Example 15: The greatest integer function $[x]$ breaks at every integer

Basics you need

Solution

1. Domain and the step shape

   Every real number lies in exactly one interval $[n, n+1)$ with $n$ an integer, so $[x]$ is defined for every real $x$⁠: the domain is $\R$⁠. On each such interval the function is constant:

   $$[x] = n \quad \text{for } n \le x < n + 1.$$

   For example $[x] = 0$ on $[0, 1)$⁠, $[x] = 1$ on $[1, 2)$⁠, and $[x] = -1$ on $[-1, 0)$⁠.

   Writing $[x]$ as "the constant $n$ on $[n, n+1)$" turns it into a piecewise function with infinitely many pieces. Then continuity can be studied exactly like the piecewise examples before: inside the pieces and at the junctions, which here are the integers.
   The left end $n$ belongs to the step ($n \le x$⁠), the right end $n + 1$ does not ($x < n + 1$⁠). That is why each step has a filled dot on the left and an open dot on the right. Basics: piecewise functions

2. Case 1: $c$ is not an integer

   Let $n = [c]$⁠. Because $c$ is not an integer, $c \ne n$⁠, so $n < c < n + 1$ with strict inequalities. Let $\delta$ be the smaller of the two gaps:

   $$\delta = \min(c - n,\ n + 1 - c) > 0.$$

   Every $x$ with $|x - c| < \delta$ satisfies $n < x < n + 1$⁠, so $[x] = n$⁠. Thus $f$ is constant near $c$⁠, and

   $$\lim_{x\to c} [x] = n = [c] = f(c).$$

   So $f$ is continuous at every real number that is not an integer.

   The book says this is "evident from the graph": near a non-integer, the graph is a flat piece of one step. The $\delta$ makes that precise. For $c = 2.3$⁠: $n = 2$⁠, the gaps are $0.3$ and $0.7$⁠, $\delta = 0.3$⁠, and every $x$ in $(2, 2.6)$ has $[x] = 2$⁠.
   A function that is constant on an interval around $c$ has that constant as its limit at $c$ (Basic 2). Basics: what a limit is

3. Case 2: $c$ is an integer, the value

   If $c$ is an integer, the greatest integer $\le c$ is $c$ itself:

   $$f(c) = [c] = c.$$
   $c \le c$⁠, and no larger integer is $\le c$⁠. In the graph this is the filled dot at the left end of the step at height $c$⁠.

4. Case 2: the left-hand limit at an integer $c$

   For $c - 1 \le x < c$ we have $[x] = c - 1$ (the step at height $c - 1$⁠). So for every $x$ in the interval $(c - 1, c)$ just to the left of $c$⁠, $f(x) = c - 1$⁠, and

   $$\lim_{x\to c^-} [x] = c - 1.$$
   Inputs approaching $c$ from the left, like $c - 0.1$⁠, $c - 0.01$⁠, $c - 0.001$⁠, all lie on the step below. The book phrases this as: for a sufficiently small $r > 0$⁠, $[c - r] = c - 1$⁠. In fact any $r$ with $0 < r \le 1$ works.
   Basics: left-hand and right-hand limits

5. Case 2: the right-hand limit at an integer $c$

   For $c \le x < c + 1$ we have $[x] = c$⁠. So for every $x$ in $(c, c + 1)$⁠, $f(x) = c$⁠, and

   $$\lim_{x\to c^+} [x] = c.$$
   Inputs just to the right of $c$ lie on the step that starts at $c$⁠. In the book's words, $[c + r] = c$ for small $r > 0$ (any $0 < r < 1$ works).

6. Case 2: compare

   $$\begin{aligned} \lim_{x\to c^-} [x] &= c - 1, \\ \lim_{x\to c^+} [x] &= c, \\ f(c) &= c. \end{aligned}$$

   Since $c - 1 \ne c$ for every $c$⁠, the two one-sided limits never agree, $\lim_{x\to c} [x]$ does not exist, and $f$ is discontinuous at every integer.

   The two one-sided limits always differ by exactly 1, the height of each step. Notice that the right-hand limit does equal $f(c)$⁠: the graph is "continuous from the right" at each integer (the filled dot sits on the right-hand step). That is not enough, because Definition 1 needs the left-hand limit as well.
   Lesson: continuity at a point

7. Conclusion

   Every real number is either a non-integer (Case 1, continuous) or an integer (Case 2, discontinuous). Hence the points of discontinuity of $[x]$ are exactly the integers, $\dots, -2, -1, 0, 1, 2, \dots$⁠, that is, the set $\Z$⁠.

   The two cases cover all of $\R$⁠, so the list is complete: no integer is missed, and no non-integer is a point of discontinuity.

Every fact used, and where it comes from