Is every well-based space with countable pseudocharacter first-countable?

Question

First some definitions. Here $X$ is a topological space.

• $X$ is first countable if there is a countable local base (= nbhd base) at every point.
• $X$ has countable pseudocharacter if every point is the intersection of a countable number of open sets, i.e., if every point is a $G_\delta$-set. Such spaces are necessarily $T_1$.
• $X$ is well-based if every point has a local base that is totally ordered by inclusion; or equivalently, that is well-ordered by reverse inclusion.

It seems that the following should be true:

Proposition: Every well-based space with countable pseudocharacter is first countable.

Can anyone provide a proof?

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Some context: Assuming $T_1$, the first countable property implies well-based and implies countable pseudocharacter. But each of these two properties in isolation is strictly weaker than first countable. How about if we assume both properties together?

Answer 1

Let $X$ be a well-based space with countable pseudocharacter.

Let $x$ be a point of $X$. As $X$ is well-based, then $x$ has a neighborhood base, let's denote as $\mathcal{B}$, totally ordered by reverse inclusion. And, as $X$ has countable pseudocharacter, then exists $\{A_n\}_{n \in \mathbb{N}}$ collection of open sets such that $\bigcap_{n \in \mathbb{N}}A_n = \{x\}$.

For each, $n \in \mathbb{N}$, exists $B_n \in \mathcal{B}$ such that $B_n \subseteq A_n$. $\{B_n\}_{n \in \mathbb{N}}$ is countable, totally ordered by reverse inclusion and it verifies that $\bigcap_{n \in \mathbb{N}}B_n = \{x\}$.

Let $A$ be a neighborhood of $x$. Then exists $B \in \mathcal{B}$ such that $B \subseteq A$. Suppose that doesn't exist $B_n \in \{B_n\}_{n \in \mathbb{N}}$ such that $B_n \subseteq B$. Then, $B \subsetneq B_n$, for all $n \in \mathbb{N}$, and $B \subseteq \bigcap_{n \in \mathbb{N}}B_n = \{x\}$. As a result, we conclude that $B = \{x\}$.

Therefore, $X$ is first countable.

Edit:

The idea has been the following one:

Given a point of the space, considering a neighborhood base that is totally ordered by reverse inclusion, with the aim of taking a countable subcollection of elements from it verifying that their intersection is such point. This can be done as the space is well-based and has countable pseudocharacter. If this subfamily is a neighborhood base of the point, we are done. But, if it isn't, it means that there is a neighborhood of the point that doesn't contain any element of this subfamily. However, it would contain an element of the base that we have considered at the beginning. As the base is totally ordered by reverse inclusion and this neighborhood doesn't contain any element of the subfamily, then the neighborhood is contained in each element of the family and, therefore, it is also contained in the intersection of all the elements of the subfamily, which is the point. We conclude that the neighborhood is the point, i.e, the point is isolated. So, the point admits a countable neighborhood base, for example, the basis that has the point itself as its single element.

Answer 2

Same idea as Almanzoris, just in my own words for my own understanding (posted in case it helps others).

Every isolated point has a countable local base, so let $x$ be a non-isolated point with a local basis $\{B_\alpha:\alpha<\kappa\}$ for some limit ordinal $\kappa$ with $\alpha \leq \beta$ implying $B_\beta \subseteq B_\alpha$ (by well-based), and let $\{x\}=\bigcap\{A_n:n<\omega\}$ with each $A_n$ open (by points $G_\delta$ aka countable pseudocharacter).

Choose $\alpha_n<\kappa$ for each $n<\omega$ such that $B_{\alpha_n}\subseteq A_n$. I claim that $\{\alpha_n:n<\omega\}$ is cofinal in $\kappa$, and thus $\{B_{\alpha_n}:n<\omega\}$ is a countable local base at $x$. To see this, suppose by contradiction that $\sup \alpha_n < \gamma < \kappa$. Then $B_\gamma\subseteq B_{\alpha_n}\subseteq A_n$ for all $n<\omega$, and it follows that $$\{x\}\subseteq B_\gamma\subseteq\bigcap\{B_{\alpha_n}:n<\omega\}\subseteq\bigcap\{A_n:n<\omega\}=\{x\}.$$ Thus $B_\gamma=\{x\}$ is open, and $x$ is isolated, a contradiction.