# Topology and locally closed subsets

Let $X$ be a topological space and let $A\subseteq X$.

Supose that for each $x\in A$ there exists a neighbourhood of $x$, $V_x$, in $X$ such that $A\cap V_x$ is closed in $V_x$.

Prove that $\Bigl(\bigcup _{x \in A}int(V_x)\Bigr)\cap cl(A)\subseteq A$.

A few notes:

$A\cap V_x$ is closed in $V_x$ $\iff$ There exists a subset $F$ of $X$ which is closed in $X$ and $A\cap V_x=F\cap V_x$, by definition.

$int(A)$ is the interior of A.

$cl(A)$ is the closure of A.

Let $x\in \Bigl(\bigcup _{x \in A}int(V_x)\Bigr)\cap cl(A)$. Please proceed.

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I think you mean $x \in \left(\bigcup_{x \in A} int(V_x)\right) \cap cl(A)$ at the end. – Austin Mohr Dec 13 '12 at 19:55

## 2 Answers

Let $x\in \Bigl(\bigcup _{x \in A}int(V_x)\Bigr) \cap cl(A)$. Then $x\in \Bigl(\bigcup _{x \in A}int(V_x)\Bigr)$ and $x \in cl(A)$. It follows that $x \in int(V_{x_o})$ for some $V_{x_o}$ in the collection of $V_x$ and $x \in A$.

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Yes, that's what I meant, thanks. Alas it seems to me that you're assuming that $V_{x_0}$ is a subset of $A$ which isn't necessarily true. – Manuel Patrício Dec 13 '12 at 20:39

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Let $y\in\bigcup_{x\in A}\int V_x\cap\cl A$; then $y\in\int V_x$ for some $x\in A$, and of course $y\in\cl A$. Fix a closed $F\subseteq X$ such that $V_x\cap A=V_x\cap F$. If $y\notin A$, then $y\in\int V_x\setminus A=\int V_x\setminus F$. But then $\int V_x\setminus F$ is an open nbhd of $y$ disjoint from $A$, which is a contradiction.

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