Proving upper bounds? This is a relatively simple question.  Say you are given a set that exists in the reals.  How would you go about formulating a proof to prove that some number that is not given is in fact an upper bound?

This is a very broad question and therefore the answer can hardly be anything but: "It depends"
So here are a few examples that should illustrate that various ideas may be needed to show that a certain number is a bound.


Let $A=\{x\in \mathbb R\mid x^6<40+2x^3\}$. Show that $A$ is bounded.

Direct computation: Note that $x^6<40+2x^3$ implies $(x^3-1)^2=x^6-2x^3+1<41<7^2$, hence $|x^3-1|<7$, i.e. $x^3<8$ and finally $x<2$. Therefore $2$ is an upper bound for $A$.


Let $B=\{\frac {F_{n+1}}{F_{n}}\mid n\in\mathbb N\}$, where $F_n$ is the $n$th Fibonacchi number, i.e. $F_1=1, F_2=2$ and $F_n=F_{n-1}+F_{n-2}$ otherwise. 

We have (by induction, if necessary) $F_n>0$ for $n\in\mathbb N$.
Therefore $F_{n+1}=F_{n}+F_{n-1}>F_{n}$ for $n>1$ (but trivially also for $n=1$), i.e. the sequence is strictly increasing.
Therefore $F_{n+1}=F_n+F_{n-1}<2F_n$ for $n>1$. Together with $F_2\le 2F_1$, which makes $2$ an upper bound for $B$.


Let $c\in \mathbb R$ and $C$ a bounded (from above) set with $0\in C$ and for all $x\in C$, also $x^2+c\in C$. Show that $2$ is an upper bound.

Assume $2$ is not an upper bound. Let $x_0$ be an element of $C$ that is $>2$ and define recursively $x_n=x_{n-1}^2+c$ so that $x_n\in C$. 
Note that 
$$\tag1x^2+c\ge x^2-x=x(x-1)>x\quad\text{if }x\ge-c\text{ and }x>2$$


*

*If $c\ge-2$, then the sequence $(x_n)$ is strictly increasing because of $(1)$. 

*If $c<-2$, $0\in C$ implies $c\in C$ implies $c^2+c\in C$. As $c^2+c=(-c)\cdot(-1-c)>(-c)\cdot 1=-c>2$, we may assume that we have chosen $x_0=c^2+c$ above and again see from $(1)$ that $(x_n)$ is strictly increasing.
As $C$ is bounded, the strictly increasing sequence $(x_n)$ converges against some limit $\xi$ (not necessarily in $C$), for which $\xi^2+c=\xi$ holds. As already $x_0\ge -c$ and $x_0>2$, this hold even more so for $\xi$ and we get a contradiction to $(1)$.
Therefore, $2$ is an upper bound for $C$.


Let $D=\{\sin(p)+3\cos(p^2)|p\text{ is a Fermat prime}\}$.

We need not know anything about Fermat primes to show that $4$ is an upper bound, simply because $\sin x\le 1$ and $\cos x\le 1$ for all $x$.


Let $E=\left\{\frac{1}{(x-u)^2+(y-v)^2}\;\bigg|\; x,y,u,v\in \mathbb R, xy=uv=1, xu<0\right\}$. Show that $\frac18$ is an upper bound for $E$

How would you proceed?


Let $F = \left\{n\in\mathbb N\;\bigg|\; \exists \frac ab\in\mathbb Q\setminus \mathbb Z\colon\left|\frac ab-\sqrt 5\right|<\frac1{q^n}\right\}$

Any idea if there is a bound? What the bound could be? How to prove it?


Let $G=\{n\in\mathbb N\mid$ There exists a binary twelve state Turing machine that halts after writing $n$ "1"s$\}$.

Finding and proving an upper bound is only for the brave.
