# Does every point in a $T_{3}$-space have a clopen base set or infinite nested base sets, none of which are clopen?

Let $X$ be a $T_{3}$-space. Then for every open set $A$ containing $x\in X$, there is an open subset $E$ such that $Cl(E)\subseteq A$.

My hypothesis is: For every point $x\in X$, there is at least one clopen base set $B_{i}(x)$ containing $x$, or there are infinite nested base sets containing $x$, none of which are clopen.

Motivation: Let us take a base set $E$ containing $x$. By definition, it is also an open set. By the $T_{3}$ axiom, it contains another open set $E'$, the closure of which is $\subseteq E$. If $E$ is clopen, we can take $E'$ to equal $E$. If $E$ is not clopen, and $E'$ is, then we don't have to consider any further open subset of $E'$. If $E'$ too is not clopen, then we come to $E''\subset E'$. And so on. Either some $E^{n}$ is clopen, or we have an infinite number of nested base sets, none of which are clopen.

Is this argument true?

The main idea of your argument is correct, but some care in writing it out might be beneficial.

Let $X$ be T$_3$, $x \in X$, and let $\mathcal{N}$ be a (open) neighbourhood base at $x$. Suppose that no set in $\mathcal{N}$ is clopen. We can construct a sequence $\langle U_n \rangle_{n=0}^\infty$ in $\mathcal{N}$ such that $U_{n+1} \subsetneq \overline{U_{n+1}} \subsetneq U_n$ as follows:

• Pick $U_0 \in \mathcal{N}$ arbitrarily.
• Given $U_n$, as $x \in U_n$ by regularity there is a $U_{n+1} \in \mathcal{N}$ such that $x \in U_{n+1} \subseteq \overline{U_{n+1}} \subseteq U_n$. Note that since $U_{n+1}$ is not clopen, then $U_{n+1} \neq \overline{U_{n+1}}$ and since $U_n$ is not clopen we have $\overline{U_{n+1}} \neq U_n$.

Just be warned that the family $\{ U_n : n \in \mathbb{N} \}$ constructed above need not itself be a neighbourhood base at $x$. In particular, if $X$ is a connected T$_3$-space which is not first countable, such as the "extended long line" (or the "extended long ray" as it is called on Wikipedia), and $x \in X$ does not have a countable neighbourhood base, then the family above cannot be a neighbourhood base at $x$.

The basic idea is okay, but the details need quite a bit of work. I’d organize it like this:

Let $x\in X$, and let $\mathscr{G}=\{G\subseteq X:x\in U\text{ and }G\text{ is clopen}\}$. If $\mathscr{G}$ is a base at $x$, we’re done, so suppose not. Then there is an open nbhd $U_0$ of $x$ such that that $G\nsubseteq U_0$ for all $G\in\mathscr{G}$. Since $x$ is not isolated (why?), there is an $x_0\in U_0\setminus\{x\}$, and there is an open $U_1$ such that $$x\in U_1\subseteq\operatorname{cl}U_1\subseteq U_0\setminus\{x_0\}\;.$$ In general, given an open nbhd $U_n$ of $x$, we can choose a point $x_n\in U_n\setminus\{x\}$ and use the fact that $X$ is $T_3$ to find an open $U_{n+1}$ such that $$x\in U_{n+1}\subseteq\operatorname{cl}U_{n+1}\subseteq U_n\setminus\{x_n\}\;.$$ Let $\mathscr{U}=\{U_n:n\in\Bbb N\}$; clearly is a nested family of open nbhds of $x$: $$U_0\supsetneqq\operatorname{cl}U_1\supseteq U_1\supsetneqq\operatorname{cl}U_2\supseteq U_2\supsetneqq\ldots\;;.$$ Suppose that some $U_n$ is clopen; then $U_n\subseteq U_0$ with $U_n\in\mathscr{G}$, contradicting the choice of $U_0$. Thus, no member of $\mathscr{U}$ is clopen, and we can actually write $$U_0\supsetneqq\operatorname{cl}U_1\supsetneqq U_1\supsetneqq\operatorname{cl}U_2\supsetneqq U_2\supsetneqq\ldots\;;.$$ That is, $\mathscr{U}$ is a strongly nested family of open nbhds of $x$, none of them clopen.

Note, though, that $\mathscr{U}$ definitely does not have to be a local base at $x$ unless $X$ is first countable.

• It is a Good job. +1) – Paul Jun 19 '13 at 0:23
• @Brian- I really had no other way of communicating this to you. I'm sorry this is completey off-topic. It would be incrediby kind of you if you could give this a look. I don't know anyone who could answer it better. Thanks! – fierydemon Jul 1 '13 at 14:18

YOU are right. If $X$ is zero-dimensional, then $x$ will always have a clopen nbhd. If $X$ is not zero-dimensional, it maybe not have such clopen nbhd. However, it always have infinite nested base sets containing itself, since $X$ is regular.