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

  1. Jul 3, 2011 #1
    Hi, All:
    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


    $d(b_{n_k},a_{n_k})$ <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.
  2. jcsd
  3. Jul 3, 2011 #2
    Your proof seems good. An other way is the following: prove that the map [tex]x\mapsto \inf_{b\in B}\, d(x,b)[/tex] is continuous, and use the fact that a continuous function on a compact gets its min. Since A and B are disjoint, this min is positive and you can conclude.
  4. Jul 3, 2011 #3
    Certainly you meant [itex]d(A,B)\neq 0[/itex]??

    Your proof seems to be ok...
  5. Jul 3, 2011 #4
    Thanks, Both; a followup, please:

    I wonder if we can extend the argument to show that if A,B are disjoint compact subsets of X metric, and d(A,B)=c >0 , then there are points a,b in A,B respectively,
    with d(a,b)=c, i.e., the distance value is actually reached:

    We use again the sequences {a_n}, {b_n}, so that d(a_n,b_n)<1/n , and then there are respective subsequences {a_n_k} and {b_n_j} of {a_n},{b_n} respectively, with
    {a_n}->a and {b_n}->b , so that d(a,B)=d(A,b)=c.
  6. Jul 3, 2011 #5
    That's good!!
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