Countable basis and first countable

Solution 1:

Consider a "really complicated" metric space $(M,d)$. If I pick a point $x\in M$, there are probably lots of open neighborhoods of $x$. However, the metric $d$ provides a canonical family of neighborhoods of $x$ which are particularly nice, namely the open balls $B_\epsilon(x):=\{y: d(x,y)<\epsilon\}$ for $\epsilon>0$.

Regardless of how big $M$ is, there are only as many open balls with center $x$ as there are real numbers. Moreover, if $U\ni x$ is open then $U\supseteq B_\epsilon(x)$ for some $\epsilon>0$. So even the neighborhoods which aren't actually open balls centered on $x$ are still "no sharper than" open balls centered on $x$ (think of an open neighborhood of a point as a guess at that point: intuitively, smaller = better).

And we can do even better by considering only rational radii: since the rationals are dense in the reals, the set $$RatBall(x):=\{B_\epsilon(x):\epsilon\in\mathbb{Q}_{>0}\}$$ of rational-radius open balls centered on $x$ has the same "sharpening" property as the set of all open balls centered on $x$. Since $\mathbb{Q}$ is countable, $RatBall(x)$ (which, remember, is a set of sets) is also countable. This shows that every metric space, no matter how complicated, is first countable.

Note that $RatBall(x)$ depends on $x$ itself: even though there's a sense of uniformity here, we're not getting a countable base for $(M,d)$ itself. First countability is much weaker than second countability (= "there is a countable base for the whole space").

Solution 2:

Imagine the metric space with underlying set $\mathbb{R}^2$, with metric the usual one. As every metric space, it can be seen as a topological space. Now, given a point $(x,y)$ in $\mathbb{R}^2$, the neighborhoods of $(x,y)$ are the unions of open balls that contain $(x,y)$. This is a very "large" collection of open neighborhoods of the point; compare this with the collection of open balls with rational radius centered at the point.

This is also a collection of neighborhoods of the point, but, as the rational numbers, is a numerable set.

Can you see why this numerable family of neighborhoods (let's denote it by $ \mathcal{F}=\{V_{n}: n \in \mathbb{N}\}$) is something we prefer to work with, compared with the first one? We can "list" their element (more properly, there exists a bijection between the natural numbers and this family of neighborhoods), and in some sense they are "enough neighborhoods of $(x, y)$", since any other neighborhood $U$ of $(x,y)$ contains at least one of the elements $V_{n}$ of this family (this is because $(x,y)$ would be an interior point of $U$, thus we can find $r>0$ such that $B_{r}((x,y))$ is contained in $U$, and since the rationals are dense on the reals, nothing stops us from taking $r$ rational), we can avoid working with $U$, and to take $V_{n}$ instead.