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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:

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

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Looks like Thomae's function: en.wikipedia.org/wiki/Thomae's_function –  Austin Mohr Jan 25 '13 at 13:35
    
Actually a later part of this question seems to involve that function. –  Joe Jan 25 '13 at 13:38
3  
That's not cheating at all, as long as the function is well defined (it certainly is) and it verifies the requisites (does it?) –  leonbloy Jan 25 '13 at 14:11
    
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. –  Joe Jan 25 '13 at 16:38
1  
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. –  Feanor Apr 8 '13 at 8:11

1 Answer 1

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

Question/Problem
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 set 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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