Proof of the extreme value theorem without using subsequences I am preparing a lecture on the Weierstrass theorem (probably best known as the Extreme Value Theorem in english-speaking countries), and I would propose a proof that does not use the extraction of converging subsequences. I did not explain subsequences in my calculus course, and I must choose between skipping the proof of the theorem and finding some proof which works only for functions $\mathbb{R} \to \mathbb{R}$.

I remember I once read a proof based on some bisection technique, but I can't find a reference right now.
I would be grateful for any reference to books, papers, web sites about this alternative proof.

Edit: since somebody modified my question, I will write down the precise theorem I want to prove.

Theorem. Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. Then $f$ has at least a maximum and a minimum point.
Note: This answers the question as it was originally asked, which unfortunately wasn’t what was intended. I’ve added a sketch of an answer to the intended question below it.

I don’t have a reference at hand, but the proof is straightforward. Suppose that $f$ is continuous on $[a_0,b_0]$ with $f(a_0)&lt;0&lt;f(b_0)$. Recursively construct a sequence of intervals $[a_k,b_k]$ as follows.


If $f\left(\frac{a_k+b_k}2\right)=0$, stop: you’ve found the desired point.
If $f\left(\frac{a_k+b_k}2\right)&lt;0$, let $a_{k+1}=\dfrac{a_k+b_k}2$ and $b_{k+1}=b_k$.
If $f\left(\frac{a_k+b_k}2\right)&gt;0$, let $a_{k+1}=a_k$ and $b_{k+1}=\dfrac{a_k+b_k}2$.


Then $\langle a_k:k\in\Bbb N\rangle$ is non-decreasing, $\langle b_k:k\in\Bbb N\rangle$ is non-increasing, $a_k&lt;b_m$ for all $k,m\in\Bbb N$, and $b_k-a_k=(b-a)2^{-k}$ for all $k\in\Bbb N$, so $\sup_ka_k=\inf_kb_k$, and by continuity the function must be $0$ at this point.

Obviously you’d want to fill in a lot of detail and show how it generalizes to the full intermediate value theorem, but that’s it in a nutshell.



To use bisection to prove the extreme value theorem, let $f$ be a continuous real-valued function on $[a,b]$. First use bisection to show that $f$ is bounded above: if not, at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)\ge k$, and otherwise choose the right half and a point $x_k$ in it such that $f(x_k)\ge k$. Then $a_k\le x_k\le b_k$ for all $k\in\Bbb N$, so $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f$ cannot be defined at $\sup_ka_k$.

Now let $y=\sup_{x\in[a,b]}f(x)$, and repeat the bisection process, but this time at stage $k$ choose the left half if it contains a point $x_k$ such that $f(x_k)&gt;y-2^{-k}$; if not, the right half contains such a point $x_k$, and we choose the right half. Then $\langle x_k:k\in\Bbb N\rangle\to\sup_ka_k$, and $f(\sup_ka_k)=y$.
Here's an outline of a "bisection" proof.

Suppose $\sup_{x \in [a,b]} f = \alpha$.  Define a sequence of intervals $I_k = [a_k,b_k]$ recursively as follows: $I_0 = [a,b]$ and for $k &gt; 0$

$
I_k = 
\begin{cases} [a_{k-1}, \frac{a_{k-1} + b_{k-1}}{2}] \quad \text{if } 
\sup_{x \in [a_{k-1},\frac{a_{k-1} + b_{k-1}}{2}]} = \alpha \\
[\frac{a_{k-1} + b_{k-1}}{2},b_{k-1}] \quad \text{otherwise}.
\end{cases}
$

Then there exists $x \in \cap_{k} I_k$ (though this requires some care to show, and is a theorem by itself), and you can show $f(x) = \alpha$.