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

$U$ is a topological space. $X$ is an open subset of $U$, and $Y$ is a closed subset. Let $Z = X \cap Y$. Does $\bar{Z} = \bar{X} \cap \bar{Y}$.

Here, $\bar{X}$ denote the closure of $X$, and $\bar{Y}$, $\bar{Z}$ respectively. (So, $\bar{Y}=Y$.)

It is clear that $\bar{Z} \subseteq \bar{X} \cap \bar{Y}$, but is it true in the reversed direction?

share|cite|improve this question
Think of an open and a closed ball of radius $1$ around $(\pm 1,0)$ in the plane, for example. – t.b. Jul 29 '11 at 17:58
One-dimensional example: think of $X = (0,1)\subset \mathbb{R}$ and $Y = \{0\}$. – Willie Wong Jul 29 '11 at 18:02
Yet another example: the open upper half plane in $\mathbb{R}^2$ and the open lower half plane. – Mark Jul 29 '11 at 20:55
Thanks everyone. – ShinyaSakai Jul 31 '11 at 17:15
up vote 10 down vote accepted

Let $X$ be open and $Y = U \smallsetminus X$. Then $X \cap Y = \emptyset$. However, $\overline{X} \cap Y = \partial X$ won't be empty in general. Take $X$ to be an open ball in $\mathbb{R}^n$, for example.

share|cite|improve this answer
What is the delta-x in English? "Boundary"?? – The Chaz 2.0 Jul 29 '11 at 19:21
$\partial X$ is a notation often used to mean the boundary of $X$, and it seems clear from the context that that's what it means in this case. But it's coded as "\partial X", not as "\delta X", since that looks like this: $\delta X$. – Michael Hardy Jul 29 '11 at 19:35
@Michael: I was reading/writing from my phone and sometimes the markup doesn't render! I should have just reloaded and looked it up :) Thanks, though! – The Chaz 2.0 Jul 29 '11 at 22:29
@Theo: Thanks very much for this example. I was wrong only to think of proofs without looking for counterexamples. – ShinyaSakai Jul 31 '11 at 17:12
@Shinya: Well, I must say that I was a bit ashamed for posting this counterexample as an answer without elaborating at all, I'm glad it helped nevertheless. Take it as a general strategy: If you don't manage to prove something, look for a simple counterexample. This will help in any case: If the statement turns out to be true, if you don't see how to concoct a counterexample, you might manage to extract the reason why you fail. Qiaochu wrote an answer recently with some good advice on that. – t.b. Jul 31 '11 at 17:21

A word on intuition. At least for metric spaces (and more generally for first-countable spaces), you can think of $\bar{X}$ as the collection of all points which are limits of sequences of points in $X$ (and in general you can replace "sequence" with "net" or "filter"). Then $\overline{X \cap Y}$ is obviously contained in both $\overline X$ and $\overline Y$ (as sequences of points in $X \cap Y$ are sequences of points in $X$ and also sequences of points in $Y$), but on the other hand a point in $\bar{X} \cap \bar{Y}$ is

  • a limit of a sequence of points in $X$ and
  • a limit of a different sequence of points in $Y$

and there's no obvious way to use either of these sequences to cook up a sequence of points in $X \cap Y$; the two sequences above may be disjoint, and in fact $X \cap Y$ may be empty while $\bar{X} \cap \bar{Y}$ is non-empty. With that in mind it isn't hard to find a counterexample.

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.