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 Y be an open subset of $\mathbb{R}^n$. If X is a closed subset of Y, disjoint from the boundary of Y, is it true that X is a closed subset of $\mathbb{R}^n$? How do I show this?

Edit: Let X be contained in a closed set B of $\mathbb{R}^n$ which is contained in Y and which is disjoint from the boundary of Y. Then X is closed in $\mathbb{R}^n.$

share|cite|improve this question
how do I accept if people answer via comments or if the comments were more helpful than the answers? – user38404 Sep 11 '12 at 19:34
There is a checkbox to the left of the answers. – AD. Sep 11 '12 at 19:38
@Vivek Ask the commenter to post the comment as a answer and accept it. – Ayman Hourieh Sep 11 '12 at 19:38
Maybe I'm misunderstanding the question, but how about $n=1$, $Y=(0,1)$, and $X=[1/2,1)$? You might ask the same question, but demand that the boundary of $Y$ be disjoint from the boundary of $X$. – David Mitra Sep 11 '12 at 19:54
up vote 2 down vote accepted


If a set $A$ is not closed in $\mathbb{R}^n$, then there is a sequence $x_n\in A$ converging to $x\in \mathbb{R}^n\setminus A$. (Why?)

Do you see where to go from there?

share|cite|improve this answer
thanks I can work it out now – user38404 Sep 11 '12 at 19:52

Your Answer


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