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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: It's no problem, happens all the time :) And good catch on the double pun!2011-10-31

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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 $\overline{A \cap B} =\varnothing\neq \{1\}=\overline{A} \cap \overline{B}.$

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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.2011-10-31