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$U$ 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?

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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
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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
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2 Answers 2

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.

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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
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@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
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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.

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