Real Number is Integer iff equals Ceiling

Theorem
Let $x \in \R$.

Then:
 * $x = \ceiling x \iff x \in \Z$

where $\ceiling x$ is the ceiling of $x$.

Proof
Let $x = \ceiling x$.

As $\ceiling x \in \Z$, then so must $x$ be.

Now let $x \in \Z$.

We have:
 * $\ceiling x = \inf \set {m \in \Z: m \ge x}$

As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
 * $x = \ceiling x$