x})$. Since $\beta$ was arbitrary, this holds for all indices $\alpha$.
\end{proof} | theorem exercise_19_6a
{n : β}
{f : fin n β Type*} {x : β β Ξ a, f a}
(y : Ξ i, f i)
[Ξ a, topological_space (f a)] :
tendsto x at_top (π y) β β i, tendsto (Ξ» j, (x j) i) at_top (π (y i)) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_21_6a | Define $f_{n}:[0,1] \rightarrow \mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\left(f_{n}(x)\right)$ converges for each $x \in[0,1]$. | \begin{proof}
If $0 \leq x<1$ is fixed, then $f_n(x) \rightarrow 0$ as $n \rightarrow \infty$. As $f_n(1)=1$ for all $n, f_n(1) \rightarrow 1$. Thus $\left(f_n\right)_n$ converges to $f:[0,1] \rightarrow \mathbb{R}$ given by $f(x)=0$ if $x=0$ and $f(1)=1$. The sequence
\end{proof} | theorem exercise_21_6a
(f : β β I β β )
(h : β x n, f n x = x ^ n) :
β x, β y, tendsto (Ξ» n, f n x) at_top (π y) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_21_8 | Let $X$ be a topological space and let $Y$ be a metric space. Let $f_{n}: X \rightarrow Y$ be a sequence of continuous functions. Let $x_{n}$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $\left(f_{n}\right)$ converges uniformly to $f$, then $\left(f_{n}\left(x_{n}\right)\right)$ converges to $f(x)$. | \begin{proof}
Let $d$ be the metric on $Y$. Let $V$ be a neighbourhood of $f(x)$, and let $\varepsilon>0$ be such that $f(x) \in B_d(f(x), \varepsilon) \subset V$. Since $\left(f_n\right)_n$ converges uniformly to $f$, there exists $N_1 \in \mathbb{Z}_{+}$such that $d\left(f_n(x), f(x)\right)<\varepsilon / 2$ for all $x \in X$ and all $n \geq N_1$, so that $d\left(f_n\left(x_n\right), f\left(x_n\right)\right)<\varepsilon / 2$ for all $n \geq N_1$. Moreover, $f$ is continuous, so there exists $N_2 \in \mathbb{Z}_{+}$such that $d\left(f\left(x_n\right), f(x)\right)<\varepsilon / 2$ for all $n \geq N_2$. Thus, if $N>\max \left\{N_1, N_2\right\}$, then
$$
d\left(f_n\left(x_n\right), f(x)\right) \leq d\left(f_n\left(x_n\right), f\left(x_n\right)\right)+d\left(f\left(x_n\right), f(x)\right)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
for all $n \geq N$, so $f_n\left(x_n\right) \in V$ for all $n \geq N$. It follows that $\left(f_n\left(x_n\right)\right)_n$ converges to $f(x)$.
\end{proof} | theorem exercise_21_8
{X : Type*} [topological_space X] {Y : Type*} [metric_space Y]
{f : β β X β Y} {x : β β X}
(hf : β n, continuous (f n))
(xβ : X)
(hx : tendsto x at_top (π xβ))
(fβ : X β Y)
(hh : tendsto_uniformly f fβ at_top) :
tendsto (Ξ» n, f n (x n)) at_top (π (fβ xβ)) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_22_2b | If $A \subset X$, a retraction of $X$ onto $A$ is a continuous map $r: X \rightarrow A$ such that $r(a)=a$ for each $a \in A$. Show that a retraction is a quotient map. | \begin{proof}
The inclusion map $i: A \rightarrow X$ is continuous and $r \circ i=1_A$ is the identity. Thus $r$ is a quotient map by (a).
\end{proof} | theorem exercise_22_2b {X : Type*} [topological_space X]
{A : set X} (r : X β A) (hr : continuous r) (h : β x : A, r x = x) :
quotient_map r := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_23_2 | Let $\left\{A_{n}\right\}$ be a sequence of connected subspaces of $X$, such that $A_{n} \cap A_{n+1} \neq \varnothing$ for all $n$. Show that $\bigcup A_{n}$ is connected. | \begin{proof}
Suppose that $\bigcup_n A_n=B \cup C$, where $B$ and $C$ are disjoint open subsets of $\bigcup_n A_n$. Since $A_1$ is connected and a subset of $B \cup C$, by Lemma $23.2$ it lies entirely within either $B$ or $C$. Without any loss of generality, we may assume $A_1 \subset B$. Note that given $n$, if $A_n \subset B$ then $A_{n+1} \subset B$, for if $A_{n+1} \subset C$ then $A_n \cap A_{n+1} \subset B \cap C=\emptyset$, in contradiction with the assumption. By induction, $A_n \subset B$ for all $n \in \mathbb{Z}_{+}$, so that $\bigcup_n A_n \subset B$. It follows that $\bigcup_n A_n$ is connected.
\end{proof} | theorem exercise_23_2 {X : Type*}
[topological_space X] {A : β β set X} (hA : β n, is_connected (A n))
(hAn : β n, A n β© A (n + 1) β β
) :
is_connected (β n, A n) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_23_4 | Show that if $X$ is an infinite set, it is connected in the finite complement topology. | \begin{proof}
Suppose that $A$ is a non-empty subset of $X$ that is both open and closed, i.e., $A$ and $X \backslash A$ are finite or all of $X$. Since $A$ is non-empty, $X \backslash A$ is finite. Thus $A$ cannot be finite as $X \backslash A$ is infinite, so $A$ is all of $X$. Therefore $X$ is connected.
\end{proof} | theorem exercise_23_4 {X : Type*} [topological_space X] [cofinite_topology X]
(s : set X) : set.infinite s β is_connected s := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_23_9 | Let $A$ be a proper subset of $X$, and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X \times Y)-(A \times B)$ is connected. | \begin{proof}
This is similar to the proof of Theorem 23.6. Take $c \times d \in(X \backslash A) \times(Y \backslash B)$. For each $x \in X \backslash A$, the set
$$
U_x=(X \times\{d\}) \cup(\{x\} \times Y)
$$
is connected since $X \times\{d\}$ and $\{x\} \times Y$ are connected and have the common point $x \times d$. Then $U=\bigcup_{x \in X \backslash A} U_x$ is connected because it is the union of the connected spaces $U_x$ which have the point $c \times d$ in common. Similarly, for each $y \in Y \backslash B$ the set
$$
V_y=(X \times\{y\}) \cup(\{c\} \times Y)
$$
is connected, so $V=\bigcup_{y \in Y \backslash B} V_y$ is connected. Thus $(X \times Y) \backslash(A \times B)=U \cup V$ is connected since $c \times d$ is a common point of $U$ and $V$.
\end{proof} | theorem exercise_23_9 {X Y : Type*}
[topological_space X] [topological_space Y]
(Aβ Aβ : set X)
(Bβ Bβ : set Y)
(hA : Aβ β Aβ)
(hB : Bβ β Bβ)
(hA : is_connected Aβ)
(hB : is_connected Bβ) :
is_connected ({x | β a b, x = (a, b) β§ a β Aβ β§ b β Bβ} \
{x | β a b, x = (a, b) β§ a β Aβ β§ b β Bβ}) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_24_2 | Let $f: S^{1} \rightarrow \mathbb{R}$ be a continuous map. Show there exists a point $x$ of $S^{1}$ such that $f(x)=f(-x)$. | \begin{proof}
Let $f: S^1 \rightarrow \mathbb{R}$ be continuous. Let $x \in S^1$. If $f(x)=f(-x)$ we are done, so assume $f(x) \neq f(-x)$. Define $g: S^1 \rightarrow \mathbb{R}$ by setting $g(x)=f(x)-f(-x)$. Then $g$ is continuous. Suppose $f(x)>f(-x)$, so that $g(x)>0$. Then $-x \in S^1$ and $g(-x)<0$. By the intermediate value theorem, since $S^1$ is connected and $g(-x)<0<g(x)$, there exists $y \in S^1$ such that $g(y)=0$. i.e, $f(y)=f(-y)$. Similarly, if $f(x)<f(-x)$, then $g(x)<0<g(-x)$ and again the intermediate value theorem gives the result.
\end{proof} | theorem exercise_24_2 {f : (metric.sphere 0 1 : set β) β β}
(hf : continuous f) : β x, f x = f (-x) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_25_4 | Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected. | \begin{proof}
Let $U$ be a open connected set in $X$. By Theorem 25.4, each path component of $U$ is open in $X$, hence open in $U$. Thus, each path component in $U$ is both open and closed in $U$, so must be empty or all of $U$. It follows that $U$ is path-connected.
