# Set of points of discontinuity of $f(x)=\lfloor x \rfloor$

Question: Give the set of points of discontinuity of $f(x)= \lfloor x \rfloor$.

Answer: When $x=0.9999, y=0$ but when $x=1, y=1$, so for a very small change in the value of $x$, the change in the value of $y$ is big and this happens at all the integers (positive as well as negative), therefore the set of all the integers is the answer. Am I correct?

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Yes, you are correct. – Josué Tonelli-Cueto Oct 8 '11 at 12:29
Formally, you could use the definition of continuity, say, with epsilon = 1/2 – The Chaz 2.0 Oct 8 '11 at 12:33
Yes Its true for $x\in \mathbb{Z}$ – Ramana Venkata Oct 8 '11 at 12:42
Rigorously speaking, the values of f at x=0.9999 and at x=1 do not prove nor disprove the continuity of f at x=1. Hence I assume that 0.9999 stands for any value less than 1 and close to 1... in which case your idea is correct but cannot be called a proof. Yet. – Did Oct 8 '11 at 12:44
You have a clear geometric idea of what's going on. Depending on the type of course, you may be expected to write it up semi-formally (as $x$ approaches an integer $n$ from the left, $\lfloor x\rfloor$ does not approach $\lfloor n \rfloor$) or formally ($\epsilon$-$\delta$). Don't forget that you also need to show that your list of points of discontinuity is complete, that is, that if $a$ is not an integer, then the function $\lfloor x\rfloor$ is continuous at $a$. – André Nicolas Oct 8 '11 at 14:21

## 2 Answers

You can use limits/$\epsilon$-$\delta$ to show that if $a\in\mathbb{Z}$, then $f(x)=\lfloor x\rfloor$ is not continuous at $a$; and that if $a\notin\mathbb{Z}$, then $f(x)$ is continuous at $a$. So the points of discontinuity are exactly the integers.

Arguing first somewhat informally with limits:

Let $a\in\mathbb{Z}$. Then any $x\gt a$ with $0\lt x-a\lt 1$ will have $f(x)=f(a)=a$, so $\lim\limits_{x\to a^+}f(x) = f(a)=a$.

However, if $x\lt a$, $0\lt a-x\lt 1$, theN $f(x) = a-1$, since the largest integer less than or equal to $x$ is $a-1$. So $\lim\limits_{x\to a^-}f(x) = a-1$.

Since $\lim\limits_{x\to a^+}f(x) \neq \lim\limits_{x\to a^-}f(x)$, then $\lim\limits_{x\to a}f(x)$ does not exist, and a fortiori $f(x)$ is not continuous at $a$.

Same argument with $\epsilon$ and $\delta$:

Let $a\in\mathbb{Z}$, and let $\epsilon=\frac{1}{2}$. For any $\delta\gt 0$ take $x=\mathrm{max}(a-1, a-\frac{\delta}{2})$. Then $a-1\leq x \lt a$, so $\lfloor x\rfloor = a-1$. Therefore, $|f(x)-f(a)| = |(a-1)-a| = 1\gt \epsilon$. Thus, there exists $\epsilon\gt 0$ (namely, $\epsilon=\frac{1}{2}$) such that for all $\delta\gt 0$ there is an $x$ with $0\lt |x-a|\lt \delta$, and $|f(x)-f(a)|\gt \epsilon$. Hence, $f(x)$ is not continuous at $a$.

Showing continuity everywhere else somewhat informally using limits:

Now assume that $a\notin\mathbb{Z}$, and let $\delta$ be the distance from $a$ to the nearest integer (that is, $\delta=\min(a-\lfloor a\rfloor, \rceil a\lceil - a)$); since $a\notin\mathbb{Z}$, $\delta\gt 0$.

Then for any $x$ with $|x-a|\lt \delta$, we have $\lfloor a\rfloor \lt x\lt \lceil a\rceil$, so $f(x) = \lfloor a\rfloor = f(a)$. Hence, for all $x$ sufficiently close to $a$, we have $f(x)=f(a)$, so $\lim\limits_{x\to a}f(x) = f(a)$, hence $f(x)$ is continuous at $a$.

I'll leave it to you to take that "somewhat informal" argument and turn it into a forma $\epsilon$-$\delta$ proof of continuity at $a\notin \mathbb{Z}$.

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Yes, that is correct.

If you like epsilons and deltas, you can say that at any integer, if you set $ε=1/2$, then there is no $δ$ so small that if the distance from $x$ to the integer in question is $<δ$, then the distance from $f(x)$ to the value of the function at the integer in question is $<ϵ$. For some of those values of $x$, the distance will be 1. But at non-integers, no matter how small $ε$ is, it is enough to make $δ$ small enough so that the neighborhood of radius $δ$ contains no integers.

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