How to tell if a space is second-countable 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!
 A: 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.
A: 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.
A: 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.