\end{proof} | theorem exercise_25_4 {X : Type*} [topological_space X]
[loc_path_connected_space X] (U : set X) (hU : is_open U)
(hcU : is_connected U) : is_path_connected U := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_26_11 | Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed connected subsets of $X$ that is simply ordered by proper inclusion. Then $Y=\bigcap_{A \in \mathcal{A}} A$ is connected. | \begin{proof}
Since each $A \in \mathcal{A}$ is closed, $Y$ is closed. Suppose that $C$ and $D$ form a separation of $Y$. Then $C$ and $D$ are closed in $Y$, hence closed in $X$. Since $X$ is compact, $C$ and $D$ are compact by Theorem 26.2. Since $X$ is Hausdorff, by Exercise 26.5, there exist $U$ and $V$ open in $X$ and disjoint containing $C$ and $D$, respectively. We show that
$$
\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))
$$
is not empty. Let $\left\{A_1, \ldots, A_n\right\}$ be a finite subcollection of elements of $\mathcal{A}$. We may assume that $A_i \subsetneq A_{i+1}$ for all $i=1, \ldots, n-1$. Then
$$
\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)=A_1 \backslash(U \cup V) \text {. }
$$
Suppose that $A_1 \backslash(U \cup V)=\emptyset$. Then $A_1 \subset U \cup V$. Since $A_1$ is connected and $U \cap V=\emptyset, A_1$ lies within either $U$ or $V$, say $A_1 \subset U$. Then $Y \subset A_1 \subset U$, so that $C=Y \cap C \subset Y \cap V=\emptyset$, contradicting the fact that $C$ and $D$ form a separation of $Y$. Hence, $\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)$ is non-empty. Therefore, the collection $\{A \backslash(U \cup V) \mid A \in \mathcal{A}\}$ has the finite intersection property, so
$$
\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))=\left(\bigcap_{A \in \mathcal{A}} A\right) \backslash(U \cup V)=Y \backslash(U \cup V)
$$
is non-empty. So there exists $y \in Y$ such that $y \notin U \cup V \supset C \cup D$, contradicting the fact that $C$ and $D$ form a separation of $Y$. We conclude that there is no such separation, so that $Y$ is connected.
\end{proof} | theorem exercise_26_11
{X : Type*} [topological_space X] [compact_space X] [t2_space X]
(A : set (set X)) (hA : β (a b : set X), a β A β b β A β a β b β¨ b β a)
(hA' : β a β A, is_closed a) (hA'' : β a β A, is_connected a) :
is_connected (ββ A) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_27_4 | Show that a connected metric space having more than one point is uncountable. | \begin{proof}
The distance function $d: X \times X \rightarrow \mathbb{R}$ is continuous by Exercise 20.3(a), so given $x \in X$, the function $d_x: X \rightarrow \mathbb{R}$ given by $d_x(y)=d(x, y)$ is continuous by Exercise 19.11. Since $X$ is connected, the image $d_x(X)$ is a connected subspace of $\mathbb{R}$, and contains 0 since $d_x(x)=0$. Thus, if $y \in X$ and $y \neq x$, then $d_x(X)$ contains the set $[0, \delta]$, where $\delta=d_x(y)>0$. Therefore $X$ must be uncountable.
\end{proof} | theorem exercise_27_4
{X : Type*} [metric_space X] [connected_space X] (hX : β x y : X, x β y) :
Β¬ countable (univ : set X) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_28_5 | Show that X is countably compact if and only if every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed nonempty sets of X has a nonempty intersection. | \begin{proof}
We could imitate the proof of Theorem 26.9, but we prove directly each direction. First let $X$ be countable compact and let $C_1 \supset C_2 \supset \cdots$ be a nested sequence of closed nonempty sets of $X$. For each $n \in \mathbb{Z}_{+}, U_n=X \backslash C_n$ is open in $X$. Then $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable collection of open sets with no finite subcollection covering $X$, for if $U_{i_1} \cup \cdots \cup U_{1_n}$ covers $X$, then $C_{i_1} \cap \cdots \cap C_{i_n}$ is empty, contrary to the assumption. Hence $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$does not cover $X$, so there exist $x \in X \backslash \bigcup_{n \in \mathbb{Z}_{+}} U_n=\bigcap_{n \in Z_{+}}\left(X \backslash U_n\right)=\bigcap_{n \in Z_{+}} C_n$.
Conversely, assume that every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed non-empty sets of $X$ has a non-empty intersection and let $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$be a countable open covering of $X$. For each $n$, let $V_n=U_1 \cup \cdots \cup U_n$ and $C_n=X \backslash V_n$. Suppose that no finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. Then each $C_n$ is non-empty, so $C_1 \supset C_2 \supset \cdots$ is a nested sequence of non-empty closed sets and $\bigcap_{n \in \mathbb{Z}_{+}} C_n$ is non-empty by assumption. Then there exists $x \in \bigcap_{n \in \mathbb{Z}_{+}} C_n$, so that $x \notin V_n$ for all $n$, contradicting the fact that $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. It follows that there exists $N \in \mathbb{Z}_{+}$such that $C_N=\emptyset$, so that $X=V_N$ and hence some finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. We deduce that $X$ is countable compact.
\end{proof} | theorem exercise_28_5
(X : Type*) [topological_space X] :
countably_compact X β β (C : β β set X), (β n, is_closed (C n)) β§
(β n, C n β β
) β§ (β n, C n β C (n + 1)) β β x, β n, x β C n := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_29_1 | Show that the rationals $\mathbb{Q}$ are not locally compact. | \begin{proof}
First, we prove that each set $\mathbb{Q} \cap[a, b]$, where $a, b$ are irrational numbers, is not compact. Indeed, since $\mathbb{Q} \cap[a, b]$ is countable, we can write $\mathbb{Q} \cap[a, b]=\left\{q_1, q_2, \ldots\right\}$. Then $\left\{U_i\right\}_{i \in \mathbb{Z}_{+}}$, where $U_i=\mathbb{Q} \cap\left[a, q_i\right)$ for each $i$, is an open covering of $\mathbb{Q} \cap[a, b]$ with no finite subcovering. Now let $x \in \mathbb{Q}$ and suppose that $\mathbb{Q}$ is locally compact at $x$. Then there exists a compact set $C$ containing a neighbourhood $U$ of $x$. Then $U$ contains a set $\mathbb{Q} \cap[a, b]$ where $a, b$ are irrational numbers. Since this set is closed and contained in the compact $C$, it follows $\mathbb{Q} \cap[a, b]$ is compact, a contradiction. Therefore, $\mathbb{Q}$ is not locally compact.
\end{proof} | theorem exercise_29_1 : Β¬ locally_compact_space β := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_29_10 | Show that if $X$ is a Hausdorff space that is locally compact at the point $x$, then for each neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subset U$. | \begin{proof}
Let $U$ be a neighbourhood of $x$. Since $X$ is locally compact at $x$, there exists a compact subspace $C$ of $X$ containing a neighbourhood $W$ of $x$. Then $U \cap W$ is open in $X$, hence in $C$. Thus, $C \backslash(U \cap W)$ is closed in $C$, hence compact. Since $X$ is Hausdorff, there exist disjoint open sets $V_1$ and $V_2$ of $X$ containing $x$ and $C \backslash(U \cap W)$ respectively. Let $V=V_1 \cap U \cap W$. Since $\bar{V}$ is closed in $C$, it is compact. Furthermore, $\bar{V}$ is disjoint from $C \backslash(U \cap W) \supset C \backslash U$, so $\bar{V} \subset U$.
\end{proof} | theorem exercise_29_10 {X : Type*}
[topological_space X] [t2_space X] (x : X)
(hx : β U : set X, x β U β§ is_open U β§ (β K : set X, U β K β§ is_compact K))
(U : set X) (hU : is_open U) (hxU : x β U) :
β (V : set X), is_open V β§ x β V β§ is_compact (closure V) β§ closure V β U := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_30_13 | Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable. | \begin{proof}
Let $\mathcal{U}$ be a collection of disjoint open sets in $X$ and let $A$ be a countable dense subset of $X$.
Since $A$ is dense in $X$, every $U \in \mathcal{U}$ intesects $S$. Therefore, there exists a point $x_U \in U \cap S$.
Let $U_1, U_2 \in \mathcal{U}, U_1 \neq U_2$. Then $x_{U_1} \neq x_{U_2}$ since $U_1 \cap U_2=\emptyset$.
Thus, the function $\mathcal{U} \rightarrow S$ given by $U \mapsto x_U$ is injective and therefore, since $S$ is countable, it follows that $\mathcal{U}$ is countable.
\end{proof} | theorem exercise_30_13 {X : Type*} [topological_space X]
(h : β (s : set X), countable s β§ dense s) (U : set (set X))
(hU : β (x y : set X), x β U β y β U β x β y β x β© y = β
) :
countable U := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_31_2 | Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint. | \begin{proof}
Let $A$ and $B$ be disjoint closed sets. Then there exist disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively.
Since $X \backslash V$ is closed and contains $U$, the closure of $U$ is contained in $X \backslash V$ hence $B$ and closure of $U$ are disjoint.
Repeat steps 1 and 2 for $B$ and $\bar{U}$ instead of $A$ and $B$ respectively and you will have open set $V^{\prime}$ which contains $B$ and its closure doesn't intersect with $\bar{U}$.
\end{proof} | theorem exercise_31_2 {X : Type*}
[topological_space X] [normal_space X] {A B : set X}
(hA : is_closed A) (hB : is_closed B) (hAB : disjoint A B) :
β (U V : set X), is_open U β§ is_open V β§ A β U β§ B β V β§ closure U β© closure V = β
:= | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_32_1 | Show that a closed subspace of a normal space is normal. | \begin{proof}
Let $X$ be a normal space and $Y$ a closed subspace of $X$.
First we shows that $Y$ is a $T_1$-space.
Let $y \in Y$ be any point. Since $X$ is normal, $X$ is also a $T_1$ space and therefore $\{y\}$ is closed in $X$.
Then it follows that $\{y\}=\{y\} \cap Y$ is closed in $Y$ (in relative topology).
Now let's prove that $X$ is a $T_4$-space.
Let $F, G \subseteq Y$ be disjoint closed sets. Since $F$ and $G$ are closed in $Y$ and $Y$ is closed in $X$, it follows that $F$ and $G$ are closed in $X$.
