# Countable Base Of Metric Spaces

Prove that a compact metric space has a countable base. Let $$X$$ be a compact metric space. Suppose for every point $$x$$ in $$X$$ we take the open n-ball $$B(x, 1)$$ with $$x$$ as the centre and radius $$1$$ units. Then we have $$X \subset \bigcup_{x \in X}B(x, 1)$$. Now as $$X$$ is compact, so a finite number of these open n-balls will cover $$X$$ and hence let $$X \subset \bigcup_{k=1}^{m}B(x_k, 1)$$. Now in each of these finite number of open n-balls we chose our collection of open sets as follows: In each of these open n-balls chose all the points with rational coordinates and their all the neighbourhoods with rational radii. Then it is easy to see that this collection of open sets is a countable collection. Now consider any open set $$G$$ in $$X$$ such that $$x \in G$$ then, $$x \in B(x_j, 1)$$ for some $$1 \leq j \leq m$$. Now as $$x \in G$$ so there is an open n-ball of $$x$$ such that $$B(x, r) \subset G$$. Now chose a point $$a$$ with rational coordinates such that it belongs to any one of these open n-balls and $$d(a, x) < \frac{r}{2}$$. Now chose a neighbourhood of $$a$$ with rational radii less than $$\frac{r}{2}$$ which contains $$x$$. Then this open set say $$C$$ belongs to our chosen collection and also belongs to $$B(x, r)$$ and as $$B(x, r) \subset G$$, so we have an open set from our chosen collection such that $$x \in C \subset G$$ and hence our collection is a countable base for $$X$$. Is My Proof Correct??

• "In each of these open n-balls chose all the points with rational coordinates and their all the neighbourhoods with rational radii." What does rational coordinates mean in a metric space? Sep 25 '20 at 17:16

Hint Instead of covering $$X$$ with balls of radius 1, try to cover it with balls of radius $$\frac{1}{n}$$ for all $$n$$.