Answer to Question #87458 in Real Analysis for Venerette

Question

For each x ∈ R, let us denote by C(x) the least integer greater than or equal to x.
For example, C(1) = 1, C(−√2) = −1. In other words, C(x) is the unique integer
satisfying C(x) − 1 < x ≤ C(x).
(1) Draw the graph of the function C(x) for x ∈ [−2, 2].
(2) Prove that C(x) is continuous at all non-integer points of R.
(3) Prove that C(x) is discontinuous at all integer points of R.

Answer 1

Let x be a real number. The ceiling function of x, denoted by $\lceil x\rceil$, is the smallest integer that is larger than or equal to x.

For $x\in[-2, 2]$

$$x=-2, f(x)=\lceil x\rceil=-2$$

$$-2\lt x \le-1, f(x)=\lceil x\rceil=-1$$

$$-1\lt x \le0, f(x)=\lceil x\rceil=0$$

$$0\lt x \le1, f(x)=\lceil x\rceil=1$$

$$1\lt x \le2, f(x)=\lceil x\rceil=2$$

For $a_i\in(i, i+1), i\in \Z$

$$\lim\limits_{x\to a_i}f(x)=i+1, f(a)=i+1, x\in(i, i+1)$$

Therefore, the function $f(x)$ is continuous at $x\in \R, x\notin \Z$

For $a_i= i\in \Z:$

for $x\in(i-1, i)$

$$\lim\limits_{x\to a_i}f(x)=i,$$

for $x\in(i, i+1)$

$$\lim\limits_{x\to a_i}f(x)=i+1.$$

Then

$$\lim\limits_{x\to a_i}f(x)=does \ not \ exist$$

Therefore, the function $f(x)$ is discontinuous at $x\in \Z.$