Prove that every compact metric space has a countable dense subset.
Solution
Let \((X,d)\) be the given compact metric space, and write \[ B(x,r)\coloneqq\{y\in X:d(x,y)<r\}. \]
1. For every integer \(n\ge1\), there is a finite subset\[ F_n\subseteq X \]such that\[ X=\bigcup_{a\in F_n}B\left(a,\frac1n\right). \]
For fixed \(n\ge1\), the family \[ \left\{B\left(x,\frac1n\right):x\in X\right\} \] is an open cover of \(X\). Compactness gives a finite subcover. Taking \(F_n\) to be the finite set of centers of the balls in that subcover gives the asserted equality.
2. The set\[ D\coloneqq\bigcup_{n=1}^{\infty}F_n \]is countable.
Each \(F_n\) is finite by step 1, and a countable union of finite sets is countable. Hence \(D\) is countable.
3. The set \(D\) is dense in \(X\).
Fix \(x\in X\) and \(\varepsilon>0\). Choose an integer \(n\ge1\) such that \[ \frac1n<\varepsilon. \] By step 1, some \(a\in F_n\) satisfies \[ x\in B\left(a,\frac1n\right). \] Therefore \[ d(x,a)<\frac1n<\varepsilon. \] Since \(a\in F_n\subseteq D\), the ball \(B(x,\varepsilon)\) meets \(D\). Thus every open ball about every point of \(X\) meets \(D\), so \[ \overline D=X. \] Hence \(D\) is dense in \(X\).
4. Q.E.D.
Step 2 gives a countable subset \(D\subseteq X\), and step 3 proves that \(D\) is dense in \(X\).