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I have a question:

Is every second countable $T_2$ topological space a developable space?

Thanks for your help.

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  • $\begingroup$ Welcome to MSE! Generally you'll get a better response if you've shown evidence that you've worked on the problem, and provide some background information about what you know, and in particular what you're stuck with. $\endgroup$
    – pjs36
    Mar 30, 2015 at 0:25
  • $\begingroup$ Not familiar with this kind of space, but this paper might be of interest, particularly the last two theorems: ams.org/journals/proc/1980-080-01/S0002-9939-1980-0574527-3/… $\endgroup$
    – Moya
    Mar 30, 2015 at 2:07
  • $\begingroup$ Actually, to add onto my last comment, the author of the above article makes the comment that a paracompact second-countable Hausdorff space is developable. $\endgroup$
    – Moya
    Mar 30, 2015 at 2:26

1 Answer 1

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$\newcommand{\st}{\operatorname{st}}$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.

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  • $\begingroup$ @Brain M.Scott Thanks! How does baire category work in the proof? $\endgroup$
    – John
    Mar 30, 2015 at 12:57
  • $\begingroup$ @John: The irrationals are a Baire space, so they are not the union of countably many nowhere dense sets. Thus, if you decompose them into countably many sets, at least one of those sets has to be dense in some interval relative to the usual topology. And that’s all that I used in the last paragraph. $\endgroup$ Mar 30, 2015 at 13:02

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