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Let $D(A ,B)= \inf_{\alpha \in A,\beta \in B} d(\alpha ,\beta )$ denote the distance between sets in the power set of a given set where $d(\alpha ,\beta )$ is a metric. Prove that $D(A,B )=0$ doesn't imply $A\bigcap B\neq \varnothing$.

My attempt: Let $D(A,B)= 0$. Then, $\inf_{\alpha \in A,\beta \in B}(d(\alpha ,\beta ))=0$. So there's $a\in A , b\in B$ such that $d(a,b)=0$, which means $a=b$. So, $A\bigcap B\neq \varnothing$. What's the problem in my proof?

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I think you mean $D(A,B)=...$. It feels like you copied the problem wrong. –  Thomas Andrews Dec 3 '12 at 17:01
Yes , Sorry for that. –  naanwa Dec 3 '12 at 17:11
You still need to edit the other instances of $D(\alpha,\beta)$ to be $D(A,B)$. Specifically, "Prove that $D(\alpha,\beta)=0$..." –  Thomas Andrews Dec 3 '12 at 17:12

3 Answers 3

There dont have to be such elements $a$ and $b$

for example take $$A = \left\{ (x,y) \,\vert\, x=0\right\}$$

and $$B = \left\{ (x,y) \,\vert\, x=1/y\right\}$$

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The infimum of a set is not necessarily attained for any particular values of the set (so your claim "there is $a\in A$ and $b\in B$ such that $d(a,b)=0$" is an issue).

Hint: try to construct an explicit example of such sets, for instance in $\Bbb{R}$.

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This statement “Then , $\inf_{\alpha\in A,\beta \in B}(d(\alpha,\beta))=0$. So there's $\alpha \in A, \beta \in B$.” is incorrect. It would be correct if we had $\min_{\alpha\in A,\beta \in B}(d(\alpha,\beta))=0$.

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