Since $X$ is normal, $X$ is also a $T_4$-space and therefore there exist disjoint open sets $U, V \subseteq$ $X$ such that $F \subseteq U$ and $G \subseteq V$.
However, then $U \cap Y$ and $V \cap Y$ are open disjoint sets in $Y$ (in relative topology) which separate $F$ and $G$.
\end{proof} | theorem exercise_32_1 {X : Type*} [topological_space X]
(hX : normal_space X) (A : set X) (hA : is_closed A) :
normal_space {x // x β A} := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_32_2b | Show that if $\prod X_\alpha$ is regular, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty. | \begin{proof}
Suppose that $X=\prod_\beta X_\beta$ is regular and let $\alpha$ be any index.
We have to prove that $X_\alpha$ satisfies the $T_1$ and the $T_3$ axiom.
Since $X$ is regular, it follows that $X$ is Hausdorff, which then implies that $X_\alpha$ is Hausdorff. However, this implies that $X_\alpha$ satisfies the $T_1$ axiom.
Let now $F \subseteq X_\alpha$ be a closed set and $x \in X_\alpha \backslash F$ a point.
Then $\prod_\beta F_\beta$, where $F_\alpha=F$ and $F_\beta=X_\beta$ for $\beta \neq \alpha$, is a closed set in $X$ since $\left(\prod_\beta F_\beta\right)^c=\prod_\beta U_\beta$, where $U_\alpha=F^c$ and $U_\beta=X_\beta$ for $\beta \neq \alpha$, which is an open set because it is a base element for the product topology.
Since all $X_\beta$ are nonempty, there exists a point $\mathbf{x} \in X$ such that $x_\alpha=x$. Then $\mathbf{x} \notin \prod_\beta F_\beta$.
Now since $X$ is regular (and therefore satisfies the $T_3$ axiom), there exist disjoint open sets $U, V \subseteq X$ such that $\mathbf{x} \in U$ and $\prod_\beta F_\beta \subseteq V$.
Now for every $\beta \neq \alpha$ we have that $x_\beta \in X_\beta=\pi_\beta(V)$. However, since $x_\beta \in \pi_\beta(U)$, it follows that $\pi_\beta(U) \cap \pi_\beta(V) \neq \emptyset$.
Then $U \cap V=\emptyset$ implies that $\pi_\alpha(U) \cap \pi_\alpha(V)=\emptyset$.. Also, $x \in \pi_\alpha(U)$ and $F \subseteq \pi_\alpha(V)$ and $\pi_\alpha(U), \pi_\alpha(V)$ are open sets since $\pi_\alpha$ is an open map.
Therefore, $X_\alpha$ satisfies the $T_3$ axiom.
\end{proof} | theorem exercise_32_2b
{ΞΉ : Type*} {X : ΞΉ β Type*} [β i, topological_space (X i)]
(h : β i, nonempty (X i)) (h2 : regular_space (Ξ i, X i)) :
β i, regular_space (X i) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_32_3 | Show that every locally compact Hausdorff space is regular. | \begin{proof}
Let $X$ be a LCH space.
Then it follows that for every $x \in X$ and for every open neighborhood $U \subseteq X$ of $x$ there exists an open neighborhood $V \subseteq X$ of $x$ such that $\bar{V} \subseteq U$ (and $\bar{V}$ is compact, but this is not important here).
Since $X$ is a Hausdorff space, it satisfies the $T_1$ axiom.
Then it follows that $X$ is regular.
\end{proof} | theorem exercise_32_3 {X : Type*} [topological_space X]
(hX : locally_compact_space X) (hX' : t2_space X) :
regular_space X := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_33_8 | Let $X$ be completely regular, let $A$ and $B$ be disjoint closed subsets of $X$. Show that if $A$ is compact, there is a continuous function $f \colon X \rightarrow [0, 1]$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$. | \begin{proof}
Since $X$ is completely regular $\forall a \in A, \exists f_a: X \rightarrow[0,1]: f_a(a)=0$ and $f_a(B)=\{1\}$. For some $\epsilon_a \in(0,1)$ we have that $U_a:=f_a^{-1}([0, \epsilon))$ is an open neighborhood of $a$ that does not intersect $B$. We therefore have an open covering $\left\{U_a \mid a \in A\right\}$ of $A$, so since $A$ is compact we have a finite subcover $\left\{U_{a_i} \mid 1 \leq i \leq m\right\}$. For each $1 \leq i \leq m$ define
$$
\begin{aligned}
\tilde{f}_{a_i}: X & \rightarrow[0,1] \\
x & \mapsto \frac{\max \left(f_{a_i}(x), \epsilon_{a_i}\right)-\epsilon_{a_i}}{1-\epsilon_{a_i}}
\end{aligned}
$$
so that $\forall x \in U_{a_i}: \tilde{f}_{a_i}(x)=0$ and $\forall x \in B, \forall 1 \leq i \leq m: \tilde{f}_{a_i}(x)=1$, and define $f:=$ $\prod_{i=1}^m \tilde{f}_{a_i}$. Then since $A \subset \cup_{i=1}^m U_{a_i}$ we have that $f(A)=\{0\}$ and also we have $f(B)=\{1\}$.
\end{proof} | theorem exercise_33_8
(X : Type*) [topological_space X] [regular_space X]
(h : β x A, is_closed A β§ Β¬ x β A β
β (f : X β I), continuous f β§ f x = (1 : I) β§ f '' A = {0})
(A B : set X) (hA : is_closed A) (hB : is_closed B)
(hAB : disjoint A B)
(hAc : is_compact A) :
β (f : X β I), continuous f β§ f '' A = {0} β§ f '' B = {1} := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Munkres|exercise_38_6 | Let $X$ be completely regular. Show that $X$ is connected if and only if the Stone-Δech compactification of $X$ is connected. | \begin{proof}
The closure of a connected set is connected, so if $X$ is connected so is $\beta(X)$
Suppose $X$ is the union of disjoint open subsets $U, V \subset X$. Define the continuous map
$$
\begin{aligned}
& f: X \rightarrow\{0,1\} \\
& x \mapsto \begin{cases}0, & x \in U \\
1, & x \in V\end{cases}
\end{aligned}
$$
By the fact that $\{0,1\}$ is compact and Hausdorff we can extend $f$ to a surjective map $\bar{f}: \beta(X) \rightarrow\{0,1\}$ such that $\bar{f}^{-1}(\{0\})$ and $\bar{f}^{-1}(\{1\})$ are disjoint open sets that cover $\beta(X)$, which makes this space not-connected.
\end{proof} | theorem exercise_38_6 {X : Type*}
(X : Type*) [topological_space X] [regular_space X]
(h : β x A, is_closed A β§ Β¬ x β A β
β (f : X β I), continuous f β§ f x = (1 : I) β§ f '' A = {0}) :
is_connected (univ : set X) β is_connected (univ : set (stone_cech X)) := | import .common
open set topological_space filter
open_locale classical topology filter
noncomputable theory
|
Axler|exercise_1_3 | Prove that $-(-v) = v$ for every $v \in V$. | \begin{proof}
By definition, we have
$$
(-v)+(-(-v))=0 \quad \text { and } \quad v+(-v)=0 .
$$
This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.
\end{proof} | theorem exercise_1_3 {F V : Type*} [add_comm_group V] [field F]
[module F V] {v : V} : -(-v) = v := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_1_6 | Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbf{R}^2$. | \begin{proof}
\[U=\mathbb{Z}^2=\left\{(x, y) \in \mathbf{R}^2: x, y \text { are integers }\right\}\]
$U=\mathbb{Z}^2$ satisfies the desired properties. To come up with this, note by assumption, $U$ must be closed under addition and subtraction, so in particular, it must contain 0 . We need to find a set which fails scalar multiplication. A discrete set like $\mathbb{Z}^2$ does this.
\end{proof} | theorem exercise_1_6 : β U : set (β Γ β),
(U β β
) β§
(β (u v : β Γ β), u β U β§ v β U β u + v β U) β§
(β (u : β Γ β), u β U β -u β U) β§
(β U' : submodule β (β Γ β), U β βU') := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_1_8 | Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$. | \begin{proof}
Let $V_1, V_2, \ldots, V_n$ be subspaces of the vector space $V$ over the field $F$. We must show that their intersection $V_1 \cap V_2 \cap \ldots \cap V_n$ is also a subspace of $V$.
To begin, we observe that the additive identity $0$ of $V$ is in $V_1 \cap V_2 \cap \ldots \cap V_n$. This is because $0$ is in each subspace $V_i$, as they are subspaces and hence contain the additive identity.
Next, we show that the intersection of subspaces is closed under addition. Let $u$ and $v$ be vectors in $V_1 \cap V_2 \cap \ldots \cap V_n$. By definition, $u$ and $v$ belong to each of the subspaces $V_i$. Since each $V_i$ is a subspace and therefore closed under addition, it follows that $u+v$ belongs to each $V_i$. Thus, $u+v$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.
Finally, we show that the intersection of subspaces is closed under scalar multiplication. Let $a$ be a scalar in $F$ and let $v$ be a vector in $V_1 \cap V_2 \cap \ldots \cap V_n$. Since $v$ belongs to each $V_i$, we have $av$ belongs to each $V_i$ as well, as $V_i$ are subspaces and hence closed under scalar multiplication. Therefore, $av$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.
Thus, we have shown that $V_1 \cap V_2 \cap \ldots \cap V_n$ is a subspace of $V$.
