A countable compact set has infinitely many isolated points 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.
 A: 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 $K'$.
A: Will you accept a proof that does use (at least, not explicitly)  Baire's Theorem? If so, suppose the claim is false.  From the set $K=(x_n)_{n\in \mathbb N}$, extract a subsequence which converges to some $x\in K$, and which we still call $(x_n)$ for convenience. Of course, then $x$ is not isolated.
But now, if $\{y_1,\cdots, y_k\}$ are the isolated points in $K$ then consider $B:=B_r(x)$ where $0<r<\min\{d(x,y_i):1\le i\le k\}$. If $x\neq y\in \overline{B_{r/2}(x)}\cap K$ then, $y\in K'$ because it is not isolated. Then, $\overline {B_{r/2}(x)}\cap K'$ is perfect set and so uncountable. But this is impossible, so $B_{r/2}(x)\cap K= \{x\}$. But now we have a contradiction because this means that $x$ is isolated.
A: $X$ is called a Baire space if the countable intersection of open and dense subsets $D_n$ has a dense intersection. These include (this is what the Baire category theorem says) all complete metric spaces and also all locally compact Hausdorff spaces.
Lemma:
If $X$ is a Baire space that is $T_1$ without isolated points (a crowded space, this is called) then $X$ is uncountable: If not, write $X=\{x_m\mid n \in \Bbb N\}$ and note that $D_n = X\setminus \{x_n\}$ is open and dense (open as $\{x_n\}$ is closed by $X$ being $T_1$, and dense as $\{x\}$ is not open, so $D_n$ is not closed so the closure can only be $X$). But then $\bigcap_n D_n = \emptyset$ (by de Morgan, essentially) contradicting Baire-ness.
Now apply this to your $K \subseteq (X,d)$ situation: $K$ is Baire (for two reasons: it's locally compact Hausdorff, and also complete metric, being closed in $X$), so $K$ has isolated points $I(K) = K\setminus K'$. Note that $I(K)$ is open in $K$, and if it were finite (!) it would be closed too and $L = K\setminus I(K)$ would be open (and closed) in $K$ so wouldn't have isolated points ($L$ open in $K$ implies $I(L) \subseteq I(K)$, and $I(L) \cap I(K)=\emptyset$ by construction) and still be countable, Baire (still compact Hausdorff or complete) and crowded, contradicting the lemma. So $I(K)$ cannot be finite.
A: If $K$ has a finite number of isolated points, then their complement $K^\prime$ is a perfect Polish space (perfect means that every point is a limit point, Polish means complete metrizable (which a compact set in $\mathbb{R}^n$ obviously is). Then, it is a theorem (Theorem 6.2 in Kechris' "Classical descriptive set theory") that such a space contains a copy of the Cantor set, and is thus uncountable. The proof is simple: pick a subset $U_0$, then two disjoint subsets of $U_0$  of at most half the diameter of $U_0$ then for each of those two disjoint subsets of at most half the diameter of, and so on. The uncountability then follows by the Cantor (natch) diagonal argument.
