# Showing a function has only one point of continuity.

Let

$$f(x) = \begin{cases}\;\;\, x\;\;,\;\text{ if } x \in \mathbb{Q}\\ -x\;\;,\; \text{ if } x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$$

(i) Determine the point or points of continuity of $f$. (ii) Show that the point of points of continuity of $f$ are the only points.

Clearly its continuous at $0$. I'm not sure how to prove it is continuous at $0$, but I know how to prove that it has no other points of continuity.

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## 2 Answers

Let $\,\{x_n\}_{n\in\Bbb N}\,$ be a real sequence s.t. $\,x_n\xrightarrow [n\to\infty]{} 0\,$ (and thus also $\,-x_n\xrightarrow [n\to\infty]{} 0\,$) .

It's easy to see that we have

$$f(x_n)=\pm x_n\xrightarrow [n\to\infty]{}0=f(0)$$

And $\,f\,$ is thus continuous at $\,x=0\,$ as the above's true for any real sequence converging to zero.

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Let $\epsilon >0$. Then for $\delta = \epsilon$ if $|x|<\delta \Rightarrow |f(x)-f(0)|=|f(x)|=|x|<\delta=\epsilon \Rightarrow$
$f$ is continuous at $0$ .

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