\end{proof} | theorem exercise_1_8 {F V : Type*} [add_comm_group V] [field F]
[module F V] {ΞΉ : Type*} (u : ΞΉ β submodule F V) :
β U : submodule F V, (β (i : ΞΉ), (u i).carrier) = βU := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_3_1 | Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\operatorname{dim} V=1$ and $T \in \mathcal{L}(V, V)$, then there exists $a \in \mathbf{F}$ such that $T v=a v$ for all $v \in V$. | \begin{proof}
If $\operatorname{dim} V=1$, then in fact, $V=\mathbf{F}$ and it is spanned by $1 \in \mathbf{F}$.
Let $T$ be a linear map from $V$ to itself. Let $T(1)=\lambda \in V(=\mathbf{F})$.
Step 2
2 of 3
Every $v \in V$ is a scalar. Therefore,
$$
\begin{aligned}
T(v) & =T(v \cdot 1) \\
& =v T(1) \ldots .(\text { By the linearity of } T) \\
& =v \lambda
\end{aligned}
$$
Hence, $T v=\lambda v$ for every $v \in V$.
\end{proof} | theorem exercise_3_1 {F V : Type*}
[add_comm_group V] [field F] [module F V] [finite_dimensional F V]
(T : V ββ[F] V) (hT : finrank F V = 1) :
β c : F, β v : V, T v = c β’ v:= | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_4_4 | Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p^{\prime}$ have no roots in common. | \begin{proof}
First, let $p$ have $m$ distinct roots. Since $p$ has the degree of $m$, then this could imply that $p$ can be actually written in the form of $p(z)=c\left(z-\lambda_1\right) \ldots\left(z-\lambda_m\right)$, which you have $\lambda_1, \ldots, \lambda_m$ being distinct.
To prove that both $p$ and $p^{\prime}$ have no roots in commons, we must now show that $p^{\prime}\left(\lambda_j\right) \neq 0$ for every $j$. So, to do so, just fix $j$. The previous expression for $p$ shows that we can now write $p$ in the form of $p(z)=\left(z-\lambda_j\right) q(z)$, which $q$ is a polynomial such that $q\left(\lambda_j\right) \neq 0$.
When you differentiate both sides of the previous equation, then you would then have $p^{\prime}(z)=(z-$ $\left.\lambda_j\right) q^{\prime}(z)+q(z)$
Therefore: $\left.=p^{\prime}\left(\lambda_j\right)=q \lambda_j\right)$
Equals: $p^{\prime}\left(\lambda_j\right) \neq 0$
Now, to prove the other direction, we would now prove the contrapositive, which means that we will be proving that if $p$ has actually less than $m$ distinct roots, then both $p$ and $p^{\prime}$ have at least one root in common.
Now, for some root of $\lambda$ of $p$, we can write $p$ is in the form of $\left.p(z)=(z-\lambda)^n q(z)\right)$, which is where both $n \geq 2$ and $q$ is a polynomial. When differentiating both sides of the previous equations, we would then have $p^{\prime}(z)=(z-\lambda)^n q^{\prime}(z)+n(z-\lambda)^{n-1} q(z)$.
Therefore, $p^{\prime}(\lambda)=0$, which would make $\lambda$ is a common root of both $p$ and $p^{\prime}$.
\end{proof} | theorem exercise_4_4 (p : polynomial β) :
p.degree = @card (root_set p β) (polynomial.root_set_fintype p β) β
disjoint
(@card (root_set p.derivative β) (polynomial.root_set_fintype p.derivative β))
(@card (root_set p β) (polynomial.root_set_fintype p β)) := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_5_4 | Suppose that $S, T \in \mathcal{L}(V)$ are such that $S T=T S$. Prove that $\operatorname{null} (T-\lambda I)$ is invariant under $S$ for every $\lambda \in \mathbf{F}$. | \begin{proof}
First off, fix $\lambda \in F$. Secondly, let $v \in \operatorname{null}(T-\lambda I)$. If so, then $(T-\lambda I)(S v)=T S v-\lambda S v=$ $S T v-\lambda S v=S(T v-\lambda v)=0$. Therefore, $S v \in \operatorname{null}(T-\lambda I)$ since $n u l l(T-\lambda I)$ is actually invariant under $S$.
\end{proof} | theorem exercise_5_4 {F V : Type*} [add_comm_group V] [field F]
[module F V] (S T : V ββ[F] V) (hST : S β T = T β S) (c : F):
map S (T - c β’ id).ker = (T - c β’ id).ker := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_5_12 | Suppose $T \in \mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator. | \begin{proof}
For every single $v \in V$, there does exist $a_v \in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \in V\{0\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$.
Now, to show that $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \in V\{0\}$. We would now want to show that $a_v=a_w$.
First, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\left(a_v v\right)=a_v w$. This is showing that $a_v=a_w$.
Finally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\left.a_{(} v+w\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$.
That previous equation implies the following: $\left.\left.\left(a_{(} v+w\right)-a_v\right) v+\left(a_{(} v+w\right)-a_w\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\left.a_{(} v+w\right)=a_v$ and $\left.a_{(} v+w\right)=a_w$. Therefore, $a_v=a_w$.
\end{proof} | theorem exercise_5_12 {F V : Type*} [add_comm_group V] [field F]
[module F V] {S : End F V}
(hS : β v : V, β c : F, v β eigenspace S c) :
β c : F, S = c β’ id := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_5_20 | Suppose that $T \in \mathcal{L}(V)$ has $\operatorname{dim} V$ distinct eigenvalues and that $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$. | \begin{proof}
First off, let $n=\operatorname{dim} V$. so, there is a basis of $\left(v_1, \ldots, v_j\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\lambda_1 v_j$ for every single $j$.
Now, for every $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \in F$. For each $j$, we would then have $(S T) v_j=S\left(T v_j\right)=\lambda_j S v_j=a_j \lambda_j v_j$ and $(T S) v_j=T\left(S v_j\right)=a_j T v_j=a_j \lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal.
\end{proof} | theorem exercise_5_20 {F V : Type*} [add_comm_group V] [field F]
[module F V] [finite_dimensional F V] {S T : End F V}
(h1 : @card T.eigenvalues (eigenvalues.fintype T) = finrank F V)
(h2 : β v : V, β c : F, v β eigenspace S c β β c : F, v β eigenspace T c) :
S * T = T * S := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_6_2 | Suppose $u, v \in V$. Prove that $\langle u, v\rangle=0$ if and only if $\|u\| \leq\|u+a v\|$ for all $a \in \mathbf{F}$. | \begin{proof}
First off, let us suppose that $(u, v)=0$.
Now, let $a \in \mathbb{F}$. Next, $u, a v$ are orthogonal.
The Pythagorean theorem thus implies that
$$
\begin{aligned}
\|u+a v\|^2 & =\|u\|^2+\|a v\|^2 \\
& \geq\|u\|^2
\end{aligned}
$$
So, by taking the square roots, this will now give us $\|u\| \leq\|u+a v\|$.
Now, to prove the implication in the other direction, we must now let $\|u\| \leq$ $\|u+a v\|$ for all $a \in \mathbb{F}$. Squaring this inequality, we get both:
$$
\begin{gathered}
\|u\|^2 a n d \leq\|u+a v\|^2 \\
=(u+a v, u+a v) \\
=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\
=\|u\|^2+\bar{a}(u, v)+a \overline{(u, v)}+|a|^2\|v\|^2 \\
\|u\|^2+2 \Re \bar{a}(u, v)+|a|^2\|v\|^2
\end{gathered}
$$
for all $a \in \mathbb{F}$.
Therefore,
$$
-2 \Re \bar{a}(u, v) \leq|a|^2\|v\|^2
$$
for all $a \in \mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives
$$
2 t|(u, v)|^2 \leq t^2|(u, v)|^2\|v\|^2
$$
for all $t>0$.
Step 4
4 of 4
Divide both sides of the inequality above by $t$, getting
$$
2|(u, v)|^2 \leq t \mid(u, v)^2\|v\|^2
$$
for all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \neq 0$, set $t$ equal to $1 /\|v\|^2$ in the inequality above, getting
$$
2|(u, v)|^2 \leq|(u, v)|^2,
$$
which implies that $(u, v)=0$.
\end{proof} | theorem exercise_6_2 {V : Type*} [add_comm_group V] [module β V]
[inner_product_space β V] (u v : V) :
βͺu, vβ«_β = 0 β β (a : β), βuβ β€ βu + a β’ vβ := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_6_7 | Prove that if $V$ is a complex inner-product space, then $\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$ for all $u, v \in V$. | \begin{proof}
Let $V$ be an inner-product space and $u, v\in V$. Then
$$
\begin{aligned}
\|u+v\|^2 & =\langle u+v, v+v\rangle \\
& =\|u\|^2+\langle u, v\rangle+\langle v, u\rangle+\|v\|^2 \\
-\|u-v\|^2 & =-\langle u-v, u-v\rangle \\
& =-\|u\|^2+\langle u, v\rangle+\langle v, u\rangle-\|v\|^2 \\
i\|u+i v\|^2 & =i\langle u+i v, u+i v\rangle \\
& =i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle+i\|v\|^2 \\
-i\|u-i v\|^2 & =-i\langle u-i v, u-i v\rangle \\
& =-i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle-i\|v\|^2 .
\end{aligned}
$$
Thus $\left(\|u+v\|^2\right)-\|u-v\|^2+\left(i\|u+i v\|^2\right)-i\|u-i v\|^2=4\langle u, v\rangle.$
\end{proof} | theorem exercise_6_7 {V : Type*} [inner_product_space β V] (u v : V) :
βͺu, vβ«_β = (βu + vβ^2 - βu - vβ^2 + I*βu + Iβ’vβ^2 - I*βu-Iβ’vβ^2) / 4 := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_6_16 | Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$ | \begin{proof}
$V=U \bigoplus U^{\perp}$, therefore $U^\perp = \{0\}$ iff $U=V$.
