How to prove that if $S\subset \mathbb{Z}$ has a supremum $a$, then $a\in S?$

Question

This question came up in my undergraduate intro to Real Analysis class. We are given that $\mathbb{R}$ is a complete ordered field, and $\mathbb{Z}$ is a subset of the set $\mathbb{R}$.

I don't really know where to start. I thought of starting from the definition of integers, but we haven't been given one. I tried to make up a definition and the best I can think of is $\{1, 1+1, 1+1+1, ...\} \cup \{-1, -1 + (-1), -1 + (-1) + (-1),...\}$, but I'm not sure of how to make it more formal.

Answer 1

If the supremum $S$ of a set of integers $X$ was not itself an integer, there is an integer below $S$ written as $\lfloor S \rfloor$ and an integer above $S$ written $\lceil S \rceil$. Note that $\lfloor S \rfloor$ is the maximal element in $X$, because $S$ is a sup. Also, since $S$ is not an integer, $S - \lfloor S \rfloor > 0$. Now the supremum of a set $X$ has the property that for every $\epsilon > 0$, $S - \epsilon$ is not an upper bound for $X$, since $S$ is the least upper bound. But if we take $\epsilon = \frac{S - \lfloor S \rfloor}{2}$, then $S - \epsilon$ is still an upper bound.

This shows that $S$ has to actually be an integer, and in particular, it has to belong to $X$. Notice that there is nothing special about the integers here, and this in fact is true fr any discrete set of points on the real line.

Answer 2

If $a= \sup S$, then $s \leq a$ for all $s \in S$. In particular, $S$ is bounded from above.

Since $|n -m| \geq 1$ for any two different elements $n, \, m \in \mathbb{Z}$, it follows that there exists $\bar{s} \in S$ such that $s \leq \bar{s}$ for all $s \in S$.

By the definition of supremum we have $a \leq \bar{s}$. On the other hand, $\bar{s} \in S$ implies $\bar{s} \leq a$.

So we deduce $a = \bar{s}$ and we are done.

Answer 3

Suppose $s = \sup S$ is a real number. There is some integer $n$ so that $n-1 < s \le n$.

Let $d$ be so that $n-1 < d < s$. By definition of least upper bound $d$ is not an upper bound.

So there exists a $t \in S$ so that $ d < t \le s$. So $n-1 < d < t \le s \le n$. But $t$ is an integer. And $n-1 < t \le n$.

The only way that is possible is if $t = n$.

So $n = t \le s \le n$.

The only way that is possible is if $n = t = s$. So $s = t \in S$.

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Note: this all hinges on knowing i) for any real number $s$ there is an integer $n$ so that $n < s \le n+1$.

and ii) for any two real numbers $a, b$ so that $a < b$, there exists a real number $d$ so that $a < d < b$.

and iii) that for any integer $n$ there is not integer between $n$ and $n+1$.

1 and 20 are the Archimedean property. They are very basic but actually must be proven in a basic analysis course. 3) is ... well, it depends on your axioms and definitions.

===== take 2 ====

. I thought of starting from the definition of integers, but we haven't been given one.

That's actually a problem. I think the single aspect of integers you should take as a given is that if $n$ is an integer then $n$ is the only integer within the open interval $(n-1, n+1)$[$*$].

This means if $s = \sup S$ then for any interval $(s - d, s]$ must contain an element of $S$. And that element must be an integer. And this must be true no matter how small $d$ is.

But if $d < 1$ then this must be the only integer in the interval. And it must exist in all such intervals no matter which choice we choose for $d$.

The only point in all $(s -d,s]$ is $s$ itself. So $s$ must be the integer element.

[$*$]. This may have trouble but I believe this is a recursively sound definition of integer:

i) $0$ is an integer.

ii) There are no integers in $(0, 1)$.

iii) $n$ is an integer if and only if $n+1$ is an integer.