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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?

Thanks for your time.

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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 $x$.

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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.

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  • $\begingroup$ It is a Good job. +1) $\endgroup$ – Paul Jun 19 '13 at 0:23
  • $\begingroup$ @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! $\endgroup$ – fierydemon Jul 1 '13 at 14:18
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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.

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