1. ## 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.

Is this correct?

2. Originally Posted by JG89
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.

Is this correct?

Looks fine to me.

Tonio

3. Originally Posted by JG89
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.