Waste less time on Facebook — follow Brilliant.
×

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
7 months, 4 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 · 7 months, 3 weeks ago

Log in to reply

@Prasun Biswas 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 · 7 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...