Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a topological space. Let $\{x\} \subset X$ and $N(x)$ be the neighborhood filter of $x$. Is $\cap_{n \in N(x)} \bar n = \overline {\{x\}}$?

It at least that $\overline {\{x\}} \subset \cap_{n \in N(x)} \bar n $, for if $n \in N(x), \{x\} \subseteq n$ thus $n \subseteq \bar n$ and $\{x\} \subseteq \overline {\{x\}}$ and thus $\{x\} \subseteq \bar n \cap \overline {\{x\}}$. This is closed, and smaller or equal to $\overline {\{x\}}$, thus $\overline {\{x\}} \subset \cap_{n \in N(x)} \bar n$.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

$\newcommand{\cl}{\operatorname{cl}}$It’s false in general. Let $X$ be an infinite space with the cofinite topology. Then $\cl U=X$ for each non-empty $U\in\tau$, so $$\bigcap_{N\in\mathscr{N}(x)}\cl N=X\ne\{x\}=\cl\{x\}$$ for each $x\in X$.

It’s true if $X$ is $T_2$. In that case let $x\in X$, and let $y$ be any other point of $X$. There are disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$. Clearly $V\cap\cl U=\varnothing$, so $$y\notin\cl U\supseteq\bigcap_{N\in\mathscr{N}(x)}N\;,$$ and since $y\in X\setminus\{x\}$ was arbitrary, $$\bigcap_{N\in\mathscr{N}(x)}N=\{x\}\;.$$

share|improve this answer
add comment

Not necessarily. Let X be a set containing the distinct points $a$ and $b$. $(X,\{{\emptyset},\{b\},X\})$ is a topological space for which the equality does not hold.

The closure of $\{a\}$ is $X - \{b\}$ and $X$ is the only neighborhood of $a$.

But I think in all uniform spaces the equality holds.

share|improve this answer
    
you're right. I'll edit my post. –  user59671 Jan 27 '13 at 20:36
    
Looks good.${}{}$ –  Arthur Fischer Jan 27 '13 at 21:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.