Is every second countable $T_2$ topological space a developable space? I have a question:
Is every second countable $T_2$ topological space a developable space?
Thanks for your help.
 A: $\newcommand{\st}{\operatorname{st}}$No; here’s a counterexample.
Let $\mathscr{B}_0$ be the family of open intervals with rational endpoints in $\Bbb R$, and let
$$\mathscr{B}=\mathscr{B}_0\cup\{B\setminus\Bbb Q:B\in\mathscr{B}_0\}\;;$$
$\mathscr{B}$ is a countable base for a Hausdorff topology $\tau$ on $\Bbb R$. Call the resulting space $X$, and suppose that $\mathscr{G}=\bigcup_{n\in\Bbb N}\mathscr{G}_n$ is a development for $X$: each $\mathscr{G}_n$ is an open cover of $X$, and for each closed $F\subseteq X$ and $x\in X\setminus F$ there is an $n\in\Bbb N$ such that $\st(x,\mathscr{G}_n)\cap F=\varnothing$, where
$$\st(x,\mathscr{G}_n)=\bigcup\{G\in\mathscr{G}_n:x\in G\}\;.$$
$\Bbb Q$ is closed in $X$, so for each irrational $x$ there is an $n(x)\in\Bbb N$ such that $\st(x,\mathscr{G}_{n(x)})\cap\Bbb Q=\varnothing$. By the Baire category theorem there are a non-empty open interval $(a,b)$ and an $m\in\Bbb N$ such that $$D=\{x\in(a,b)\setminus\Bbb Q:n(x)=m\}$$ is dense in $(a,b)$ in the usual topology on $\Bbb R$.
Now let $q\in(a,b)\cap\Bbb Q$. $\mathscr{G}_m$ covers $X$, so there is a $G\in\mathscr{G}_m$ such that $q\in G$, and there is a $B\in\mathscr{B}$ such that $q\in B\subseteq G$. But then $B\cap D\ne\varnothing$. Fix $x\in B\cap D$; then
$$q\in G\cap\Bbb Q\subseteq\st(x,\mathscr{G}_m)\cap\Bbb Q=\varnothing\;,$$
which is obviously impossible. This contradiction shows that $X$ is not developable.
