Some definitions:
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A space is well-based if every point has a neighborhood base well-ordered by reverse inclusion (or equivalently, totally ordered by inclusion).
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A space $X$ is countably tight if for each set $A\subseteq X$ and each point $p\in\overline A$, $p$ belong to the closure of some countable subset of $A$.
Can anyone provide a proof of the following?
Proposition: Every countably tight well-based space is first countable.
Remark:
A well-based space is automatically radial, but it is not true that a countably tight radial space is always first countable. In fact, countably tight radial spaces are exactly the Fréchet Urysohn spaces. But there are many such spaces that are not first countable, as shown in this query.
Best Answer
Here is a proof:
Let $x \in X$, $\mathfrak {U}$ be a linearly ordered neighborhood base of $x$.
Define $A := \{y \in X: \exists \space U \in \mathfrak {U} \space (y \notin U)\}$.
Then for each $U \in \mathfrak {U}$ it is $X \setminus U \subset A$, hence $X \setminus \overline{A} \subset U$.
Case 1: $x \notin \overline{A}$. Then, by the above, $X \setminus \overline{A}$ is the smallest neighrborhood of $x$.
Case 2: $x \in \overline{A}$. Then, by countable tightness, there is a countable $B \subset A$, such that $x \in \overline{B}$. For each $y \in B$ choose $U_y \in \mathfrak{U}$, such that $y \notin U_y$. Then $\{U_y: y \in B\}$ is a countable neighborhhood base of $x$. [Let $U$ be a neighborhood of $x$, w.l.o.g. $U \in \mathfrak{U}$. Then there is a $y$ such that $y \in U \cap B$. Hence, $U_y \subset U$.]