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Is the closure of $ X \cap Y$ equal to $\bar{X} \cap \bar{Y}$?

I'm sorry to ask another question so soon after my last one, but my exam Introduction to Functional Analysis is near and I'm curious about this exercise :)

Exercise:

Give an example of a Metric Space $(X,\rho)$ and two sets $A, B \subset X$ such that $\overline{A \cap B} \neq \overline{A} \cap \overline{B}$.

So (the closure of the intersection of $A$ with $B$) should not equal (the intersection of the closure of $A$ with the closure of $B$).

My first idea was to test in the $\mathbb{R}^n$, but if I'm correct the inequality doesn't hold in those spaces.

Next idea was to look to the set (or space) of all polynomials on $[a,b]$, since I know that the closure (or completion, not sure what the difference is) of this space is the space of continuous functions on $[a,b]$. But the exercise states that I have to pick two sets out of my chosen Metric Space (i.e. the polynomials), not sure what to pick.

Could someone provide a hint, should I for instance look to a Sequence Space or a Function Space? Or something with a discrete Metric?

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Ailurus
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  • I am almost certain that this question was asked before. As Zev shows, it is easier to come up with a counterexample than the link to the duplicate :-) – Asaf Karagila Oct 31 '11 at 00:34
  • @Asaf: I found at least one copy of it, there may be more :) – Zev Chonoles Oct 31 '11 at 00:38
  • @Zev: I was just finding that question which you marked as a duplicate. Nice for beating me up by 18 seconds. :-) – Asaf Karagila Oct 31 '11 at 00:40
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    My apologies, I should have searched a little better. By the way, am I the only one who sees a pun in the [closed]-tag (and the square brackets)...? Thanks @ZevChonoles for both the answer and the redirecting. – Ailurus Oct 31 '11 at 00:44
  • @Ailurus: It's no problem, happens all the time :) And good catch on the double pun! – Zev Chonoles Oct 31 '11 at 00:45

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Consider $X=\mathbb{R}$ with its usual metric, and let $A=(0,1)$, and $B=(1,2)$. Then $A\cap B=\varnothing$, hence $\overline{A\cap B}=\overline{\varnothing}=\varnothing$, but $\overline{A}=[0,1]$ and $\overline{B}=[1,2]$, hence $\overline{A}\cap\overline{B}=\{1\}$. Thus $$\overline{A \cap B} =\varnothing\neq \{1\}=\overline{A} \cap \overline{B}.$$

Zev Chonoles
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  • Ah, of course! For some reason I only thought about closed sets and their intersection. Perfect solution, thanks! Now I only have to find the difference between completion and closure (my teacher mixed them up and it doesn't get very clear from the book), and then I'm ready for my exam. – Ailurus Oct 31 '11 at 00:38
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    @Ailurus Two facts: 1. The intersection of two closed sets is closed. 2. The closure of a closed set is itself. From these facts, one can deduce, for closed sets $X$ and $Y$, $\overline{X \cap Y} = X \cap Y = \overline{X} \cap \overline{Y}$. Thus, starting with two closed sets will never produce a counterexample. – Austin Mohr Oct 31 '11 at 01:34