Why does it happen that, if $A\subset \mathbb{Z}$ is bounded from above, it has a maximum value? How can it be explained?
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The main thing to prove here is that the integers possess the least upper bound property. This is not a result of analysis. See here: http://en.wikipedia.org/wiki/Least-upper-bound_property and http://en.wikipedia.org/wiki/Well-ordering_principle – Christopher A. Wong Jan 09 '14 at 09:33
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It is more convenient to show that if $A\subset \mathbb{Z}$ is bounded from below (rather than above) then there is a minimum. Namely, if $A$ is bounded from below then there is an integer $n$ such that $A+n$ is bounded from below by $0$. Thus we can assume wlog that $A$ is a collection of natural numbers. But the natural numbers are well-ordered in the sense that any subset has a least element.
Mikhail Katz
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I would go as follows:
Bounded means that there exists $N\in\Bbb Z$ such that $A\subseteq\{n\in\Bbb Z\,\mid\,n<N\}$. For eack $k\in\Bbb N$, let $$I_k=\{n\in\Bbb Z\mid N-k\leq n\leq N\}$$ and set $$ \Sigma=\{k\mid I_k\cap A\neq\emptyset\}. $$ Assuming $A\neq\emptyset$, the set $\Sigma$ is not empty because $A\subseteq\bigcup_kI_k$ and so it has a minimum $m$. But then $N-m=\max A$.
Andrea Mori
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Assume $A\ne\emptyset$ and that there is an $M\in{\mathbb Z}$ with $x\leq M$ for all $x\in A$. Put $$B:=\{y\in{\mathbb N}_{\geq0}\>|\>M-y\in A\}\ .$$ Then $B\ne\emptyset$; therefore $B$ has a minimal element $y_*$. I claim that $x_*:=M-y_*\in A$ is the maximal element of $A$. Proof: For any $x\in A$ one has $y:=M-x\in B$ and therefore $$x=M-y\leq M-y_*= x_*\ .$$
Christian Blatter
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