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A topological space is called second-countable iff it has a countable basis.

How to prove or at least make an assumption about whether a space does, or does NOT have countable basis? Which properties of a space can imply that it is, or isn´t second countable? Is second-countability intuitively something like "small cardinality" or "weaker countability"?

Of particular interest are $l^2$ Hilbert spaces - are they second countable? I assume not, but am not sure.

Thank you!

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    $\begingroup$ A metric space is second countable iff it is separable. A pre-hilbertian space is separable iff it admits a countable Hilbert basis. If X is a separable space then $L^2(X, \mathcal{B}(X),\mu)$ is also separable. $\endgroup$ Sep 26 at 8:04
  • $\begingroup$ @Eevee, isn't that basis of Kronecker deltas you gave a Hilbert basis in the sense of linear algebra, which has little to do with the topological "basis" (I prefer "base")? $\endgroup$
    – Mark S.
    Sep 26 at 12:29
  • $\begingroup$ Oh, you're right, my bad. Had some wires crossed last night; sorry about that. $\endgroup$ Sep 26 at 18:27
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Any metric space that has a countable dense subset has a countable base, this is a classic fact. I show this and more here; it could have been more succinctly stated as $w(X)=d(X)=l(X)=nw(X)$, etc. for metrisable $X$, e.g.

So $\ell^2$ and in fact all $\ell^p$ for $p\neq \infty$ are separable (so have countable weight) while $\ell^\infty$ has not, because it has a closed discrete subset of size continuum. The reasoning is: if $X$ has a countable base, this holds for all subspaces too, but a large discrete subspace does not have a countable base:

Suppose $X$ is discrete, and let $\mathcal{B}$ be a base for $X$. It's clear that for each $x \in X$, $\{x\}$ is in $\mathcal B$ because that's the only way to write to open singleton as a union of base elements. So $|\mathcal B| \ge |X|$ and so if $X$ is uncountable, $X$ does not have a countable base.

The minimal size of a base (weight) or dense set (density) etc. are just a few of the many ways to "measure" the "size" of a space. In analysis many spaces are indeed of countable base; but e.g. weak topologies on Banach spaces often are not.

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  • $\begingroup$ So is the $l^2$ second-countable, or not? You are saying only it is separable. EDIT: Separable = has countable dense subset = has countable base = is second-countable.... Sorry, I get it. $\endgroup$ Sep 26 at 11:16
  • $\begingroup$ @TerezaTizkova both because it’s metric. $\endgroup$ Sep 26 at 11:44
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Whether a space is second-countable is a purely topological property, because it's defined in terms of its open sets (whether it has a countable basis, that is, whether its open sets are generated by a countable basis). So to find out whether a given space is second-countable, you'll have to look at it's topology. Intuitively, second-countable spaces are 'smaller' and more 'well-behaved' than ones that aren't.

For example, $\mathbb{R}^n$ is second countable since it has many countable bases for its topology - for example the open balls with rational radii. Subspaces of second countable spaces are second-countable, so all subsets of $\mathbb{R}^n$ with the subspace topology are second-countable too.

For non second-countable spaces, you can consider the uncountable product of $\mathbb{R}$ (with either the box or product topology) or the Long line.

Also see Henno's answer on $\ell^p$ spaces for second-countable and non second-countable Hilbert spaces.

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If a space is second-countable, the usual way to prove this will be to exhibit a specific countable basis.

If a space is not second-countable, proving this can be more intricate, and what methods are promising will vary a lot with the kind of space you are looking at. Any second-countable space is separable, and disproving separability can be easier. However, there are plenty of separable non-second-countable spaces.

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