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Homework Help: Distance between Sets and their Closures

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose [itex](X,d)[/itex] is a metric space, and suppose that [itex]A,B\subseteq X[/itex]. Show that [itex]dist(A,B)=dist(cl(A),cl(B))[/itex].

    2. Relevant equations
    [itex]cl(A)=\partial A\cup A[/itex].
    [itex]dist(A,B)=\inf \{d(a,b):a\in A,b\in B\}[/itex]

    3. The attempt at a solution
    Its clear that [itex]dist(cl(A),cl(B))\leq \min\{dist(A,B),dist(A,\partial B),dist(\partial A,B),dist(\partial A,\partial B)\}\leq dist(A,B)[/itex]. I just can't exactly find a way to go the other direction of the inequality. I'm not looking for a direct solution, just some intuition.
  2. jcsd
  3. Oct 11, 2011 #2
    Here's my attempt at a solution:

    [itex]dist(A,B)\leq dist(A,x)+dist(x,B)\implies \inf _{x\in cl(A)}\{dist(A,B)\}\leq \inf _{x\in cl(A)}\{dist(A,x)+dist(x,B)\}=\inf _{x\in cl(A)}\{dist(x,B)\}[/itex]. And, [itex]dist(cl(A),B)\leq dist(cl(A),y)+dist(y,B)\implies \inf _{y\in cl(B)}\{dist(cl(A),B)\}\leq\inf _{y\in cl(B)}\{dist(cl(A),y)+dist(y,B)\}=\inf _{y\in cl(B)}\{dist(cl(A),y)\}=dist(cl(A),cl(B))[/itex]. Hence, [tex]dist(A,B)=\inf _{x\in cl(A)}\{\inf _{y\in cl(B)}\{dist(A,B)\}\}\leq dist(cl(A),cl(B)).[/tex]

    Does this work?
  4. Oct 11, 2011 #3


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    Homework Helper

    dist(cl(A),cl(B)) is clearly less than dist(A,B) just because A is contained in cl(A) and B is contained in cl(B). Do you agree with that? If so why? To go farther you need to deal with the definition of what cl(A) or boundary(A) in a metric space is. What is it?
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