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Thread: Metric space proof

  1. #1
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    Metric space proof

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

    Proof:

    Suppose that $\displaystyle Dist(A,B) = inf\{ d(a,b) : a \in A, b \in B \} $ = 0. Then either 1) $\displaystyle \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 $\displaystyle a = b $, contradicting the disjointedness of $\displaystyle A $ and $\displaystyle B $.

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

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

    Thus $\displaystyle Dist(A,B) > 0 $. QED.


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

    Proof:

    Suppose that $\displaystyle Dist(A,B) = inf\{ d(a,b) : a \in A, b \in B \} $ = 0. Then either 1) $\displaystyle \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 $\displaystyle a = b $, contradicting the disjointedness of $\displaystyle A $ and $\displaystyle B $.

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

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

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