\end{proof} | theorem exercise_6_16 {K V : Type*} [is_R_or_C K] [inner_product_space K V]
{U : submodule K V} :
U.orthogonal = β₯ β U = β€ := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_7_6 | Prove that if $T \in \mathcal{L}(V)$ is normal, then $\operatorname{range} T=\operatorname{range} T^{*}.$ | \begin{proof}
Let $T \in \mathcal{L}(V)$ to be a normal operator.
Suppose $u \in \operatorname{null} T$. Then, by $7.20$,
$$
0=\|T u\|=\left\|T^* u\right\|,
$$
which implies that $u \in \operatorname{null} T^*$.
Hence
$$
\operatorname{null} T=\operatorname{null} T^*
$$
because $\left(T^*\right)^*=T$ and the same argument can be repeated.
Now we have
$$
\begin{aligned}
\text { range } T & =\left(\text { null } T^*\right)^{\perp} \\
& =(\text { null } T)^{\perp} \\
& =\operatorname{range} T^*,
\end{aligned}
$$
where the first and last equality follow from items (d) and (b) of 7.7.
Hence, range $T=$ range $T^*$.
\end{proof} | theorem exercise_7_6 {V : Type*} [inner_product_space β V]
[finite_dimensional β V] (T : End β V)
(hT : T * T.adjoint = T.adjoint * T) :
T.range = T.adjoint.range := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_7_10 | Suppose $V$ is a complex inner-product space and $T \in \mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$. | \begin{proof}
Based on the complex spectral theorem, there is an orthonormal basis of $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues. Therefore,
$$
T e_1=\lambda_j e_j
$$
for $j=1 \ldots n$.
Next, by applying $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\left(\lambda_j\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\left(\lambda_j\right)^8 e_j$, which implies that $\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint.
Now, by applying $T$ to both sides of the equation above, we get
$$
\begin{aligned}
T^2 e_j & =\left(\lambda_j\right)^2 e_j \\
& =\lambda_j e_j \\
& =T e_j
\end{aligned}
$$
which is where the second equality holds because $\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal.
\end{proof} | theorem exercise_7_10 {V : Type*} [inner_product_space β V]
[finite_dimensional β V] (T : End β V)
(hT : T * T.adjoint = T.adjoint * T) (hT1 : T^9 = T^8) :
is_self_adjoint T β§ T^2 = T := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Axler|exercise_7_14 | Suppose $T \in \mathcal{L}(V)$ is self-adjoint, $\lambda \in \mathbf{F}$, and $\epsilon>0$. Prove that if there exists $v \in V$ such that $\|v\|=1$ and $\|T v-\lambda v\|<\epsilon,$ then $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda-\lambda^{\prime}\right|<\epsilon$. | \begin{proof}
Let $T \in \mathcal{L}(V)$ be a self-adjoint, and let $\lambda \in \mathbf{F}$ and $\epsilon>0$.
By the Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues.
Choose an eigenvalue $\lambda^{\prime}$ of $T$ such that $\left|\lambda^{\prime}-\lambda\right|^2$ is minimized.
There are $a_1, \ldots, a_n \in \mathbb{F}$ such that
$$
v=a_1 e_1+\cdots+a_n e_n .
$$
Thus, we have
$$
\begin{aligned}
\epsilon^2 & >|| T v-\left.\lambda v\right|^2 \\
& =\left|\left\langle T v-\lambda v, e_1\right\rangle\right|^2+\cdots+\left|\left\langle T v-\lambda v, e_n\right\rangle\right|^2 \\
& =\left|\lambda_1 a_1-\lambda a_1\right|^2+\cdots+\left|\lambda_n a_n-\lambda a_n\right|^2 \\
& =\left|a_1\right|^2\left|\lambda_1-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda_n-\lambda\right|^2 \\
& \geq\left|a_1\right|^2\left|\lambda^{\prime}-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda^{\prime}-\lambda\right|^2 \\
& =\left|\lambda^{\prime}-\lambda\right|^2
\end{aligned}
$$
where the second and fifth lines follow from $6.30$ (the fifth because $\|v\|=1$ ). Now, we taking the square root.
Hence, $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda^{\prime}-\lambda\right|<\epsilon$
\end{proof} | theorem exercise_7_14 {π V : Type*} [is_R_or_C π]
[inner_product_space π V] [finite_dimensional π V]
{T : End π V} (hT : is_self_adjoint T)
{l : π} {Ξ΅ : β} (he : Ξ΅ > 0) : β v : V, βvβ= 1 β§ (βT v - l β’ vβ < Ξ΅ β
(β l' : T.eigenvalues, βl - l'β < Ξ΅)) := | import .common
open set fintype complex polynomial submodule linear_map finite_dimensional
open module module.End inner_product_space
open_locale big_operators
|
Ireland-Rosen|exercise_1_27 | For all odd $n$ show that $8 \mid n^{2}-1$. | \begin{proof}
We have $n^2-1=(n+1)(n-1)$. Since $n$ is odd, both $n+1, n-1$ are even, and moreso, one of these must be divisible by 4 , as one of the two consecutive odd numbers is divisible by 4 . Thus, their product is divisible by 8 . Similarly, if 3 does not divide $n$, it must divide one of $n-1, n+1$, otherwise it wouldn't divide three consecutive integers, which is impossible. As $n$ is odd, $n+1$ is even, so $(n+1)(n-1)$ is divisible by both 2 and 3 , so it is divisible by 6 .
\end{proof} | theorem exercise_1_27 {n : β} (hn : odd n) : 8 β£ (n^2 - 1) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_1_31 | Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$. | \begin{proof}
We have $(1+i)^2=1+2 i-1=2 i$, so $2=-i(1+i)^2$.
\end{proof} | theorem exercise_1_31 : (β¨1, 1β© : gaussian_int) ^ 2 β£ 2 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_2_21 | Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$. | \begin{proof}
$$
\left\{
\begin{array}{cccl}
\land(n)& = & \log p & \mathrm{if}\ n =p^\alpha,\ \alpha \in \mathbb{N}^* \\
& = & 0 & \mathrm{otherwise }.
\end{array}
\right.
$$
Let $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ the decomposition of $n$ in prime factors. As $\land(d) = 0$ for all divisors of $n$, except for $d = p_j^i, i>0, j=1,\ldots t$,
\begin{align*}
\sum_{d \mid n} \land(d)&= \sum_{i=1}^{\alpha_1} \land(p_1^{i}) + \cdots+ \sum_{i=1}^{\alpha_t} \land(p_t^{i})\\
&= \alpha_1 \log p_1+\cdots + \alpha_t \log p_t\\
&= \log n
\end{align*}
By Mobius Inversion Theorem,
$$\land(n) = \sum_{d \mid n} \mu\left (\frac{n}{d}\right ) \log d.$$
\end{proof} | theorem exercise_2_21 {l : β β β}
(hl : β p n : β, p.prime β l (p^n) = log p )
(hl1 : β m : β, Β¬ is_prime_pow m β l m = 0) :
l = Ξ» n, β d : divisors n, moebius (n/d) * log d := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_3_1 | Show that there are infinitely many primes congruent to $-1$ modulo 6 . | \begin{proof}
Let $n$ any integer such that $n\geq 3$, and $N = n! -1 = 2 \times 3 \times\cdots\times n - 1 >1$.
Then $N \equiv -1 \pmod 6$. As $6k +2, 6k +3, 6k +4$ are composite for all integers $k$, every prime factor of $N$ is congruent to $1$ or $-1$ modulo $6$. If every prime factor of $N$ was congruent to 1, then $N \equiv 1 \pmod 6$ : this is a contradiction because $-1 \not \equiv 1 \pmod 6$. So there exists a prime factor $p$ of $N$ such that $p\equiv -1 \pmod 6$.
If $p\leq n$, then $p \mid n!$, and $p \mid N = n!-1$, so $p \mid 1$. As $p$ is prime, this is a contradiction, so $p>n$.
Conclusion :
for any integer $n$, there exists a prime $p >n$ such that $p \equiv -1 \pmod 6$ : there are infinitely many primes congruent to $-1$ modulo $6$.
\end{proof} | theorem exercise_3_1 : infinite {p : primes // p β‘ -1 [ZMOD 6]} := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_3_5 | Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers. | \begin{proof}
If $7x^2 + 2 = y^3,\ x,y \in \mathbb{Z}$, then $y^3 \equiv 2 \pmod 7$ (so $y \not \equiv 0 \pmod 7$)
From Fermat's Little Theorem, $y^6 \equiv 1 \pmod 7$, so $2^2 \equiv y^6 \equiv 1 \pmod 7$, which implies $7 \mid 2^2-1 = 3$ : this is a contradiction. Thus the equation $7x^2 + 2 = y^3$ has no solution in integers.
\end{proof} | theorem exercise_3_5 : Β¬ β x y : β€, 7*x^3 + 2 = y^3 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_3_14 | Let $p$ and $q$ be distinct odd primes such that $p-1$ divides $q-1$. If $(n, p q)=1$, show that $n^{q-1} \equiv 1(p q)$. | \begin{proof}
As $n \wedge pq = 1, n\wedge p=1, n \wedge q = 1$, so from Fermat's Little Theorem
$$n^{q-1} \equiv 1 \pmod q,\qquad n^{p-1} \equiv 1 \pmod p.$$
$p-1 \mid q-1$, so there exists $k \in \mathbb{Z}$ such that $q-1 = k(p-1)$.
