Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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 $x \in A$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


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.