So, I'm trying to prove that the Cantor Set is uncountable using some compactness argument, but I have some doubts about particular details. My proof attempt is as follows:
"Proof": Let $P$ denote the Cantor Set, and suppose for a contradiction that $P$ is countable. It is not hard to see that $P$ must be infinite since every real number, say of the form $\frac{1}{3^m}$ with $m \in \mathbb{Z}_{\geq 0}$, is an element of $P$. Now, since $P$ is countable, we may let $\{r_n\}$ be an enumeration of $P$. In fact, require that $r_i < r_{i+1}$ for each $i \in \mathbb{Z}_{\geq 0}$. Then, for each $i > 0$, define $d_i = \min(r_{i+1} - r_i, r_i - r_{i-1})$, and let $d_0 = r_1 - r_0$. Intuitively, the numbers $d_i$ give us the distance between an element of $P$ and its nearest neighbor in the enumeration. Then, for each $i$, let $V_i$ be an open ball of radius $d_i/2$ centered at $r_i$. Each $V_i$ then contains exactly one element of $P$, namely $r_i$, so $\{V_i\}$ is an open cover of $P$. Since $P$ is compact, there is a finite subcover $\{V_{p_1}, \ldots, V_{p_k}\}$ of $\{V_i\}$ for $P$. Then, $$P \subset \bigcup_{j=1}^k V_{p_j}$$ But the union on the RHS contains at most $k$ elements of $P$, a contradiction.
At first, I was happy by this proof, but then a small question caught my attention. Is it reasonable to have the requirement $r_i < r_{i+1}$? I'm thinking now that the answer is definitive no, since if it was reasonable, knowing that $1 \in P$, we would have trouble finding $1$ in the enumeration $\{r_n\}$. Perhaps I'm wrong, but if my suspicion is correct and the ordering requirement is not reasonable, would there be a way to salvage this proof? Any ideas or suggestions would be greatly appreciated.
Your suspicion is correct. For instance, there’s no enumeration of $\mathbb Q$ that fulfils that requirement, since between $r_i$ and $r_{i+1}$ there are further rational numbers that could neither come before $r_i$ nor after $r_{i+1}$.
The intention of the proof is not entirely clear to me. You say you want to prove uncountability by compactness, but then the first thing you do is to prove infinitude by specific examples – if you’re going to use specific examples, why not just continue by saying that the uncountably many numbers with ternary expansions that use only $0$ and $2$ are in the set?