Equivalence of Definitions of Floor Function

Theorem
Let $x$ be a real number.

Definition 1 equals Definition 2
Follows from Supremum of Set of Integers equals Greatest Element.

Definition 1 equals Definition 3
Let $S$ be the set:
 * $S = \set {m \in \Z: m \le x}$

Let $n = \sup S$.

By Supremum of Set of Integers is Integer, $n \in \Z$.

By Supremum of Set of Integers equals Greatest Element, $n\in S$.

Because $n \in S$, we have $n \le x$.

Because $n + 1 > n$, we have by definition of supremum:
 * $n + 1 \notin S$

Thus $n + 1 > x$.

Thus $n$ is an integer such that:
 * $n \le x < n + 1$

So $n$ is the floor function by definition 3.

Definition 3 equals Definition 2
Let $n$ be an integer such that:
 * $n \le x < n + 1$

We show that $n$ is the greatest element of the set:
 * $S = \set {m \in \Z: m \le x}$

Let $m \in \Z$ such that $m \le x$.

We show that $n \ge m$.

$m > n$.

By Weak Inequality of Integers iff Strict Inequality with Integer plus One:
 * $m \ge n + 1$

and so from the definition of $g$ it follows that $m > x$.

By Proof by Contradiction it follows that $m \le n$.

Because $m \in S$ was arbitrary, $n$ is the greatest element of $S$.

Thus $n$ is the floor function by definition 2.