Continuity of piecewise function involving rationals/irrationals and Cantor set

Question

I'm struggling to determine the continuity of the following function: $$ f(x)=\begin{cases}0 \quad \text{if $x \in \mathbb{Q}\cap D,$}\\ x^3 \quad \text{if $x \notin \mathbb{Q}\cap D$;} \end{cases}$$ where $D$ is the Cantor set over $[0,1]$.

I'm thinking that $f(x)$ is discontinuous for every $x\notin [0,1]\setminus (\mathbb{Q}\cap D)$ - the set of all irrational numbers in $[0,1]$ together with the set of all numbers in $[0,1]$ that are not element of $D$.

1. Can you guys send me some help? Please!
2. For these type of piecewise functions (which involving rationals and irrationals), is there any kind of general steps to find theirs discountinuous points?
3. I know about using $\epsilon -\delta$ definition to show that the limit at a specific point does not exist and so it is a discontinuous point, but it depends a lot on the function to choose such $\epsilon -\delta$. But I'm not sure how to use this method efficiently when I'm trying to prove a set contains all discontinuous points of a function, any tips?.

Thank you for your time.

Answer 1

If $x\notin\Bbb Q\cap D$, then $f(x)=x^3\ne0$. There are two possibilities now:

• $x\in D$: then, for any $\delta>0$, there s some $y\in\Bbb Q\cap D$ such that $|y-x|<\delta$. But $f(y)=0$. So, this proves, by the $\varepsilon-\delta$ definition (I took $\varepsilon=x^3$) that $f$ is discontinuous at $x$.
• $x\notin D$: then, since $D$ is a closed subset of $[0,1]$, there is some $\delta>0$ such that$$(x-\delta,x+\delta)\subset[0,1]\setminus D,$$and $f(y)=y^3$ for each $y\in(x-\delta,x+\delta)$. So, $f$ is continuous at $x$.

If $x=0(\in\Bbb Q\cap D)$, then $f(x)=0$. And, for any $y\in[0,1]$, $|f(y)|\leqslant y^3$, and therefore $\bigl|f(y)-f(0)\bigr|\leqslant y^3$. So, by the squeeze theorem, $\lim>_{y\to0}f(y)=0=f(0)$, and therefore $f$ is continuous at $0$.

And if $x\in\Bbb Q\cap D$ and $x\ne0$, then $f(x)=0$. But, for each $\delta>0$, there is some $y\notin\Bbb Q\cap D$ such that $|y-x|<\delta$. Take $y$ such that $y>\frac x2$, and therefore $f(y)>\frac{x^3}8$. So, if you take $\varepsilon=\frac{x^3}8$, this proves that, for every $\delta>0$, there is some $y$ such that $|y-x|<\delta$ and that $\bigl|f(y)-f(x)\bigr|\geqslant\varepsilon$. So, $f$ is discontinuous at $x$.