Exercise 5.1 · Q19: $x-[x]$ is discontinuous at every integer

Basics you need

Solution

1. The domain, and an arbitrary integer

   $[x]$ is defined for every real $x$⁠, so $g(x)=x-[x]$ is defined on all of $\R$⁠. In particular every integer is in the domain, so it makes sense to ask about continuity there. Fix an arbitrary integer $n$⁠.

   Continuity is only discussed at points of the domain. Working with a general $n$⁠, not $n=1$ or $n=2$⁠, is what turns the argument into a proof for all integers at once.
   An arbitrary element proves a "for all" statement; examples do not. Lesson: continuity at a point

2. Pin down $[x]$ near $n$

   Let $0\lt h\lt1$⁠.

   Left of $n$⁠: $n-1\lt n-h\lt n$⁠. The number $n-1$ is an integer $\le n-h$⁠, and the next integer, $n$⁠, is bigger than $n-h$⁠. So

   $$[\,n-h\,]=n-1.$$

   At $n$⁠: $n$ is itself an integer $\le n$⁠, so $[n]=n$⁠.

   Right of $n$⁠: $n\lt n+h\lt n+1$⁠, so $n\le n+h\lt n+1$ and

   $$[\,n+h\,]=n.$$
   This is exactly the definition $m\le x\lt m+1$ applied with $m=n-1$ and with $m=n$⁠. For example, with $n=-1$ and $h=0.1$⁠: $[-1.1]=-2$ and $[-0.9]=-1$⁠. The value of $[x]$ drops by $1$ as soon as $x$ goes below an integer.
   The same reasoning appears in Example 15, Case 2, for the function $[x]$ itself. Example 15

3. The value $g(n)$

   $$g(n)=n-[n]=n-n=0.$$
   At an integer, $x$ sits exactly on the integer below-or-equal to it, so its fractional part is $0$⁠.

4. The left-hand limit at $n$

   Using $[n-h]=n-1$ from Step 2:

   $$g(n-h)=(n-h)-(n-1)=1-h.$$ $$\lim_{x\to n^-}g(x)=\lim_{h\to0}\,(1-h)=1.$$
   The $n$'s cancel, so the answer is the same for every integer: just left of any integer, the fractional part is $1-h$⁠, close to $1$⁠.
   Left-hand limit: $x=n-h$⁠, $h\gt0$⁠, $h\to0$⁠. Basics: left-hand and right-hand limits

5. The right-hand limit at $n$

   Using $[n+h]=n$ from Step 2:

   $$g(n+h)=(n+h)-n=h.$$ $$\lim_{x\to n^+}g(x)=\lim_{h\to0}h=0.$$
   Just right of an integer the fractional part is the small distance $h$ past it, which shrinks to $0$⁠.

6. Compare: a jump at every integer

   From Steps 3 to 5:

   $$\begin{aligned} \lim_{x\to n^-}g(x)&=1,\\ g(n)&=0,\\ \lim_{x\to n^+}g(x)&=0. \end{aligned}$$

   The two one-sided limits are different ($1\ne0$⁠), so $\lim_{x\to n}g(x)$ does not exist and $g$ is not continuous at $n$⁠. Since $n$ was an arbitrary integer, $g$ is discontinuous at every integral point.

   Definition 1 needs the limit at $n$ to exist and equal $g(n)$⁠. A limit exists only when the left-hand and right-hand limits agree, and here they never do. (The right-hand limit does equal $g(n)$⁠: the graph is continuous from the right at each integer, and the break is on the left.)
   Different one-sided limits mean the limit does not exist. Basics: left-hand and right-hand limits

7. A second proof, from Theorem 1

   Suppose, to reach a contradiction, that $g$ were continuous at some integer $n$⁠. The identity function $x$ is continuous everywhere (Example 6). Then by Theorem 1 (2), the difference

   $$x-g(x)=x-\left(x-[x]\right)=[x]$$

   would be continuous at $n$⁠. But Example 15 shows that $[x]$ is discontinuous at every integer. This contradiction shows $g$ is discontinuous at $n$⁠.

   Theorem 1 says "continuous minus continuous is continuous". If $g$ were continuous at $n$⁠, we could rebuild $[x]$ from two continuous pieces, which is impossible. This proof is shorter but relies on Example 15; the direct proof in Steps 2 to 6 does not.
   Theorem 1 (2): if $f$ and $g$ are continuous at $c$⁠, so is $f-g$⁠. Lesson: algebra of continuous functions

8. For completeness: what happens between integers

   The question asks only about integers, but the picture is clearer with the rest. If $c$ is not an integer, let $m=[c]$⁠, so $m\lt c\lt m+1$⁠. For all $x$ in the open interval $(m,m+1)$ around $c$⁠, $[x]=m$⁠, so there

   $$g(x)=x-m,$$

   a polynomial. Hence $\lim_{x\to c}g(x)=c-m=g(c)$⁠, and $g$ is continuous at every non-integer.

   This matches Example 15, Case 1: $[x]$ is constant near a non-integer. So the integers are exactly the points of discontinuity of $g$⁠.

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