Thus
$$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$
$p \mid n^{q-1} - 1, q \mid n^{q-1} - 1$, and $p\wedge q = 1$, so $pq \mid n^{q-1} - 1$ :
$$n^{q-1} \equiv 1 \pmod{pq}.$$
\end{proof} | theorem exercise_3_14 {p q n : β} (hp0 : p.prime β§ p > 2)
(hq0 : q.prime β§ q > 2) (hpq0 : p β q) (hpq1 : p - 1 β£ q - 1)
(hn : n.gcd (p*q) = 1) :
n^(q-1) β‘ 1 [MOD p*q] := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_4_5 | Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$. | \begin{proof}
Let $a$ a primitive root modulo $p$.
As $a^{p-1} \equiv 1(\bmod p), p \mid\left(a^{(p-1) / 2}-1\right)\left(a^{(p-1) / 2}+1\right)$, so $p \mid a^{(p-1) / 2}-1$ or $p \mid$ $a^{(p-1) / 2}+1$. As $a$ is a primitive root modulo $p, a^{(p-1) / 2} \not \equiv 1(\bmod p)$, so
$$
a^{(p-1) / 2} \equiv-1 \quad(\bmod p) .
$$
Hence $(-a)^{(p-1) / 2}=(-1)^{2 t+1} a^{(p-1) / 2} \equiv(-1) \times(-1)=1(\bmod p)$.
Suppose that $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$.
Then $a^{2 n}=(-a)^{2 n} \equiv 1(\bmod p)$, so $p-1\left|2 n, \frac{p-1}{2}\right| n$.
So $-a$ has order $(p-1) / 2$ modulo $p$.
Conversely, suppose that $-a$ has order $(p-1) / 2=2 t+1$ modulo $p$. Let $2, p_1, \ldots p_k$ the prime factors of $p-1$, where $p_i$ are odd.
$a^{(p-1) / 2}=a^{2 t+1}=-(-a)^{2 t+1}=-(-a)^{(p-1) / 2} \equiv-1$, so $a^{(p-1) / 2} \not \equiv 1(\bmod 2)$.
As $p-1$ is even, $(p-1) / p_i$ is even, so $a^{(p-1) / p_i}=(-a)^{(p-1) / p_i} \not \equiv 1(\bmod p)($ since $-a$ has order $p-1)$.
So the order of $a$ is $p-1$ (see Ex. 4.8) : $a$ is a primitive root modulo $p$.
\end{proof} | theorem exercise_4_5 {p t : β} (hp0 : p.prime) (hp1 : p = 4*t + 3)
(a : zmod p) :
is_primitive_root a p β ((-a) ^ ((p-1)/2) = 1 β§ β (k : β), k < (p-1)/2 β (-a)^k β 1) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_4_8 | Let $p$ be an odd prime. Show that $a$ is a primitive root modulo $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$. | \begin{proof}
$\bullet$ If $a$ is a primitive root, then $a^k \not \equiv 1$ for all $k, 1\leq k < p-1$, so $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
$\bullet$ In the other direction, suppose $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
Let $\delta$ the order of $a$, and $p-1 = q_1^{a_1}q_2^{a_2}\cdots q_k^{a_k}$ the decomposition of $p-1$ in prime factors. As $\delta \mid p-1, \delta = q_1^{b_1}p_2^{b_2}\cdots q_k^{b_k}$, with $b_i \leq a_i, i=1,2,\ldots,k$. If $b_i < a_i$ for some index $i$, then $\delta \mid (p-1)/q_i$, so $a^{(p-1)/q_i} \equiv 1 \pmod p$, which is in contradiction with the hypothesis. Thus $b_i = a_i$ for all $i$, and $\delta = q-1$ : $a$ is a primitive root modulo $p$.
\end{proof} | theorem exercise_4_8 {p a : β} (hp : odd p) :
is_primitive_root a p β (β q β£ (p-1), q.prime β Β¬ a^(p-1) β‘ 1 [MOD p]) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_5_13 | Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 . | \begin{proof}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
$\bullet$ As $a^6 +1 = (a^2+1)(a^4-a^2+1)$, $p\mid a^4 - a^2+1$ implies $p \mid a^6 + 1$, so $\legendre{-1}{p} = 1$ and $p\equiv 1 \pmod 4$.
$\bullet$ $p \mid 4a^4 - 4 a^2 +4 = (2a-1)^2 + 3$, so $\legendre{-3}{p} = 1$.
As $-3 \equiv 1 \pmod 4$, $\legendre{-3}{p} = \legendre{p}{3}$, so $\legendre{p}{3} = 1$, thus $p \equiv 1 \pmod 3$.
$4 \mid p-1$ and $3 \mid p-1$, thus $12 \mid p-1$ : $$p \equiv 1 \pmod {12}.$$
\end{proof} | theorem exercise_5_13 {p x: β€} (hp : prime p)
(hpx : p β£ (x^4 - x^2 + 1)) : p β‘ 1 [ZMOD 12] := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_5_37 | Show that if $a$ is negative then $p \equiv q(4 a) together with p\not | a$ imply $(a / p)=(a / q)$. | \begin{proof}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Write $a = -A, A>0$. As $p \equiv q \pmod {4a}$, we know from Prop. 5.3.3. (b) that $(A/p) = (A/q)$.
Moreover,
\begin{align*}
\legendre{a}{p}&= \legendre{-A}{p} = (-1)^{(p-1)/2} \legendre{A}{p}\\
\legendre{a}{q}&= \legendre{-A}{q} = (-1^{(q-1)/2} \legendre{A}{q}
\end{align*}
As $p \equiv q \pmod {4a}$, $ p = q + 4ak, k\in \mathbb{Z}$, so
$$(-1)^{(p-1)/2} = (-1)^{(q+4ak-1)/2} = (-1)^{(q-1)/2},$$
so $(a/p) = (a/q)$.
\end{proof} | theorem exercise_5_37 {p q : β} [fact(p.prime)] [fact(q.prime)] {a : β€}
(ha : a < 0) (h0 : p β‘ q [ZMOD 4*a]) (h1 : Β¬ ((p : β€) β£ a)) :
legendre_sym p a = legendre_sym q a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Ireland-Rosen|exercise_18_4 | Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways. | \begin{proof}
Let $n=a^3+b^3$, and suppose that $\operatorname{gcd}(a, b)=1$. If a prime $p \mid a^3+b^3$, then
$$
\left(a b^{-1}\right)^3 \equiv_p-1
$$
Thus $3 \mid \frac{p-1}{2}$, that is, $p \equiv_6 1$.
If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as
$$
\begin{aligned}
& n=(a+b)\left(a^2-a b+b^2\right) \\
& n=(c+d)\left(c^2-c d+d^2\right)
\end{aligned}
$$
Thus we need $n$ to have atleast 3 disctinct prime factors, and so the smallest taxicab number is on the form
$$
n=(6 k+1)(12 k+1)(18 k+1)
$$
\end{proof} | theorem exercise_18_4 {n : β} (hn : β x y z w : β€,
x^3 + y^3 = n β§ z^3 + w^3 = n β§ x β z β§ x β w β§ y β z β§ y β w) :
n β₯ 1729 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open zsqrtd gaussian_int char_p nat.arithmetic_function
open_locale big_operators
noncomputable theory
|
Shakarchi|exercise_1_13a | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Re}(f)$ is constant, then $f$ is constant. | \begin{proof}
Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
Since $\operatorname{Re}(f)=$ constant,
$$
\frac{\partial u}{\partial x}=0, \frac{\partial u}{\partial y}=0 .
$$
By the Cauchy-Riemann equations,
$$
\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=0 .
$$
Thus, in $\Omega$,
$$
f^{\prime}(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0+0=0 .
$$
3
Thus $f(z)$ is constant.
\end{proof} | theorem exercise_1_13a {f : β β β} (Ξ© : set β) (a b : Ξ©) (h : is_open Ξ©)
(hf : differentiable_on β f Ξ©) (hc : β (c : β), β z β Ξ©, (f z).re = c) :
f a = f b := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_1_13c | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $|f|$ is constant, then $f$ is constant. | \begin{proof}
Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
We first give a mostly correct argument; the reader should pay attention to find the difficulty. Since $|f|=\sqrt{u^2+v^2}$ is constant,
$$
\left\{\begin{array}{l}
0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\
0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .
\end{array}\right.
$$
Plug in the Cauchy-Riemann equations and we get
$$
\begin{gathered}
u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\
-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 \\
(1.14) \Rightarrow \frac{\partial v}{\partial x}=\frac{v}{u} \frac{\partial v}{\partial y}
\end{gathered}
$$
Plug (1.15) into (1.13) and we get
$$
\frac{u^2+v^2}{u} \frac{\partial v}{\partial y}=0 .
$$
So $u^2+v^2=0$ or $\frac{\partial v}{\partial y}=0$.
If $u^2+v^2=0$, then, since $u, v$ are real, $u=v=0$, and thus $f=0$ which is constant.
Thus we may assume $u^2+v^2$ equals a non-zero constant, and we may divide by it. We multiply both sides by $u$ and find $\frac{\partial v}{\partial y}=0$, then by (1.15), $\frac{\partial v}{\partial x}=0$, and by Cauchy-Riemann, $\frac{\partial u}{\partial x}=0$.
$$
f^{\prime}=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0 .
$$
Thus $f$ is constant.
