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.
Thank you in advance
Best Answer
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 $X.$