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