Why is the above only mostly a proof? The problem is we have a division by $u$, and need to make sure everything is well-defined. Specifically, we need to know that $u$ is never zero. We do have $f^{\prime}=0$ except at points where $u=0$, but we would need to investigate that a bit more.
Let's return to
$$
\left\{\begin{array}{l}
0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\
0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .
\end{array}\right.
$$
Plug in the Cauchy-Riemann equations and we get
$$
\begin{array}{r}
u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\
-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 .
\end{array}
$$
We multiply the first equation $u$ and the second by $v$, and obtain
$$
\begin{aligned}
u^2 \frac{\partial v}{\partial y}+u v \frac{\partial v}{\partial x} & =0 \\
-u v \frac{\partial v}{\partial x}+v^2 \frac{\partial v}{\partial y} & =0 .
\end{aligned}
$$
Adding the two yields
$$
u^2 \frac{\partial v}{\partial y}+v^2 \frac{\partial v}{\partial y}=0,
$$
or equivalently
$$
\left(u^2+v^2\right) \frac{\partial v}{\partial y}=0 .
$$
We now argue in a similar manner as before, except now we don't have the annoying $u$ in the denominator. If $u^2+v^2=0$ then $u=v=0$, else we can divide by $u^2+v^2$ and find $\partial v / \partial y=0$. Arguing along these lines finishes the proof.
\end{proof} | theorem exercise_1_13c {f : β β β} (Ξ© : set β) (a b : Ξ©) (h : is_open Ξ©)
(hf : differentiable_on β f Ξ©) (hc : β (c : β), β z β Ξ©, abs (f z) = c) :
f a = f b := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_1_19b | Prove that the power series $\sum zn/n^2$ converges at every point of the unit circle. | \begin{proof}
Since $\left|z^n / n^2\right|=1 / n^2$ for all $|z|=1$, then $\sum z^n / n^2$ converges at every point in the unit circle as $\sum 1 / n^2$ does ( $p$-series $p=2$.)
\end{proof} | theorem exercise_1_19b (z : β) (hz : abs z = 1) (s : β β β)
(h : s = (Ξ» n, β i in (finset.range n), i * z / i ^ 2)) :
β y, tendsto s at_top (π y) := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_1_26 | Suppose $f$ is continuous in a region $\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant. | \begin{proof}
Suppose $F_1$ adn $F_2$ are primitives of $F$. Then $(F_1-F_2)^\prime = f - f = 0$, therefore $F_1$ and $F_2$ differ by a constant.
\end{proof} | theorem exercise_1_26
(f Fβ Fβ : β β β) (Ξ© : set β) (h1 : is_open Ξ©) (h2 : is_connected Ξ©)
(hFβ : differentiable_on β Fβ Ξ©) (hFβ : differentiable_on β Fβ Ξ©)
(hdFβ : β x β Ξ©, deriv Fβ x = f x) (hdFβ : β x β Ξ©, deriv Fβ x = f x)
: β c : β, β x, Fβ x = Fβ x + c := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_2_9 | Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_{0} \in \Omega$ such that $\varphi\left(z_{0}\right)=z_{0} \quad \text { and } \quad \varphi^{\prime}\left(z_{0}\right)=1$ then $\varphi$ is linear. | \begin{proof}
When $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n \geq 2$ and $a_n \neq 0$, such that
$$
\varphi(z)=z+a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .
$$
As you have noticed, by induction, it follows that for every $k \geq 1$,
$$
\varphi^k(z)=z+k a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .
$$
Let $r>0$ be such that when $\left|z-z_0\right| \leq r$, then $z \in \Omega$. Then by (1),
$$
k a_n=\frac{1}{2 \pi i} \int_{\left|z-z_0\right|=r} \frac{\varphi^k(z)}{\left(z-z_0\right)^{n+1}} d z .
$$
Since $\varphi^k(\Omega) \subset \Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $\left|\varphi^k\right| \leq M$ on $\Omega$. Then by (2),
$$
k\left|a_n\right| \leq M r^{-n} .
$$
Since $k$ is arbitrary, $a_n=0$, a contradiction.
\end{proof} | theorem exercise_2_9
{f : β β β} (Ξ© : set β) (b : metric.bounded Ξ©) (h : is_open Ξ©)
(hf : differentiable_on β f Ξ©) (z β Ξ©) (hz : f z = z) (h'z : deriv f z = 1) :
β (f_lin : β βL[β] β), β x β Ξ©, f x = f_lin x := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_3_3 | Show that $ \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + a^2} dx = \pi \frac{e^{-a}}{a}$ for $a > 0$. | \begin{proof}
$\cos x=\frac{e^{i x}+e^{-i x}}{2}$. changing $x \rightarrow-x$ we see that we can just integrate $e^{i x} /\left(x^2+a^2\right)$ and we'll get the same answer. Again, we use the same semicircle and part of the real line. The only pole is $x=i a$, it has order 1 and the residue at it is $\lim _{x \rightarrow i a} \frac{e^{i x}}{x^2+a^2}(x-i a)=\frac{e^{-a}}{2 i a}$, which multiplied by $2 \pi i$ gives the answer.
\end{proof} | theorem exercise_3_3 (a : β) (ha : 0 < a) :
tendsto (Ξ» y, β« x in -y..y, real.cos x / (x ^ 2 + a ^ 2))
at_top (π (real.pi * (real.exp (-a) / a))) := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_3_9 | Show that $\int_0^1 \log(\sin \pi x) dx = - \log 2$. | \begin{proof}
Consider
$$
\begin{gathered}
f(z)=\log \left(1-e^{2 \pi z i}\right)=\log \left(e^{\pi z i}\left(e^{-\pi z i}-e^{\pi z i}\right)\right)=\log (-2 i)+\pi z i+\log \\
(\sin (\pi z))
\end{gathered}
$$
Then we have
$$
\begin{aligned}
\int_0^1 f(z) d z & =\log (-2 i)+\frac{i \pi}{2}+\int_0^1 \log (\sin (\pi z)) d z \\
& =\int_0^1 \log (\sin (\pi z)) d z+\log (-2 i)+\log (i) \\
& =\log (2)+\int_0^1 \log (\sin (\pi z)) d z
\end{aligned}
$$
Now it suffices to show that $\int_0^1 f(z) d z=0$. Consider the contour $C(\epsilon, R)$ (which is the contour given in your question) given by the following.
1. $C_1(\epsilon, R)$ : The vertical line along the imaginary axis from $i R$ to $i \epsilon$.
2. $C_2(\epsilon)$ : The quarter turn of radius $\epsilon$ about 0 .
3. $C_3(\epsilon)$ : Along the real axis from $(\epsilon, 1-\epsilon)$.
4. $C_4(\epsilon)$ : The quarter turn of radius $\epsilon$ about 1 .
5. $C_5(\epsilon, R)$ : The vertical line from $1+i \epsilon$ to $1+i R$.
6. $C_6(R)$ : The horizontal line from $1+i R$ to $i R$.
$f(z)$ is analytic inside the contour $C$ and hence $\oint_C f(z)=0$. This gives us
$$
\begin{aligned}
\int_{C_1(\epsilon, R)} f d z+\int_{C_2(\epsilon)} f d z+\int_{C_3(\epsilon)} f d z+\int_{C_4(\epsilon)} f d z+\int_{C_5(\epsilon, R)} f d z+\int_{C_6(R)} f d z \\
=0
\end{aligned}
$$
Now the integral along 1 cancels with the integral along 5 due to symmetry. Integrals along 2 and 4 scale as $\epsilon \log (\epsilon)$. Integral along 6 goes to 0 as $R \rightarrow \infty$. This gives us
$$
\lim _{\epsilon \rightarrow 0} \int_{C_3(\epsilon)} f d z=0
$$
which is what we need.
\end{proof} | theorem exercise_3_9 : β« x in 0..1, real.log (real.sin (real.pi * x)) = - real.log 2 := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Shakarchi|exercise_3_22 | Show that there is no holomorphic function $f$ in the unit disc $D$ that extends continuously to $\partial D$ such that $f(z) = 1/z$ for $z \in \partial D$. | \begin{proof}
Consider $g(r)=\int_{|z|=r} f(z) d z$. Cauchy theorem implies that $g(r)=0$ for all $r<1$. Now since $\left.f\right|_{\partial D}=1 / z$ we have $\lim _{r \rightarrow 1} \int_{|z|=r} f(z) d z=\int_{|z|=1} \frac{1}{z} d z=\frac{2}{\pi i} \neq 0$. Contradiction.
\end{proof} | theorem exercise_3_22 (D : set β) (hD : D = ball 0 1) (f : β β β)
(hf : differentiable_on β f D) (hfc : continuous_on f (closure D)) :
Β¬ β z β (sphere (0 : β) 1), f z = 1 / z := | import .common
open complex filter function interval_integral metric
open_locale big_operators
open_locale filter
open_locale topology
|
Putnam|exercise_2020_b5 | For $j \in\{1,2,3,4\}$, let $z_{j}$ be a complex number with $\left|z_{j}\right|=1$ and $z_{j} \neq 1$. Prove that $3-z_{1}-z_{2}-z_{3}-z_{4}+z_{1} z_{2} z_{3} z_{4} \neq 0 .$ | \begin{proof}
It will suffice to show that for any $z_1, z_2, z_3, z_4 \in \mathbb{C}$ of modulus 1 such that $|3-z_1-z_2-z_3-z_4| = |z_1z_2z_3z_4|$, at least one of $z_1, z_2, z_3$ is equal to 1.
