As every $F_\sigma$ subspace of a $T_2$ or regular paracompact space is again paracompact, the required whole space must be non-$T_2$ as well as non-regular.


1 Answer 1


Let $S=\{0,1\}$ with the topology $\big\{\varnothing,\{0\},S\big\}$. Let $Z=S\times\Bbb N$, where $\Bbb N$ has the discrete topology. Let $p$ be a point not in $Z$, and let $Y=Z\cup\{p\}$. $Z$ is an open subset of $Y$, and basic open nbhds of $p$ are the sets of the form

$$B(F)=\{p\}\cup\big(\{0\}\times(\Bbb N\setminus F)\big)$$

for finite $F\subseteq\Bbb N$.

Let $\mathscr{U}=\{B(\varnothing)\}\cup\big\{S\times\{n\}:n\in\Bbb N\big\}$; this is an open cover of $Y$. Let $\mathscr{R}$ be an open refinement of $\mathscr{U}$; then $S\times\{n\}\in\mathscr{R}$ for each $n\in\Bbb N$, so $\mathscr{R}$ is not locally finite at $p$. Thus, $Y$ is not paracompact.

Now let $X$ be the Alexandroff extension of $Y$. $X$ is compact, hence paracompact. The sets $S\times\{n\}$ for $n\in\Bbb N$ are closed in $X$, as is $\{p\}$, so $Y$ is an $F_\sigma$ in $X$.


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