Let $C$ be the Cantor set. Prove $g(x)=\left\{\begin{array}{ll}1 & x\in C \\ 0 & x\notin C\end{array}\right.$ not continuous at any $c\in C$.

Question

Is this proof sound? Also it feels like this proof is unnecessarily confusing, but I'm not sure how to improve it.

Proof: Let $c \in C$ and $\epsilon = 1$. Let $\delta > 0$. Since $\displaystyle \lim_{n \to \infty} \frac{1}{3^{n}} = 0$ (we are skipping the proof of this), there exists $N \in \mathbb{R}$ such that for all $n \in \mathbb{N}$, if $n > N$ then $|\frac{1}{3^{n}} - 0| = \frac{1}{3^{n}} < \delta$. Let $n > N$. We know $C_{n}$ is the union of $2^{n}$ closed intervals, each of length $\frac{1}{3^{n}} < \delta$. Define $A=[a_{1}, a_{2}]$ to be the interval of $C_{n}$ that contains $c$. Since $A$ has length $\frac{1}{3^{n}}$, we know $c - \delta < a_{1}$ and $c + \delta > a_{2}$. For any $b \in C_{n}$ with $b \notin A$, we know $b \geq a_{2} + \frac{1}{3^{n}}$ or $b \leq a_{1} - \frac{1}{3^{n}}$. If $1 \notin A$, there exists $x \in [0,1]$ such that $-(c + \delta) < a_{2} < x <$ min$\{c + \delta, a_{2} + \frac{1}{3^{n}}\}$. If $1 \in A$, there exists $y \in [0,1]$ such that max$\{c - \delta, a_{1} - \frac{1}{3^{n}}\} < y < a_{1} < -(c - \delta)$. We will assume that such an $x$ exists. The proof is similar for the case where such a $y$ exists. Since $x > a_{2}$, we know $x \notin A$. Since $x < a_{2} + \frac{1}{3^{n}}$, we know $x \notin C_{n}$. Thus $x \notin C$ and so $g(x) = 0$. Then we have $|x - c| < \delta$ and $|g(x) - g(c)|=|0-1|=1 = \epsilon$. Since we chose $\delta$ arbitrarily, $g$ is not continuous at $c$.

Answer 1

A simpler proof: Suppose $g$ is continuous at $c \in C$. Then there exits an interval $(c-r,c+r)$ such that $|g(x)-g(c)| <\frac 1 2$ for all $x \in (c-r,c+r)$. But then $g(x) >g(c)-\frac 1 2=\frac 1 2>0$ which gives $g(x)=1$ for all $x \in (c-r,c+r)$. But the $(c-r,c+r)\subseteq C$ which is a contradiction since $C$ has no interior.

Answer 2

If $x\in C$ exists a sequence $(x_n)_n$ of points in excluded intervals (in the construction of cantor set) such that $x_n\to x$.

The above affirmation is easily to see because, if you sum the lengths of excluded intervals, this is equal to 1. If exists a point of cantor set and $0<\epsilon<1/2$, such that $$(x-\epsilon,x+\epsilon)\cap \{\mbox{union of excluded intervals of [0,1]}\}=\emptyset$$ this imply that

$$1< \mbox{sum of lengths of excluded intervals}+2\epsilon\le 1\,\,\,\,(*)$$

Because $(x-\epsilon,x+\epsilon)\cup \{\mbox{excluded intervals}\} \subseteq [0,1]$, the sum of lengths of excluded intervals are equals to 1, and $2\epsilon>0.$

Clearly (*) is an absurd. (because of this, on every neighborhood of x, there is a point of excluded intervals, and you can construct such sequence).

Using the above proved affirmation we have that $f(x_n)\to 0\neq 1=f(x)$.