$\def\C{\mathcal{C}}$ $\def\N{\mathbf{N}}$ $\def\R{\mathbf{R}}$ Let $C_0 := [0,1]$. Having already defined $C_n$, we define $$C_{n+1} := \frac{C_n}{3} \cup \left( \frac{2}{3} + \frac{C_n}{3} \right)$$ The Cantor Set $\C :=\bigcap_{n=0}^{\infty}{C_n}$.
I already know and understand why the Cantor Set is perfect and totally disconnected. However every proof I've seen so far appeals to the ternary number system, or other $``$obvious$"$ facts about the Cantor Set eg.
- Each $C_n$ can be expressed as the disjoint union of $2^n$ closed intervals each of length $3^{-n}$
- The boundary points of each $C_n$ lie in the Cantor Set.
However obvious they are, I want to be as rigorous as possible and phrase my arguements without appealing directly to these facts, at least not without a rigorous proof of them. I want to prove from scratch: using nothing but the definition of the Cantor Set given above and a few choice theorems, that it is perfect and totally disconnected. I have so far managed to prove the following propositions - all by induction arguements:
- For each $n \in \N$, we have $C_{n+1} \subseteq C_{n}$.
- For each $n \in \N$, the set $C_n$ is compact and non-empty.
- (Follows from 1&2 by the Nested Compact Set Theorem) $\C$ is non-empty and compact.
- $0,1 \in \C$
- For all $n \in \N$, $x \in C_n$ implies $\frac{1}{3}x \in C_{n+1}$
- For all $n \in \N$, $x \in C_n$ if and only if $1-x \in C_n$
- (Follows from 4 & 5 & 6 and induction) $\frac{1}{3^k}, \frac{1}{4},\frac{3}{4} \in \C$, where $k\in\N$
My proof ideas to prove that $\C$ is Perfect:
Let $A_0 := \left\{0,1\right\}$. Having already defined $A_n$, we define $$A_{n+1} := \frac{A_n}{3} \cup \left( \frac{2}{3} + \frac{A_n}{3} \right)$$ Informally, $A_n$ is the set of boundary points of the intervals of $C_n$. By induction I have proved that for each $n \in \N$, we have $A_{n} \subseteq A_{n+1} \subseteq \C$. If I can prove that for every $n \in \N$ and $x \in C_n$, there exists $y \in A_n$ such that $0<|x-y|\leq\frac{1}{3^n}$ then that'd be enough to prove that $\C$ is perfect, since $\frac{1}{3^n}$ can get arbitrarily small.
Another proof idea: Let $\epsilon>0$. Construct an open cover of $\C$ via $O = \bigcup_{x \in \C}{V_{\epsilon}(x)}$. Since $\C$ is compact, there exists a finite subcover $ \bigcup_{i=1}^{n}{V_{\epsilon}(x_i)}$. So for every $x \in \C$ with $x \neq x_i$, there exists an $x_i \in \C$ that satisfies $0<|x-x_i|\leq \epsilon$. That takes care of all but finitely many points - namely the $x_i$, which I don't know what to do with now. I know I have only used the compactness of $\C$ and not its definition, but I can't make progress.
Final Remark. Like I said, I've seen and understood other proofs of the above facts before. Please do not reply with proofs that begin: "Since each $C_n$ has intervals of length $1/3^n$". If you do use such a fact, then you're going to have to prove it rigorously, without relying on geometrical intuition. For the above propositions that I claimed to have proven, I have no knowledge of metric spaces or topology other than in the real numbers $\R$. This is my first StackExchange post; I have tried to follow the guidelines as much as possible, and giving as much detail about my background and my purpose for asking this question. Please be kind, thank you for reading.