1. ## 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.

Is this correct?

2. 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

3. 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.