I have this question:

Find a function $f :\mathbb R \to\mathbb R$ which is discontinuous at the points of the set $\{\frac1n : n \text{ a positive integer}\} \cup \{0\}$ but is continuous everywhere else.

I really don't know what to do. I was thinking maybe: $$ f(x) = \begin{cases} 1 \quad&\text{if }x=0 \\ 0 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\\ x &\text{otherwise} \end{cases} $$ But that kind of seems like 'cheating'. Is there a better example?

EDIT: Would it be better to have:

$$ f(x) = \begin{cases} 1 &\text{if } x \text{ is in } \{\tfrac1n : n \text{ a positive integer}\}\cup \{0\}\\ 0 &\text{otherwise} \end{cases} $$

  • $\begingroup$ Looks like Thomae's function: en.wikipedia.org/wiki/Thomae's_function $\endgroup$ – Austin Mohr Jan 25 '13 at 13:35
  • $\begingroup$ Actually a later part of this question seems to involve that function. $\endgroup$ – Joe Jan 25 '13 at 13:38
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    $\begingroup$ That's not cheating at all, as long as the function is well defined (it certainly is) and it verifies the requisites (does it?) $\endgroup$ – leonbloy Jan 25 '13 at 14:11
  • $\begingroup$ It does verify the requisites right? f(x) = 0 as x tends to 0 but f(0)=1 which is not the same, so it's discontinuous at 0, Same for all the 1/n as well, unless I've got this very wrong. $\endgroup$ – Joe Jan 25 '13 at 16:38
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    $\begingroup$ I fully agree with @leonbloy. Since the function satisfies the assumptions, it is a correct answer. It is good in that it is simple, so you immediately see what happens. That possibly makes it better than any "natural" example. $\endgroup$ – Jakub Konieczny Apr 8 '13 at 8:11

You can modify the fractional part function: $ \{x\} = \lceil x \rceil - x $, which is discontinuous at integers; to

$$ f(x) = \begin{cases} 0 & ; x = 0 \\ \left\{\dfrac{1}{x}\right\} & ; otherwise \end{cases} $$


I'm going to assume your true question is finding an answer that you do not consider "cheating."

Find a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is discontinuous at each point in $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$ and $f$ is continuous at each point in the complement of $K$ which is denoted $(\mathbb{R}\setminus K)$

General Answer
Let $g:\mathbb{R}\to\mathbb{R}$ be an arbitrary continuous function. Let $\epsilon>0$ be an arbitrary positive real number.

Your edited answer has $g$ be the zero function and $\epsilon$ be $1$

Define $f:\mathbb{R}\to\mathbb{R}$ by $$f(x)=\begin{cases} g(x)+\epsilon&\text{if }x\in K\\ g(x)&\text{if }x\in(\mathbb{R}\setminus K) \end{cases}$$ for every $x\in\mathbb{R}$ where $K\overset{\text{def}}{=}\{\frac{1}{n}:n\in\mathbb{N}\text{ and }n\ne 0\}\cup\{0\}$.

The reason why I introduce "$K$" is because this method can be used for any given nowhere dense set $K$ where you want discontinuities. The set you were given is not special in any way for this problem. However it is a classic example of a compact set, but that's not relevant to your problem.

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    $\begingroup$ Your reason for introducing $K$ is misleading. The given set is special: it contains no open intervals. If $K = [0, \infty)$ your construction yields a function with only one point of discontinuity: $0$. $\endgroup$ – Unit Jun 1 '15 at 15:35
  • $\begingroup$ Post has been edited. (thanks) $\endgroup$ – Alberto Takase Jun 5 '15 at 8:39

Put $A = \{\frac{1}{n}\mid n \geq 1\}$ and

$$f: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases} 0 \quad x \notin A\cup \{0\} \\ 1 \quad x \in A \cup \{0\} \end{cases}$$


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