One can show the following: if $I$ is a directed set and $(A_i)_i$ is a surjective inverse system of compact Hausdorff spaces such that the inverse limit $A=\lim_{i\in I} A_i$ is second countable, then the limit is essentially countable, i.e. there is a countable directed $J\subseteq I$ such that $A=\lim_{j\in J} A_j$.
I am curious how sharp this result is. In particular, does it fail if:
- We no longer assume that $A_i$ are compact?
- Alternatively, we no longer assume that $(A_i)_i$ is surjective?
- Putting those two together, if neither $A_i$ nor $A$ is compact and the system is not surjective?
Best Answer
Here's a counterexample where the maps are not surjective. Let $S$ be an uncountable set and let $I$ be the set of cofinite subsets of $S$ ordered by reverse inclusion. Let $A_i=[0,1]^i$; if $i\subseteq j$ let the map $A_i\to A_j$ be the inclusion that is $0$ on all the coordinates of $j\setminus i$. Then the inverse limit is just a singleton (the intersection of all the $A_i$ inside $[0,1]^S$), but the inverse limit of any countable subsystem is $[0,1]^T$ for some cocountable $T\subseteq S$.
And here is a counterexample where the spaces are not compact. Let $I=\omega_1$ and let $A_i$ be the set of injections $i\to\mathbb{N}$ whose image has infinite complement, with the discrete topology. The natural restriction map $A_j\to A_i$ is then surjective when $j\geq i$. Any element of the inverse limit determines an injection $\omega_1\to\mathbb{N}$, and so the inverse limit is empty and hence second countable. But the inverse limit of any countable subsystem is nonempty, since if $i\in\omega_1$ is the supremum of indices of the subsystem, the inverse limit naturally contains $A_i$.