A non-empty subset of integers bounded above has a maximum

Question

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 $s$ $\in$ $\mathrm A$ ?

Answer 1

Let $A \subseteq \Bbb N$ be bounded above (say by $M \in \Bbb N$: bounded above is unspecified, so by default it will mean bounded in $\Bbb N$. If bounded above in $\Bbb R$ is what is meant (by some bound $B \in \Bbb R$), find an integer larger or equal to $B$ by the archimean property of $\Bbb R$, but such an $M$ exists in either case).

Then $B=\{n \in \Bbb N: \forall a \in A: a \le n\}$ (the set of all upperbounds of $A$) is a non-empty subset of $\Bbb N$ so it has a minimum $m_0$ by the well-foundedness of $\Bbb N$. If $m_0 \notin A$, then $m_0 -1$ is also in $B$ (if $n \in A$ then $n < m_0$ ($n \le m_0$ is by definition, but $m_0= n$ is not the case by assumption, or $m_0 \in A$) and so $n \le m_0 -1$, as $n$ is arbitrary $m_0 -1 \in B$. But this contradicts the minimality of $m_0$. So $m_0 \in A$ and is equal to $\max(A)$.

So all we need is the well-foundedness of $\Bbb N$ and $n < m \to n \le m-1$ as an essential order property of $\Bbb N$. Completeness of $\Bbb R$ is "overkill".

Answer 2

Let non-empty subset of integers be B.

$\mathtt \enspace Assume \enspace that \enspace B \enspace has \enspace no \enspace maximum.$

B is non-empty and B is bounded above → B has a supremum. (By the completeness axiom). Let sup(B)=s.

So $$∀x ∈ B , s>x$$(s can't be equal to x.Because if s=x then x will be the maximum).

Also $$∀ϵ>0, ∃x_1 \enspace s.t\enspace x_1 + ϵ > s → x_1 > s - ϵ$$ So $$∃x_2 ∈ B \enspace s.t\enspace x_2>x_1 $$ (Because B has no maximum. So there can be element higher than $x_1$).

So $$s> x_2 > x_1 > s- ϵ $$
$\mathtt Let \enspace ϵ = 1.$ Then $$s> x_2 > x_1 > s- 1$$

So observe this inequality. You can see that there are two integer numbers s.t there's difference is lower than 1(Two integers difference must be equal to 1 or higher). So that can't happen. So it is a contradiction. So our assume is false. Because of that B has a maximum element.

Answer 3

There is a more direct proof by using well ordering property of integers or induction,But as in the question use of R is suggested, I am taking that route. First let's construct the supremum, we can do it by taking the supremum in R and then taking the ceiling function of it.Why it must be the supremum of the set? Because it is clearly an upper bound and when we take any int less than it, it must be smaller than the supremum in $R$,hence not an upper bound(otherwise in $R$ the sup will be smaller than this number).Let us call that supremum $s \space$. Suppose $s\space$ is not in the set $A$, then $s-1 < s \space$, it means there exists p in $A$ such that $s-1<p<=s$, otherwise $s-1$ will be the supremum by definition, which is not the case. But we know that between two consecutive integers, we can't find another integer. Hence our assumption is false. So, $s\in A$.

Answer 4

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.