- Show that every compact metrizable space has a countable basis.
Solution
1. For each \(n \ge 1\), the open balls \(\{B(x, 1/n) : x \in X\}\) form an open cover of \(X\).
every point lies in its own ball.
2. Since \(X\) is compact, there is a finite subcover \(\{B(x_{n,1}, 1/n), \ldots, B(x_{n,k_n}, 1/n)\}\).
compactness.
3. Let \(\mathcal B = \{B(x_{n,j}, 1/n) : n \ge 1,\ 1 \le j \le k_n\}\).
collect all these balls.
4. \(\mathcal B\) is countable.
it is a countable union of finite sets.
5. \(\mathcal B\) is a basis for the topology of \(X\).
5.1. Let \(U\) be open and \(x \in U\).
take an arbitrary point of an arbitrary open set.
5.2. There is \(\epsilon > 0\) with \(B(x, \epsilon) \subseteq U\).
\(U\) is open.
5.3. Choose \(n\) with \(1/n < \epsilon/2\).
Archimedean property.
5.4. Since \(\{B(x_{n,j}, 1/n)\}\) covers \(X\), there is \(j\) with \(x \in B(x_{n,j}, 1/n)\).
2.
5.5. Then \(B(x_{n,j}, 1/n) \subseteq B(x, \epsilon) \subseteq U\).
if \(y \in B(x_{n,j}, 1/n)\), then \(d(x,y) \le d(x, x_{n,j}) + d(x_{n,j}, y) < 1/n + 1/n = 2/n < \epsilon\).
5.6. Hence every point of \(U\) lies in a member of \(\mathcal B\) contained in \(U\).
<2>1–<2>5.
6. Therefore \(\mathcal B\) is a countable basis.
4 and 5.
7. Q.E.D.
6.