Floor Function is Continuous on Right at Integer

Theorem
Let $f$ be the real function defined as:
 * $\map f x = \floor x$

where $\floor{\, \cdot \,}$ denotes the floor function.

Let $n \in \Z$ be an integer.

Then $\map f x$ is continuous on the right at $n$.

Proof
From Real Number is Integer iff equals Floor:
 * $\floor n = n$

By definition $\floor x$ is the unique integer such that:
 * $\floor x \le x < \floor x + 1$

Consider the open real interval:
 * $\II = \openint n {n + 1}$

By definition of $\floor x$ we have that:
 * $\forall x \in \II: \floor x = n$

That is:
 * $\forall x \in \II: \floor x - n = 0$

Let $\epsilon \in \R_{>0}$.

Let $\delta \in \openint n {n + 1}$.

Then:
 * $\forall x \in \II: n < x < n + \delta: \floor x < \size {\floor x - n} < \epsilon$

That is:
 * $\ds \lim_{x \mathop \to n^+} \floor x = \floor n$

Hence, by definition, $\map f x$ is continuous on the right at $n$.