Metric space proof

**JG89** · March 24th 2010, 07:04 AM

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.

Is this correct?

**tonio** · March 24th 2010, 10:56 AM

Originally Posted by JG89

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.

Is this correct?

Looks fine to me.

Tonio

**Drexel28** · March 26th 2010, 10:27 PM

Originally Posted by JG89

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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