To this end, let $z_1=e^{\alpha i}, z_2=e^{\beta i}, z_3=e^{\gamma i}$ and
\[
f(\alpha, \beta, \gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^2.
\]
A routine calculation shows that
\begin{align*}
f(\alpha, \beta, \gamma)&=
10 - 6\cos(\alpha) - 6\cos(\beta) - 6\cos(\gamma) \\
&\quad + 2\cos(\alpha + \beta + \gamma) + 2\cos(\alpha - \beta) \\
&\quad + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha).
\end{align*}
Since the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\nabla f=(0,0,0)$. A routine calculation now shows that
\begin{align*}
\frac{\partial f}{\partial \alpha} + \frac{\partial f}{\partial \beta} + \frac{\partial f}{\partial \gamma} &=
6(\sin(\alpha) +\sin(\beta)+\sin(\gamma)- \sin(\alpha + \beta + \gamma)) \\
&=
24\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\beta+\gamma}{2}\right)
\sin\left(\frac{\gamma+\alpha}{2}\right).
\end{align*}
Hence every critical point of $f$ must satisfy one of $z_1z_2=1$, $z_2z_3=1$, or $z_3z_1=1$. By symmetry, let us assume that $z_1z_2=1$. Then
\[
f = |3-2\mathrm{Re}(z_1)-z_3|^2-|1-z_3|^2;
\]
since $3-2\mathrm{Re}(z_1)\ge 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$.
\end{proof} | theorem exercise_2020_b5 (z : fin 4 β β) (hz0 : β n, βz nβ < 1)
(hz1 : β n : fin 4, z n β 1) :
3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) β 0 := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
Putnam|exercise_2018_b2 | Let $n$ be a positive integer, and let $f_{n}(z)=n+(n-1) z+$ $(n-2) z^{2}+\cdots+z^{n-1}$. Prove that $f_{n}$ has no roots in the closed unit disk $\{z \in \mathbb{C}:|z| \leq 1\}$. | \begin{proof}
Note first that $f_n(1) > 0$, so $1$ is not a root of $f_n$.
Next, note that
\[
(z-1)f_n(z) = z^n + \cdots + z - n;
\]
however, for $\left| z \right| \leq 1$, we have
$\left| z^n + \cdots + z \right| \leq n$ by the triangle inequality;
equality can only occur if $z,\dots,z^n$ have norm 1 and the same argument, which only happens for $z=1$.
Thus there can be no root of $f_n$ with $|z| \leq 1$.
\end{proof} | theorem exercise_2018_b2 (n : β) (hn : n > 0) (f : β β β β β)
(hf : β n : β, f n = Ξ» z, (β (i : fin n), (n-i)* z^(i : β))) :
Β¬ (β z : β, βzβ β€ 1 β§ f n z = 0) := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
Putnam|exercise_2017_b3 | Suppose that $f(x)=\sum_{i=0}^{\infty} c_{i} x^{i}$ is a power series for which each coefficient $c_{i}$ is 0 or 1 . Show that if $f(2 / 3)=3 / 2$, then $f(1 / 2)$ must be irrational. | \begin{proof}
Suppose by way of contradiction that $f(1/2)$ is rational. Then $\sum_{i=0}^{\infty} c_i 2^{-i}$ is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers $m,n$ such that
$c_i = c_{m+i}$ for all $i \geq n$. We may then write
\[
f(x) = \sum_{i=0}^{n-1} c_i x^i + \frac{x^n}{1-x^m} \sum_{i=0}^{m-1} c_{n+i} x^i.
\]
Evaluating at $x = 2/3$, we may equate $f(2/3) = 3/2$ with
\[
\frac{1}{3^{n-1}} \sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \frac{2^n 3^m}{3^{n+m-1}(3^m-2^m)} \sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-1-i};
\]
since all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction.
\end{proof} | theorem exercise_2017_b3 (f : β β β) (c : β β β)
(hf : f = Ξ» x, (β' (i : β), (c i) * x^i))
(hc : β n, c n = 0 β¨ c n = 1)
(hf1 : f (2/3) = 3/2) :
irrational (f (1/2)) := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
Putnam|exercise_2010_a4 | Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime. | \begin{proof}
Put
\[
N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.
\]
Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer.
For any nonnegative integer $j$,
\[
10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}.
\]
Since $10^n \geq n \geq 2^m \geq m+1$, $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$,
and similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$. It follows that
\[
N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}.
\]
Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$, it follows that $N$ is composite.
\end{proof} | theorem exercise_2010_a4 (n : β) :
Β¬ nat.prime (10^10^10^n + 10^10^n + 10^n - 1) := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
Putnam|exercise_2000_a2 | Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers. | \begin{proof}
It is well-known that the equation $x^2-2y^2=1$ has infinitely
many solutions (the so-called ``Pell'' equation). Thus setting
$n=2y^2$ (so that $n=y^2+y^2$, $n+1=x^2+0^2$, $n+2=x^2+1^2$)
yields infinitely many $n$ with the desired property.
\end{proof} | theorem exercise_2000_a2 :
β N : β, β n : β, n > N β§ β i : fin 6 β β, n = (i 0)^2 + (i 1)^2 β§
n + 1 = (i 2)^2 + (i 3)^2 β§ n + 2 = (i 4)^2 + (i 5)^2 := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
Putnam|exercise_1998_a3 | Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that | theorem exercise_1998_a3 (f : β β β) (hf : cont_diff β 3 f) :
β a : β, (f a) * (deriv f a) * (iterated_deriv 2 f a) * (iterated_deriv 3 f a) β₯ 0 := | import .common
open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal
|
|
Pugh|exercise_2_12a | Let $(p_n)$ be a sequence and $f:\mathbb{N}\to\mathbb{N}$. The sequence $(q_k)_{k\in\mathbb{N}}$ with $q_k=p_{f(k)}$ is called a rearrangement of $(p_n)$. Show that if $f$ is an injection, the limit of a sequence is unaffected by rearrangement. | \begin{proof}
Let $\varepsilon>0$. Since $p_n \rightarrow L$, we have that, for all $n$ except $n \leq N$, $d\left(p_n, L\right)<\epsilon$. Let $S=\{n \mid f(n) \leq N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall n>n_0 f(n)>N$ which implies that $p_{f(n)} \rightarrow L$, as required.
\end{proof} | theorem exercise_2_12a (f : β β β) (p : β β β) (a : β)
(hf : injective f) (hp : tendsto p at_top (π a)) :
tendsto (Ξ» n, p (f n)) at_top (π a) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Pugh|exercise_2_29 | Let $\mathcal{T}$ be the collection of open subsets of a metric space $\mathrm{M}$, and $\mathcal{K}$ the collection of closed subsets. Show that there is a bijection from $\mathcal{T}$ onto $\mathcal{K}$. | \begin{proof}
The bijection given by $x\mapsto X^C$ suffices.
\end{proof} | theorem exercise_2_29 (M : Type*) [metric_space M]
(O C : set (set M))
(hO : O = {s | is_open s})
(hC : C = {s | is_closed s}) :
β f : O β C, bijective f := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
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ProofNet
Dataset Summary
ProofNet is a benchmark for autoformalization and formal proving of undergraduate-level mathematics. The ProofNet benchmarks consists of 371 examples, each consisting of a formal theorem statement in Lean 3, a natural language theorem statement, and a natural language proof. The problems are primarily drawn from popular undergraduate pure mathematics textbooks and cover topics such as real and complex analysis, linear algebra, abstract algebra, and topology. We intend for ProofNet to be a challenging benchmark that will drive progress in autoformalization and automatic theorem proving.
Citation:
@misc{azerbayev2023proofnet,
title={ProofNet: Autoformalizing and Formally Proving Undergraduate-Level Mathematics},
author={Zhangir Azerbayev and Bartosz Piotrowski and Hailey Schoelkopf and Edward W. Ayers and Dragomir Radev and Jeremy Avigad},
year={2023},
eprint={2302.12433},
archivePrefix={arXiv},
primaryClass={cs.CL}
}
Leaderboard
Statement Autoformalization
| Model | Typecheck Rate | Accuracy |
|---|---|---|
| Davinci-code-002 (prompt retrieval) | 45.2 | 16.1 |
| Davinci-code-002 (in-context learning) | 23.7 | 13.4 |
| proofGPT-1.3B | 10.7 | 3.2 |
Statement Informalization
| Model | Accuracy |
|---|---|
| Code-davinci-002 (in-context learning) | 62.3 |
| proofGPT-6.7B (in-context learning) | 6.5 |
| proofGPT-1.3B (in-context learning) | 4.3 |
Data Fields
id: Unique string identifier for the problem.nl_statement: Natural language theorem statement.nl_proof: Natural language proof, in LaTeX. Depends onamsthm, amsmath, amssymbpackages.formal_statement: Formal theorem statement in Lean 3.src_header: File header including imports, namespaces, and locales required for the formal statement. Note that local import of common.lean, which has to be manually downloaded and place in the same directory as your.leanfile containing the formal statement.
Authors
Zhangir Azerbayev, Bartosz Piotrowski, Jeremy Avigad
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Repository:
zhangir-azerbayev/ProofNet
Paper:
ProofNet
Point of Contact:
Zhangir Azerbayev
Size of downloaded dataset files:
539 kB
Size of the auto-converted Parquet files:
211 kB
Number of rows:
371
Models trained or fine-tuned on hoskinson-center/proofnet