Question about proof that every bounded subset of the integers has a largest element

Question

Supposing we have a nonempty subset $A_0$ of $\mathbb{Z}$ bounded from above, I'm trying to prove that $A_0$ has a largest element.

I've seen a few complicated proofs that are nothing like mine, and I'm a little concerned about a particular step of my proof, so I'd like to check if my argument is valid.

Proof:

Since $A_0$ is bounded from above and $A_0\subset\mathbb{Z}\subset\mathbb{R}$, we have that $s:=\sup(A_0)$ exists in $\mathbb{R}$ and is such that $s\geq a$ for all $a\in A_0$.

If $s\in A_0$, we are done since then $s$ is larger than all the elements of $A_0$, yet is a member of $A_0$ itself, so $s$ is the largest element of $A_0$.

If $s\notin A_0$, then still by definition $s>a$ for all $a\in A_0$, but not equal to any elements $a$ of $A_0$ since $s$ is not in $A_0$. Thus $\exists t=(s+a)/2$ such that $s>t>a$ for all $a\in A_0$, so $t$ is an upper bound of $A_0$ less than $s$, contradicting that $s$ is the least upper bound of $A_0$. So $s\in A_0$ after all. $\blacksquare$

I guess I'm concerned with the last paragraph and the naive construction of $t$ for all $a$. Any check would be appreciated.

Answer 1

It depends on what you are allowed to use in your proof. If you are allowed to use the fact that $\mathbb N$ is well-ordered, then the argument is simple. First, replace your set $A_0$ by $-A_0$ so as to have a subset of $\mathbb Z$ bounded from below rather than above. Then use a lower bound $L$ to convert the set to one contained in $\mathbb N$, by adding $L$ (or $L+1$ depending on your definition of $\mathbb N$). Now your set is contained in $\mathbb N$, and you can apply the well-order of $\mathbb N$ to conclude.

Answer 2

When you define $t \in \mathbb{R}$ as the arithmetic mean between $s = \sup A_0$ and $a \in A_0,$ you're choosing a particular element of $A_0$ to do this, which doesn't necessarily imply that $t$ is an upper bound for $A_0.$ For $t$ to be an upper bound for $A_0$ you would need to pick the largest $a \in A_0,$ which defeats the purpose of your proof.

To arrive at a contradiction, I think you could argue as follows. First of all, note that $s$ has to be an integer, otherwise $\lfloor s \rfloor$ would be an upper bound for $A_0$ (there are no integers strictly between $\lfloor s \rfloor$ and $s$ if $s \notin \mathbb{Z}$), which contradicts the definition of $s.$
Now, if $s \notin A_0$, the number $s-1$ has to be in $A_0,$ otherwise it would be an upper bound and this again contradicts the definition of $s.$ But if $s-1 \in A_0$ then $s = \sup A_0 = s-1,$ which is impossible.