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Let X be a metric space and let A be a compact subset of X, B a closed subset of X. I am trying to show this implies that d(A,B)=0.

Please critique my proof:

First, we define d(A,B) as inf{d(a,b): a in A, b in B}. We then show that compactness of A forces the existence of a in A with d(a,B)=0 (note that A,B both closed is not enough; a counterexample is given by S={(x,0)} and S':={(x,1/x):x>0}, both in $\mathbb R$.

If d(A,B)=0, then there are sequences ${a_n}$ in A and ${b_n}$ in B such that

$d(a_n,b_n)<1/n$. Since A is compact+metric (as a closed subset of X-metric), there

exists a subsequence $a_{n_k}$ of ${a_n}$ with $a_{n_k}$-->a (since A is compact and X is Hausdorff--on golf :) -- A is also closed, so that a is in A ). Then a_{n_k} is Cauchy, there is an integer j with :

$d(a_{n_k}, a)$< 1/2n for j>k,m

and

$d(b_{n_k},a_{n_k})$ <1/2n , j>k

By triangle inequality, for every index a_{n_k} with k> j , we have $d(b_{n_k},a)<1/n

so that d(a,B)=0 .

But in a compact metric space X, the closure of a subset B consists of the set of points {x} in X with d(x,B)=0 , so a is in the closure of B. But B is closed to start with, so a belongs to both A,B, contradicting the assumption that A,B are disjoint.

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# A compact, B closed Disjoint subsets of Metric Space then d(A,B)=0

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