# How does one prove this?

I have come across with a function, which, allegedly, is only continuous at one point.

Consider this function defined over $$\mathbb{R}$$.

$f(x) = \begin{cases} x \: \text{for} \: x \in \mathbb{Q} \\ 0 \: \text{for} \: x \notin \mathbb{Q} \end{cases}$

How does one prove its continuity at $$x=0$$ (and the fact that it is only continuous at that point) using the delta-epsilon definition of continuity? Thank you!

Note by Efren Medallo
1 year, 11 months ago

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Continuity at $$x=0$$ :

We first show that $$\lim\limits_{x\to 0}f(x)$$ exists and equals $$0$$. To do this, we need to show that for all $$\epsilon\gt 0$$, there exists a $$\delta\gt 0$$ such that $$|x|\lt\delta\implies |f(x)|\lt\epsilon$$. If we choose $$\delta=\epsilon$$, then for all rationals $$x$$ satisfying $$|x|\lt\delta$$, we have $$|f(x)|=|x|\lt\delta=\epsilon$$, i.e., $$|f(x)|\lt\epsilon$$. For the irrationals $$x$$ satisfying $$|x|\lt\delta$$, we have $$|f(x)|=0\lt\epsilon$$. So, choosing $$\delta=\epsilon$$, we conclude using the epsilon-delta definition of limit that $$\lim\limits_{x\to 0}f(x)=0$$. Also, we note that $$f(0)=0$$ from the definition of $$f$$. Hence, we conclude that $$f$$ is continuous at $$x=0$$.

Discontinuity at all other points :

We need to show that $$f$$ is discontinuous at $$x=r\neq 0$$, i.e., we need to show that there exists $$\epsilon\gt 0$$ such that for all $$\delta\gt 0$$, there exists $$x\in\Bbb R$$ such that $$|x-r|\lt\delta\implies |f(x)-f(r)|\not\lt\epsilon$$.

Consider $$\epsilon=|r|$$ and $$\delta\gt 0$$.

Let $$r\notin\Bbb Q$$. Since $$\Bbb Q$$ is dense in $$\Bbb R$$, there exist $$r_1,r_2\in\Bbb Q$$ such that $$r-\delta\lt r_1\lt r\lt r_2\lt r+\delta$$. If $$r\lt 0$$, choose $$x=r_1$$ and if $$r\gt 0$$, choose $$x=r_2$$. In both cases, we have $$|f(x)-f(r)|=|f(x)|\not\lt\epsilon$$. This shows that $$f$$ is discontinuous at $$x=r$$ for all irrationals $$r$$.

Now, let $$r\in\Bbb Q\setminus\{0\}$$. Since $$\Bbb R\setminus\Bbb Q$$ is also dense in $$\Bbb R$$, there exist $$i_1\in\Bbb R\setminus\Bbb Q$$ such that $$r-\delta\lt i_1\lt r\lt r+\delta$$. Choose $$x=i_1$$ and then we have $$|f(x)-f(r)|=|0-r|=|r|\not\lt\epsilon$$. This shows that $$f$$ is discontinuous at $$x=r$$ for all non-zero rationals $$r$$.

So, we conclude that $$f$$ is discontinuous at $$x=r$$ for all $$r\in\Bbb R\setminus\{0\}$$.

- 1 year, 11 months ago

Thanks for that! I knew I was on the right track on trying to prove its continuity at $$x = 0$$, but I could never find the mathematical backing in proving its discontinuity elsewhere!

- 1 year, 11 months ago