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Math Help - Metric space proof

  1. #1
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    Metric space proof

    Problem: Let  (M, d) be a metric space that satisfies the Bolzano-Weierstrass Property. Also, let  A and  B be disjoint, compact subsets of  (M, d) .  Dist(A,B) is defined as  inf\{d(a,b): a \in A, b \in B\} . Show that  Dist(A,B) > 0 .

    Proof:

    Suppose that  Dist(A,B) = inf\{ d(a,b) : a \in A, b \in B \} = 0. Then either 1)  \exists a \in A, b \in B : d(a,b) = 0 or 2) Given any positive number, there exists a point in A and a point in B whose distance apart from eachother is less than this positive number. If it is 1), then  a = b , contradicting the disjointedness of  A and  B .

    So assume 2). We will construct two sequences. First take the number  1 . There exists points in  A and  B , let's call them  a_1 and  b_1 , such that  d(a_1, b_1) < 1 . Now consider the number  \frac{1}{2} . There exists points in  A and  B , let's call them  a_2 and  b_2 , such that  d(a_2, b_2) < \frac{1}{2} . Now consider the number  \frac{1}{3} . There are points in  A and  B , let's call them  a_3 and  b_3 , such that  d(a_3, b_3) < \frac{1}{3} . We can continue on like this and construct two sequences,  a_n and  b_n such that  d(a_n, b_n) < \frac{1}{n} .

    By compactness of  A , there is a subsequence  a_{n_k} of  a_n converging to  L \in A . Since  d(a_{n_k}, b_{n_k}) \rightarrow 0 , then we see that the subsequence  b_{n_k} also approaches  L .  B is closed, so this point  L must also lie in  B . Thus  L \in A \cap B , contradicting the disjointedness of  A and  B .

    Thus  Dist(A,B) > 0 . QED.


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  2. #2
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    Quote Originally Posted by JG89 View Post
    Problem: Let  (M, d) be a metric space that satisfies the Bolzano-Weierstrass Property. Also, let  A and  B be disjoint, compact subsets of  (M, d) .  Dist(A,B) is defined as  inf\{d(a,b): a \in A, b \in B\} . Show that  Dist(A,B) > 0 .

    Proof:

    Suppose that  Dist(A,B) = inf\{ d(a,b) : a \in A, b \in B \} = 0. Then either 1)  \exists a \in A, b \in B : d(a,b) = 0 or 2) Given any positive number, there exists a point in A and a point in B whose distance apart from eachother is less than this positive number. If it is 1), then  a = b , contradicting the disjointedness of  A and  B .

    So assume 2). We will construct two sequences. First take the number  1 . There exists points in  A and  B , let's call them  a_1 and  b_1 , such that  d(a_1, b_1) < 1 . Now consider the number  \frac{1}{2} . There exists points in  A and  B , let's call them  a_2 and  b_2 , such that  d(a_2, b_2) < \frac{1}{2} . Now consider the number  \frac{1}{3} . There are points in  A and  B , let's call them  a_3 and  b_3 , such that  d(a_3, b_3) < \frac{1}{3} . We can continue on like this and construct two sequences,  a_n and  b_n such that  d(a_n, b_n) < \frac{1}{n} .

    By compactness of  A , there is a subsequence  a_{n_k} of  a_n converging to  L \in A . Since  d(a_{n_k}, b_{n_k}) \rightarrow 0 , then we see that the subsequence  b_{n_k} also approaches  L .  B is closed, so this point  L must also lie in  B . Thus  L \in A \cap B , contradicting the disjointedness of  A and  B .

    Thus  Dist(A,B) > 0 . QED.


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    Looks fine to me.

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JG89 View Post
    Problem: Let  (M, d) be a metric space that satisfies the Bolzano-Weierstrass Property. Also, let  A and  B be disjoint, compact subsets of  (M, d) .  Dist(A,B) is defined as  inf\{d(a,b): a \in A, b \in B\} . Show that  Dist(A,B) > 0 .
    Also looks good to me. A simpler way would be to notice that Tychonoff's theorem implies that A\times B is compact and since d:A\times B\to\mathbb{R} given by (x,y)\mapsto d(x,y) is continuous we know that it assumes a minimum on A\times B. Namely, \inf\text{ }d(A\times B)=d(a,b) for some a\in A,b\in B. But, since A\cap B=\varnothing and compact subspaces of metric spaces are closed we know that d(a,b)>0. The conclusion follows.
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