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Let $X$ be a compact Hausdorff space such that for every $x \in X$ , there exist a nested sequence of open sets $\{U_n\}$ such that $\{x\}=\bigcap_{n=1}^\infty U_n$ , then is it true that $X$ is first countable?

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Yes.

For every $n$ we can construct $V_{n}\subseteq U_{n}$ such that $V_{n}$ is a closed neighborhood of $x$.

This because disjoint open sets $A_n,B_n$ exist with $x\in A_n$ and $U_n^c\subseteq B_n$. We then take $V_n=B_n^c$.

Then automatically $\left\{ x\right\} =\bigcap_{n=1}^{\infty}V_{n}$.

Let $x\in W$ where $W$ is open.

Then $W^{c}$ is closed hence compact with $W^{c}\subseteq\bigcup_{n=1}^{\infty}U_{n}^{c}\subseteq\bigcup_{n=1}^{\infty}V_{n}^{c}$.

The sets $V_{n}^c$ are open so that $W^{c}\subseteq\bigcup_{n=1}^{m}V_{n}^{c}$ for some $m$.

Then we have $x\in\bigcap_{n=1}^{m}V_{n}\subseteq W$.

Here $\bigcap_{n=1}^{m}V_{n}$ is a neighbourhood of $x$ and apparantly the countable collection: $$\left\{ \bigcap_{n=1}^{m}V_{n}\mid m\in\mathbb{N}\right\} $$ serves as neighborhood basis for $x$.

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  • $\begingroup$ but the $V_n$'s are not open ... am I missing something ? $\endgroup$ – user228168 Sep 6 '16 at 4:19
  • $\begingroup$ Indeed, they are not open but that is no objection. If you are practicizing the definition that neighborhoods are open (Munkres) and want to see a local basis of open neighborhoods then you can take the interiors of the $V_n$. $\endgroup$ – drhab Sep 6 '16 at 7:05
  • $\begingroup$ but what if the interior is empty ? $\endgroup$ – user228168 Sep 6 '16 at 15:32
  • $\begingroup$ $A_n$ and $B_n$ are disjoint sets so that $x\in A_n\subseteq B_n^c=V_n$. So the open set $A_n$ will be a subset of the interior of $V_n$. That makes $V_n$ a neighborhood of $x$. Moreover as complement of the open set $B_n$ it is a closed neighborhood. $\endgroup$ – drhab Sep 6 '16 at 17:21
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It’s not necessary that the open sets be nested. In fact, the general theorem is that if $X$ is a compact Hausdorff space, $x\in X$, and $\mathscr{U}$ is an infinite family of open sets such that $\bigcap\mathscr{U}=\{x\}$, then $x$ has a local base of cardinality at most $|\mathscr{U}|$; your result is the special case in which $\mathscr{U}$ is countable. The proof is essentially the same as the one given by drhab for that special case.

$X$ is compact Hausdorff, so for each $U\in\mathscr{U}$ there are disjoint open sets $V_U$ and $H_U$ such that $x\in V_U$ and $X\setminus U\subseteq H_U$; let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$; clearly $|\mathscr{V}|\le|\mathscr{U}|$.

Let $G$ be an arbitrary open nbhd of $x$, and let $K=X\setminus G$. For each $y\in K$ there is a $U(y)\in\mathscr{U}$ such that $y\notin U(y)$. Moreover,

$$x\in V_{U(y)}\subseteq\operatorname{cl}V_{U(y)}\subseteq X\setminus H_{U(y)}\subseteq U(y)\;,$$

so $V_{U(y)}$ and $X\setminus\operatorname{cl}V_{U(y)}$ are disjoint open nbhds of $x$ and $y$, respectively. $\{X\setminus\operatorname{cl}V_{U(y)}:y\in K\}$ is an open cover of the compact set $K$, so there is a finite $F\subseteq K$ such that

$$K\subseteq\bigcup_{y\in F}\big(X\setminus\operatorname{cl}V_{U(y)}\big)\;,$$

and therefore

$$x\in\bigcap_{y\in F}V_{U(y)}\subseteq G\;.$$

Let

$$\mathscr{W}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{V}\text{ and }\mathscr{F}\text{ is finite}\right\}\;;$$

then we’ve just shown that $\mathscr{W}$ is a local base at $x$, and $|\mathscr{W}|\le|\mathscr{U}|$, since an infinite set $S$ has $|S|$ finite subsets.

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  • $\begingroup$ Nice generalization. $\endgroup$ – drhab Sep 6 '16 at 7:08

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