The Cantor set is the set of all numbers that can be written in the following form:
$$\mathfrak{C} = \sum_{i = 1}^{\infty} \frac{a_i}{3^i}$$
for an infinite sequence $a_i$ such that $a_i = 0$ or $2$ for all $a_i$, such that each $a_i$ defines a unique element of $\mathfrak{C}$.
Now consider the metric space consisting of the Cantor set equipped with the Euclidean metric $(\mathfrak{C},|\bullet-\bullet|)$. The induced topology $\tau$ on this metric space is just the collection of all open balls $B_{\epsilon}(x)$ around any and all elements $x \in \mathfrak{C}$, as well as unions of them.
I would like to show that every open ball in $(\mathfrak{C},|\bullet-\bullet|)$ can be represented as a union of subsets of $\mathfrak{C}$, where every element in a subset possesses the same $n$ initial symbols in their $a_i$ sequence for some finite number $n$. Ideally, I would like to do this constructively—given some specific $B_{\epsilon}(x)$, I would like to find (at least one) representation for it in terms of the subsets described above.
MY ATTEMPTS SO FAR: My efforts have focused on trying to show that, because the diameter of such subsets becomes smaller and smaller as $n$ increases, that each open ball will contain at least one such subset. However, I'm not sure how to show that the open ball will exclusively contain such sets—perhaps they don't?
Ok, let $B$ be an open ball in $\mathfrak{C}$. For $x\in\mathfrak{C}$, let $U_{x,n}\subseteq\mathfrak{C}$ denote the set of those elements in the Cantor set whose first $n$ digits in their base 3 representation agree with those of $x$. I surmise you've already shown that, for each $x\in B$, we can choose a large enough $n_x\gg0$, such that $U_{x,n_x}\subseteq B$. Then, I claim that $B=\bigcup_{x\in B}U_{x,n_x}$. This is the desired representation. To verify the claim, note that if $y\in B$, then $y\in U_{y,n_y}\subseteq\bigcup_{x\in B}U_{x,n_x}$. This proves one inclusion. On the other hand, $U_{x,n_x}\subseteq B$ for each $x\in B$ by construction, whence $\bigcup_{x\in B}U_{x,n_x}\subseteq B$, which is the other inclusion.