Is second-countability invariant under homotopy equivalence?

Question

I am wondering if second-countability is invariant under homotopy equivalence. If I had to guess I would say so. Intuitively, if we have a countable basis of a space, and then stretch, contract, bend the space, I don't see how we could get an uncountable basis from this. However, I have failed at proving it formally and I would like to know if it is true at all. If so, why?

Answer 1

In fact, it's not. Intuition about stretchings and bendings isn't of much use for thinking about bases of a topology, simply because homotopies don't have to preserve distinctness of open sets. For an example, observe that for any set $I$, the wedge $X=\vee_{i\in I} [0,1]_i$ of $I$ copies of the unit interval (the intervals are glued together and $0$) is contractible, via the obvious homotopy $[0,1]_i\ni x_i\mapsto tx_i$. However, if $I$ is uncountable, $X$ certainly can't be second countable, as for instance by removing the wedge point we see it contains a subspace homeomorphic to an uncountable disjoint union of copies of $(0,1]$.

Answer 2

Take any space $X$ which is not second countable and consider the cone $CX=(X\times [0,1])/{\sim}$ where $(x,0)\sim (x',0)$ for all $x,x'\in X$. The space $CX$ is not second countable because $X$ is not, but $CX$ is contractible via a deformation retraction onto the 'cone point' $X\times\{0\}$.

Answer 3

Consider the NLS of all bounded sequences $l^{\infty}$ , it is not separable , hence not second countable . But being a Normed Linear space , $l^{\infty}$ is contractible , so homotopically equivalent to a point space which is obviously second countable