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:

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

That is:

$\ds \lim_{x \mathop \to n^+} \floor x = \floor n$

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

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $2$