PMEA makes normal low-character spaces collectionwise normal

If $X=\mathrm{\varnothing }$ or the discrete family is empty, empty expansions suffice. Otherwise use AC to choose one local base of the stated size at each point, and replace its members by their interiors. This is an image of the original base, so its cardinality does not increase; each interior contains the point, and the image remains a local base. Fix the discrete family $\mathcal{F}$ and the open neighbourhood bases ${\mathcal{U}}_x$ from the Given line, with $\mathrm{\mid }{\mathcal{U}}_x\mathrm{\mid }<\mathfrak{c}$, or with $\mathrm{\mid }{\mathcal{U}}_x\mathrm{\mid }\le \omega$ in the first countable case.

Each ${\mathcal{U}}_x$ is nonempty, by applying its base property to $X$. For $U,V\in {\mathcal{U}}_x$, the open intersection $U\cap V$ contains $x$, so [F2] supplies $W\in {\mathcal{U}}_x$ with $W\subseteq U\cap V$. This is precisely downward directedness; the original base need not be closed under finite intersections and is not enlarged. If $x\in F_i\subseteq G$ with $G$ open, [F2] likewise supplies a base member inside $G$. Apply [F1] to $\mathcal{F}$ and the bases ${\mathcal{U}}_x$, obtaining $U^x\in {\mathcal{U}}_x$ with $U^x\cap U^y=\mathrm{\varnothing }$ whenever $x\in F_i$, $y\in F_j$, $i\mathrm{\ne }j$.

For each $i$ put $G_i:=\bigcup \{U^x:x\in F_i\}$. Each selected $U^x$ belongs to the open base ${\mathcal{U}}_x$, so $G_i$ is open; it contains $F_i$ because $x\in U^x$, and distinct $G_i,G_j$ are disjoint by step 2.1. Hence $\mathcal{F}$ is separated and $X$ is collectionwise normal.