A non-empty subset of integers bounded above has a maximum Suppose the set $\mathrm A$ $\neq$ $\emptyset$ , $\mathrm A$ $\subseteq$ $\Bbb Z$ is bounded above. Then since $\Bbb Z$ $\subseteq$ $\Bbb R$, I know that by the completeness axiom there exists a $supremum$ for the set $\mathrm A$, say $s$ $=$ $sup$($\mathrm A $). But, I need to show this is in fact the maximum of this set $\mathrm A$. For that, I know it has to be an element of the set $\mathrm A$. How can we show that $s$ $\in$ $\mathrm A$ ?

If $A$ is your subset of $\mathbb{Z}$, consider the identity function, $i:A\rightarrow \mathbb{R}$.  $f(A)$ is closed since $i$ is continuous and $A$ is closed, and $f(A)$ is bounded since there is an $M$ such that each term of the sequence is bounded, so $|x_n| \le |i(x_n)| \le M$ for all $n$. Therefore, $f(A)$ is compact in $\mathbb{R}$, so it has a greatest and least element that are in the set, and the greatest and least element are the maximizer and minimizer, since $i$ is the identity function.
