Recently I stumbled across the fact that any compact metric space is the union of some countable set and a subset that, when given the induced topology, is a perfect space.
Can anyone provide a proof or reference to a proof of this fact? Thanks.
[Math] Why is any compact metric space the union of a countable set a subset which is a perfect space under the induced topology
general-topology
Best Answer
A compact metric space is second countable (i.e. has a countable basis): an open cover consisting of all open balls with radius $1/n$ has a finite subcover $\mathcal{B}_n$ by compactness, and one can show that the union of all these $\mathcal{B}_n,n\in\mathbb{N}$ is a countable basis.
Now that we know that a compact metric space is second countable, we can use the following proposition which can be proven using the hint given in the comments by George Lowther.
Proof: Let $X$ be a second countable space. Pick a countable basis $\mathcal{B}$ of $X$. Denote $$C=\{x\in X:x\text{ has a countable neighbourhood}\}\quad\text{and}\quad P=X\setminus C.$$ Let us show that $C$ is countable. For each $x\in C$ choose a countable neighbourhood $B_x\in\mathcal{B}$ of $x$. For each $B\in\mathcal{B}$ denote $C_B=\{x\in C:B_x=B\}$. Since $x\in B_x$ and $B_x$ is countable for each $x\in C$, each $C_B$ is countable too. Now $C=\bigcup_{B\in\mathcal{B}}C_B$ is countable, being a countable union of countable sets.
Note that any open and countable neighbourhood $U$ of a point $x\in C$ is a subset of $C$, since the same $U$ is a countable neighbourhood of each point of $U$. Since each point of $C$ has such a neighbourhood, $C$ is open and $P$ is closed.
Lastly let us show that $P$ does not have isolated points in itself (i.e. $P$ is dense-it-itself). Pick $x\in P$ and a neighbourhood $U$ of $x$. $U$ is uncountable by definition of $C$, so $C$ being countable, $U\setminus(C\cup\{x\})\subseteq P$ is nonempty and thus $x$ is not an isolated point in $P$.
We have shown that $X=C\cup P$, where $C$ is countable and $P$ is perfect. $\square$
By the way, a similar argument can be used to prove that every space having a basis of cardinality $\kappa$ is a union of a set of cardinality $\kappa$ and a perfect set.