Help with proof that nonempty set of integers bounded above has a max

Question

Can I get some feedback?I am studying this proof and am having trouble interpreting it.

Question:

Let $U= \mathbb{Z}$ Prove if $S \subseteq \mathbb{Z}$ and $s\neq \emptyset$ and S is bounded above

then $S$ has a maximum

Attempt By the completeness axiom $S$ has a supremum call it $s$.Assume $s \notin S$. Then $s>n$ for all $n \in S$.But $s-1$ is not an upper bound so $\exists n \in S$ such that $s-1<n$.Also since $n<s$, $\exists k \in S$ such that $k>n$.

$\implies$ $s-1<n<k<s$

$\implies k-n<1$ a contradiction

Is the contradiction the fact that we assumed that $k$ and $n$ are integers and since their difference is between $0$ and $1$ this is impossible? Also is $k-n<1$ because $n$ and $k$ have to be numbers in between $0$ and $1$, since $s-(s-1)=1$?

Answer 1

The completeness axiom says that if $S \subset of \mathbb R$ is bounded above then a real $s =\sup S$ exists.

So if $S\subset \mathbb Z\subset \mathbb R$ is bounded above then there is a real $s =\sup S$.

The proof is to prove three things.

1) $s$ is actually an integer.

2) $s$ is a member of $S$

3) $s = \max S$.

Now the definition of $\sup S$ is that for any $y < s$ then $y$ isn't an upper bound of $S$. So $s-1 < s$ so $s-1$ is not an upper bound of $S$ which means there is an element $n\in S$ so that $n > s-1$.

So we have $s-1 < n \le s$. We don't yet know that $s$ is an integer (that's our goal) but we do know that $n$ is an integer because $n\in S\subset \mathbb Z$.

With me so far?

Okay, our assumption (which is going to be proven false) is that $s \not \in S$.

That means $s-1 < n < s$. That's seems okay, that means that $n$ is between two non-integers $s-1$ and $s$. That can happen, right.

Except..... $n < s$. So $n$ is also not an upper bound. So there is another $k \in S$ so that $s-1 < n < k\le s$.

And $k \in S \subset \mathbb Z$ so $k$ is an integer.

So we have two integers $n,k$ so that $s - 1 < n < k \le s$. That's not okay.

That means $1 = s-(s-1)\ge k-(s-1) > k-n > n-n = 0$

Or,, to highlight the important stuff: $0 < k-n < 1$.

SO you ask:

Is the contradiction the fact that we assumed that k and n are integers and since their difference is between 0 and 1 this is impossible?

Yes, If $n,k$ are integers then so is $k-n$ but there are no integers between $0$ and $1$.

So that is the contradiction.

And

Also is k−n<1 because n and k have to be numbers in between 0 and 1, since s−(s−1)=1?

Yes, that's right.

....

So we have contradicted our assumption that $s\not \in S$ so $s\in S$.

1) $s$ is an integer.

2) $s\in S$.

3) if $t \in S$ than $t \le s$ because $s$ is a an upper bound so

$s = \sup S = \max S$.