I tried to prove that if $(X, d)$ is a complete metric space and $K$ a compact countable subset of $X$, then $K$ has infinite many isolated points. I tried to prove it with Baire's Theorem but I can't.
My try:
By reduction to the absurd, suppose that $ K $ has finitely many isolated points, that is, $ K \setminus K '=\{x_0, x_1, \ldots, x_n \} $. Note that,
\begin{align*}
K \setminus \{x_0, x_1, \ldots, x_n \} & = K \setminus (K \setminus K ') \\
& = K \cap (K^c \cup K') = K \cap K'
\end{align*}
On the other hand, we have that $ K $ and $ K'$ are closed, therefore, $ K \cap K' $ also is, and therefore this set is complete. For all $ i \in \mathbb{N} $ such that $ 0 \leq i \leq n $, we have that $ M_i = K \setminus \{x_i\} $ is an open set for the induced topology on $ K $, moreover, $ \overline{M_i} \cap K = K $, that is, $ M_i $ is dense in $ K $. With this, $ \{M_k \}_{k = 0} ^ n $ is a family of open and dense sets in $ K $, since $ K $ is closed, $ K $ is complete. Thus, using the Baire Category Theorem, we have that $ \overline{\displaystyle \bigcap_{k = 0}^n M_k} = K $, but $ \displaystyle \bigcap_{k = 0}^n M_k =\emptyset $, which is a contradiction.
I note that $ \overline{M_i} \cap K = K $ this could be false and I do not use that $K$ is countable. Any help will be of use, thank you very much.
Best Answer
Since $K$ is compact, $K$ is closed, and therefore $K\cap K'=K'$. $K'$ is closed, so you can apply the Baire category theorem to it: each of the sets $K'\setminus\{x\}$ for $x\in K'$ is a dense open subset of $K'$, and there are only countably many of these sets, but their intersection is empty and therefore not dense in $K'$.