# Separable metric space has a countable base

A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a $\textit{base}$ for $X$ if the following is true: For every $x \in X$ and every open set $G \subset X$ such that $x \in G$, we have $x \in V_{\alpha} \subset G$ for some $\alpha$. In other words, every open set in X is the union of a subcollection of $\{V_{\alpha}\}$ Prove that every separable metric space has a countable base.

Now, I know how to actually solve the problem. My question is, since I need to present it in a paper, how could I introduce this problem in an interesting way. In addition to that could someone give me examples and/or extensions of the above proof: as in how could this type of property be used or could be use in order to prove another property.

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Here is an example where this implication is used: As you might know, a countably compact metric space $X$ is compact. Since countably compact Lindelöf spaces are compact, it suffices to show the Lindelöf property, which follows (in any space) from the existence of a countable base. But such a base exists if the space is separable, so we have to construct a countable dense subset for $X$. This can be done as follows:
For each natural $n$ there is a ball $B_{1/n}(x_1)$ of radius $1/n$ around some point $x_1$. Then we choose a point $x_2$ outside of that ball and another point $x_3$ outside of $B_{1/n}(x_1)$ and $B_{1/n}(x_2)$. If no finite number of such balls covers the space $X$, then we can repeat this process infinitely often and obtain a sequence without a limit point, in contradiction to the countable compactness of $X$. That means that finitely many of these balls with centers $x_{n,1},...,x_{n,l(n)}$ cover $X$, and that holds for each radius $1/n$. The set $\bigcup_{n\in\Bbb N}\bigcup_{i=1}^{l(n)} x_{n,i}$ is then a countable dense subset of $X.$