A co-meagre topology on $\aleph_{\omega}$

Question

My topology professor asked us to give a topology in which every singleton is $G_{\delta}$, yet it is not first-countable. I found an example, but since I like set theory, I thought I would try being creative, and observe the co-$<\aleph_{\omega}$ topology (meaning, the open sets are the empty set and the sets who have complements of smaller cardinality than the space). It looks like a singular cardinal might work here, due to their interesting properties, and indeed every singleton is $G_{\delta}$. But I can't, for the life of me, prove or disprove whether this topology has a countable local base for every point. I first thought that a good collection of sets that could not possibly all be contained in an element from a countable collection is $\aleph_{\omega}\setminus\{\aleph_{n}+\alpha\mid n\in\mathbb N\}$ where $\alpha$ runs through the entire space, but this backfired, since $U_{i}=\aleph_{\omega}\setminus\{\aleph_{n}+j\mid j<\aleph_{i},n\in\mathbb N\}$ works (an interesting note here is that I discovered that the pidgeonhole principle "fails" for singular cardinals $\kappa$ and functions $f:\kappa\rightarrow \lambda $ when $cf(\kappa)\leq\lambda$, since we can have instead of an element of $\lambda$ with $\kappa$ preimages, an element in $\lambda$ with $\beta\in cf(\kappa)$ images for all such $\beta$). My question is, is this space first-countable? If so, what would be a local base for each point (if we show that it is not second-countable, it is the same in co-topologies)? If not, which collection of open sets in this topology refute first countability?

Answer 1

Your intuition was correct, the space you describe ist not first countable. Note that your question is equivalent to the following:

Let $S=\{a\subseteq\aleph_\omega\mid \aleph_\omega\setminus a\in\mathcal P_{\aleph_\omega}(\aleph_\omega)\}$. Does $(S, \supseteq)$ have a countable dense subset?

I will argue that the answer is no. Suppose $D$ is such a set. Wlog, we may assume that $D=\{d^n_m\mid n,m<\omega\}$ where $c^n_m:=\aleph_\omega\setminus d^n_m$ has size $\aleph_n$. As $\aleph_{n+1}$ is regular we have that $(\bigcup_{m<\omega} c^n_m)\cap\aleph_{n+1}$ is bounded in $\aleph_{n+1}$, say it is bounded by $\beta^n<\aleph_{n+1}$. Now $a=\aleph_\omega\setminus\{\beta^n\mid n<\omega\}$ provides a counterexample to the density of $C$: For any $n, m<\omega$, we have that $\beta^n\in d^n_m\setminus a$!

A slight modification of this argument shows that in fact, $(S, \supseteq)$ does not have a dense subset of size ${<}\aleph_\omega$.So I must say, this space is a really good example for your exercise: Not only does no point have a local base which is countable, no point has a local base of size smaller than $\aleph_\omega$!

Answer 2

Directly going for the question in sentence 1 that originated it all:

Let $X$ be a countable dense subset of $\{0,1\}^\mathbb{R}$ (I show its existence here, e.g.). Then $X$ is $T_4$, separable, Lindelöf and every point is a $G_\delta$ trivially (complements of singletons are open..) and all points have a size continuum minimal local base. The ultrafilter space I also constructed in that answer would also work (but has only one point of non first-countability, here we have all of them, to show it's possible).