Hi, All:(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**