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A space $X$ is said to have a countable basis at $x$ is there is a countable collection $B$ of neighbourhoods of $x$ such that each neighbourhood of $x$ contains at least one of the most elements from $B$. What does this mean? We take countable neighbourhoods and again some neighbourhood such that each contains at least one from the countable neighbourhoods. How is the second neighbourhood different from the first? What is the significance?

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  • $\begingroup$ It's about the fact that in a first-countable space there exists at most a countable system (set) of neighborhoods for each point. So this way you can be sure that for each point it suffices to work with at most $\aleph_0$ neighborhoods. $\endgroup$
    – Aelx
    Commented Nov 30, 2021 at 9:56

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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").

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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.

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You take a particular point $x\in X$ and consider all neighbourhoods of $x$. Then a countable collection $\{N_{i}\}$ is a neighbourhood base if for any neighbourhood $N$ of $x$ there exists $i\in\mathbb{N}$ such that $N_{i}\subset N$.

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