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We have $$f(x)=\left\{\begin{matrix} 1 & \exists n\in\mathbb N,\ x=\frac{1}{n}\\ 0 & \text{otherwise} \end{matrix}\right.$$ on the interval $(-1,1)$. I am asked to find the points of continuity and discontinuity.

My Attempt: First of all, on $(-1,0)$ we have $f(x)=0$ and so from the locality of the limit we can say the limit at any point there is $0$ - $f$ is continuous.

If $x_0=0$ we can look at the sequences $\frac{1}{n},\ \frac{\sqrt2}{n}$ and show easily there's no limit at $0$ by Heine- discontinuity of 2nd kind.

If $x_0=\frac{1}{n}$ we can take the neighborhood with $\delta=\min\left \{ x_0-\frac{1}{n+1},\frac{1}{n}-x_0 \right \}$ (wihout $x_0$) and show the limit is $0$ because at any point in the neighborhood, $f(x)=0$, thus there's a discontinuity of 1st kind.

Any comments on what I did so far are welcome.

My Problem: Showing $f$ is continious at any other point, $x_0\in (0,1)-\left \{\frac{1}{n}|n\in\mathbb N \right \}$. My best attempt is the idea to show that for any sequence $x_0\neq x_k\rightarrow x_0$ starting from some spot, $x_k$ cant be of the form $\frac{1}{n}$ (cough, might be revolved around partial limits). However, I can't figure out how to prove that AND I don't think I want to. there must be a better way.

To summarize- I'd like to see if my solution so far can be improved/rewritten in a completly different way, and solve what's left to prove. Thanks in advance for anyone who helps!

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    You say "locality of the limit" on $(-1,0)$. Is there a neighborhood for which the same tactic can be applied to answer your question? – 2017-01-04
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    Of course. It can be used aswell in the other points I need help with. I don't know which neighborhood can be used. I do have in mind using $n=floor(\frac{1}{x_0})$, and look at a neighborhood in $(\frac{1}{n+1},\frac{1}{n})$, but I'm not really sure exactly what to write. Dare I say that at that neighborhood, $f(x)=0$? – 2017-01-04
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    You can use `\begin{cases} ... \end{cases}` instead of `\begin{matrix} ... \end{matrix}` for MathJax. – 2017-01-04

2 Answers 2

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Hint:

If $x_0\in(0,1)\setminus\{1/n:n\in\Bbb N\}$ then $$\frac1{x_0}\in\left(\left\lfloor\frac1{x_0}\right\rfloor, \left\lceil\frac1{x_0}\right\rceil\right)$$

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    So I see my direction in the comments is kind of valid as it's almost like yours. So I look at the neighborhood with $\delta=\min\left \{ x_0-\frac{1}{\left \lfloor \frac{1}{x_0} \right \rfloor},\frac{1}{\left \lceil \frac{1}{x_0} \right \rceil}-x_0 \right \}$? – 2017-01-04
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The set $\{0, \frac 1n|n\in \mathbb N\}$ is a compact set.

If it is compact then it is closed.

the compliment of a closed set is an open set.

$X = (0,1) - \{\frac 1n|n\in \mathbb N\}$ is an open set.

Around every $x\in X$ there is a neighborhood such at everything in that neighborhood is in $X.$

$f(x) = 0$ for all $x$ in that neighborhood.

$f(x)$ is continuous when $x \in X.$

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    In this course we don't touch the topology of the reals too much (if that's even related to that topic? haha), so unfortunately this solution isn't something I understand. :/ – 2017-01-04
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    It is indeed a topology of the real numbers sort of answer, but it is a topology of the reals sort of question. So, it seemed to be fair game. – 2017-01-04