Definition of floor function as the maximum of the set $\{m\in\mathbb{Z}|m\leq x\}$

Question

On Wikipedia's article, https://en.m.wikipedia.org/wiki/Floor_and_ceiling_functions , The definition of the floor function is given by:

$\forall x\in\mathbb{R},\lfloor x \rfloor =\max\{m\in\mathbb{Z}|m\leq x\}$

Intuitively it is clear that $\max\{m\in\mathbb{Z}|m\leq x\}$ exist, But when I try to prove that the maximum is indeed exist I got stuck, Here's what I did:

Let $x\in\mathbb{R}$ be some real number, Let's denote $S(x)=\{m\in\mathbb{Z}|m\leq x\}$, It is clear that there exists an integer $m_0\in\mathbb{Z}$ such that $m_0\leq x$ and thus $m_0\in S(x)$ and so $S(x)\neq \emptyset$, Now since $S(x)$ is bounded above by $x$, We get that $S(x)$ is a non-empty set of real numbers that is bounded above and thus by the Supremum Axiom, The supremum of $S(x)$, $\sup ( S(x))$ exist, Now in order to show that the maximum of $S(x)$ also exist, we have to show that $\sup (S(x))\in S(x)$ and we can conclude by this, that $\max(S(x))= \sup(S(x))$ and thus the floor function can be defined as $\max(S(x))$.

The problem is how can we show that indeed $\sup(S(x))\in S(x)$?

By definition of supremum we know that:

1. $\forall m\in S(x), m\leq \sup(S(x))$
2. $\forall \epsilon\in (0,\infty), \exists m_0\in S(x), \sup(S(x))-\epsilon<m_0$

By definition of $S(x)$, It is enough for us to show that $\sup(S(x))\in \mathbb{Z}$, because, We know that $x$ is an upper bound of $S(x)$, And since, $\sup(S(x))$ is the least upper bound of $S(x)$ we get that $\sup(S(x))\leq x$ and thus we can conclude by this fact and the fact that $\sup(S(x))\in \mathbb{Z}$, that $\sup(S(x))\in S(x)$ and we can conclude that $\max(S(x))=\sup(S(x))$ and thus $\max(S(x))$ exist and the floor function can be defined as above.

So what is left for us to show is that indeed $\sup(S(x))\in \mathbb{Z}$ and this is where I got stuck.

Thanks for any help.

Answer 1

Put $s = \sup(S(x))$. Note that for all $\epsilon \in (0, 1/2)$, there is an $m_\epsilon \in S(x)$ such that $s - \epsilon < m_\epsilon \leq s$. Now, each such $m_\epsilon$ is an integer, and since we pick our $\epsilon$ in $(0, 1/2)$, all $m_\epsilon$ are actually the same (all $m_\epsilon$ lie between $s - 1/2$ and $s$ and there is at most one integer between $s - 1/2$ and $s$). Let $m$ be this common value. Then we see that $$ s - \epsilon < m \leq s $$ for all $\epsilon \in (0, 1/2)$, or equivalently, $|m - s| < \epsilon$ for all $\epsilon \in (0, 1/2)$. This can only happen if $m = s$.