(Prove or Disprove) If $X$ is a metric space, then the Borel $\sigma$-algebra ${\cal B}_X$ contains every countable subset of $X$.

Question

I am tasked with determining whether the following statement is true or false:

If $X$ is a metric space, then the Borel $\sigma$-algebra ${\cal B}_X$ contains every countable subset of $X$.

I understand how the Borel $\sigma$-algebra is constructed, but I have very little intuition into what sets it contains.

Answer 1

First, note that it contains every singleton subset of $X$. For take any point $x \in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.

Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons $\{x\}$ for $x\in A$ and it’s closed under countable unions, it contains $A$.

Answer 2

Show that $\{x\}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,\frac1n)$ so in either case, singletons are in the Borel $\sigma$-algebra. Countable sets are countable unions of those.