Let $\mathcal{C}$ be the middle-third's Cantor set. For each integer $k \geq 0$, let $C_k \subset [0,1]$ denote the union of disjoint closed intervals obtained at the $k$th stage of the construction of $\mathcal{C}$. Let $\{I_{k,j}\}_{j=1}^{2^k}$ denote the collection of closed disjoint intervals that comprise $C_k$ (i.e. $C_k = \bigsqcup_{j=1}^{2^k} I_{k,j}$). Also, let
$$\tilde{\mathcal{C}} := \left\{x \in [0,1] : x \text{ is an endpoint of } I_{k,j} \text{ for some integer } k \geq 0 \text{ and some } 1 \leq j \leq 2^k \right\}.$$
I am trying to show that $\mathcal{C} = \tilde{\mathcal{C}}$, but I'm having some trouble. Here's is my attempt so far:
Proof that $\tilde{\mathcal{C}} \subseteq \mathcal{C}$:
Let $x \in \tilde{\mathcal{C}}$. Then $x$ is an endpoint of some $I_{k,j} \subset C_k$, so $x \in C_k$ for some $k \geq 0$. Since $C_0 \supset C_1 \supset C_2 \supset \cdots \supset C_k$, we have that $x \in C_m$ for all $0 \leq m \leq k$. We also know that if $x$ is an endpoint of some $I_{k,j} \subset C_k$ then $x$ is an endpoint of $I_{k+1,l} \subset C_{k+1}$ for some $1 \leq l \leq 2^{k+1}$; this is clear because in constructing $C_{k+1}$ from $C_k$, we only remove interior points of $C_k$. Thus, $x \in C_m$ for all integers $m \geq k$ as well, and so we have $x \in C_m$ for all $k \geq 0$. Thus $x \in \bigcap_{k=0}^{\infty} C_k = \mathcal{C}.$
Now this is the part I'm having trouble with:
Proof that $\mathcal{C} \subseteq \tilde{\mathcal{C}}$:
Let $x \in \mathcal{C}$. For a set $E \subseteq \mathbb{R}$ and a point $a \in \mathbb{R}$, let the distance between $a$ and $E$, denoted $d(a,E)$, be defined as
$$d(a,E) := \inf\{|a-b| : b \in E \}.$$
My strategy is to first show that $d(x,\tilde{\mathcal{C}}) = 0$. It follows by the Triangle Inequality that \begin{align*}
d(x,\tilde{\mathcal{C}}) & \leq d(x,C_k) + d(C_k, \tilde{\mathcal{C}}),
\end{align*}
where $d(C_k, \tilde{\mathcal{C}}) := \inf\{|a-b|: a \in C_k, b \in \tilde{\mathcal{C}} \}$. We note that $C_k \supseteq \mathcal{C}$ for all $k$, and $\mathcal{C} \supseteq \tilde{\mathcal{C}}$ which we just proved. Therefore, $C_k \supseteq \mathcal{C}$ for all $k$, which then implies $d(C_k, \tilde{\mathcal{C}}) = 0$. Thus,
$$d(x,\tilde{\mathcal{C}}) \leq d(x,C_k).$$
Next, we note that for each integer $k \geq 0$ the $I_{k,j}$'s all have the same length, and the $I_{k+1,j}$'s are each $1/3$rd the length of the $I_{k,j}$'s. So by induction, $|I_{k,j}| = \frac{1}{3^k}$ for each integer $k \geq 0$. So if $x \in C_k$, the farthest that $x$ can be from the endpoint of the closed interval $I_{j,k}$ that contains it is $|I_{j,k}|/2 = 1/(2 \cdot 3^k)$. Whence, $d(x,C_k) \leq 1/(2 \cdot 3^k)$ for all integers $k \geq 0$. Therefore,
$$d(x,\tilde{\mathcal{C}}) \leq \frac{1}{2 \cdot 3^k} \quad \text{for all } k \in \mathbb{N}.$$
Also, we must have $d(x,\tilde{\mathcal{C}}) \geq 0$. And the only nonnegative number which is less than $1/(2 \cdot 3^k)$ for all $k \in \mathbb{N}$ is $0$, so $d(x,\tilde{\mathcal{C}}) = 0$.
This is my sticking point: How do I show that $d(x,\tilde{\mathcal{C}}) = 0 \implies x \in \tilde{\mathcal{C}}$?
From this post I learned that it's enough to show that $\tilde{\mathcal{C}}$ is closed, which I haven't figured out how to show. I do see why $\mathcal{C}$ must be closed (since $\mathcal{C}^c$ is an intersection of open sets, and so it's open). But without knowing that $\mathcal{C} = \tilde{\mathcal{C}}$ a priori, I'm not sure how to prove this. Any insights would be greatly appreciated.
Best Answer
First, the bad news: $\tilde{\mathcal{C}} \subsetneq \mathcal{C}$, so you cannot prove they're equal.
There is a useful characterization of both sets that should make this evident: $$\mathcal{C} = \left\{\sum_{k=1}^\infty \frac{a_k}{3^k} \,\middle|\, \forall k, a_k \in \{0,2\}\right\} \\ \tilde{\mathcal{C}} = \left\{\sum_{k=1}^\infty \frac{a_k}{3^k}\,\middle|\,\forall k, a_k \in \{0,2\} \,\text{ and }\, \exists N, a_N = a_{N+1} = a_{N+2} = \cdots\right\}$$
This also gives a way to find, as I had given in the comments, a particular point in $\mathcal{C}\setminus \tilde{\mathcal{C}}$. For instance, $1/4$ corresponds to the sequence $$a_k = \begin{cases} 0, & k \text{ is odd} \\ 2, & k \text{ is even}\end{cases}$$
Now the good news: You've successfully shown $\tilde{\mathcal{C}} \subseteq \mathcal{C} \subseteq \operatorname{cl}\tilde{\mathcal{C}}$ where $\operatorname{cl}$ is the closure operator.
If you additionally know that $\mathcal{C}$ is closed (since it is the intersection of sets each of which is a finite union of closed intervals and hence closed), then this suffices to show $$\mathcal{C} = \operatorname{cl}\tilde{\mathcal{C}}.$$
This looks to be the strongest result you can deduce from your proofs.