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!

#Calculus

Easy Math Editor

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Comments

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 $r\in\Bbb R\setminus\{0\}$ .

Prasun Biswas - 4 years, 8 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!

Efren Medallo - 4 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$