Exercise 19, § 8

Let $X$ be a Hausdorff space which is not regular (cf. Exercise 20), and let $a$ be a point of $X$ such that there is a neighbourhood $U$ of $a$ which does not contain any closed neighbourhood of $a$. Let $\mathfrak{F}$ be the neighbourhood filter of $a$, and let $Y = \mathfrak{F} \cup \{\omega\}$ be the topological space associated with the section filter of $\mathfrak{F}$. Let $Z$ be the set $Y \times X$ with the following topology: for any point $(y, z) \neq (\omega, a)$, the neighbourhoods of $(y, z)$ are the same as for the product topology, and the neighbourhoods of $(\omega, a)$ are the sets containing sets of the form

where $V$ is a neighbourhood of $a$ in $X$. Let $A$ be the subset of $Z$ which is the union of the sets $\{V\} \times V$ where $V \in \mathfrak{F}$. Let $f$ be the mapping $(V, x) \to x$ of $A$ into $X$. Show that $f$ has a limit relative to $A$ at each point of $\overline{A}$, but that there is no continuous mapping of $\overline{A}$ into $X$ which extends $f$.