# [SOLVED]Distance between Compact Subsets

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Here's a question that I couldn't find the full answer. Any ideas will be greatly appreciated.

Let $$(X,\,d)$$ be a metric space. Let $$F_1,\,F_2$$ be two nonempty compact subsets of $$X$$. Prove that, there exists $$x_1\in F_1,\,x_2\in F_2,$$ such that,

$d(x_1,\,x_2)=\mbox{inf}\{d(x,\,y):x\in F_1,\,y\in F_2\}$
I felt that the compactness of $$F_1$$ and $$F_2$$ could be brought into the question using the following equivalency. However all my attempts to solve the question weren't successful.

Let $$(X,\,d)$$ be a metric space and $$S$$ is a compact subspace of $$X$$. Then, any sequence of points in $$S$$ has a subsequence which is convergent to a point in $$S$$.

#### Plato

##### Well-known member
MHB Math Helper
Here's a question that I couldn't find the full answer. Any ideas will be greatly appreciated.
I felt that the compactness of $$F_1$$ and $$F_2$$ could be brought into the question using the following equivalency. However all my attempts to solve the question weren't successful.
Assume that $$\displaystyle F_1\cap F_2=\emptyset$$, otherwise there is nothing to prove.
Let $$\displaystyle D(F_1,F_2)=\delta=\inf\{d(x,y):x\in F_1~\&~y\in F_2\}$$
$$\displaystyle \exists x_1\in F_1~\&~y_1\in F_2$$ such that $$\displaystyle \delta\le d(x_1,y_1)<\delta+1$$

Let $$\displaystyle \epsilon_2=\min\{.5, d(x_1,y_1)\}$$.
$$\displaystyle \exists x_2\in F_1~\&~y_2\in F_2$$ such that $$\displaystyle \delta\le d(x_2,y_2)<\delta+\epsilon_2$$.

Let $$\displaystyle \epsilon_n=\min\{n^{-1}, d(x_{n-1},y_{n-1})\}$$.
$$\displaystyle \exists x_n\in F_1~\&~y_n\in F_2$$ such that $$\displaystyle \delta\le d(x_n,y_n)<\delta+\epsilon_n$$.

Now use compactness to get limits points of the two sequences.

#### Sudharaka

##### Well-known member
MHB Math Helper
Assume that $$\displaystyle F_1\cap F_2=\emptyset$$, otherwise there is nothing to prove.
Let $$\displaystyle D(F_1,F_2)=\delta=\inf\{d(x,y):x\in F_1~\&~y\in F_2\}$$
$$\displaystyle \exists x_1\in F_1~\&~y_1\in F_2$$ such that $$\displaystyle \delta\le d(x_1,y_1)<\delta+1$$

Let $$\displaystyle \epsilon_2=\min\{.5, d(x_1,y_1)\}$$.
$$\displaystyle \exists x_2\in F_1~\&~y_2\in F_2$$ such that $$\displaystyle \delta\le d(x_2,y_2)<\delta+\epsilon_2$$.

Let $$\displaystyle \epsilon_n=\min\{n^{-1}, d(x_{n-1},y_{n-1})\}$$.
$$\displaystyle \exists x_n\in F_1~\&~y_n\in F_2$$ such that $$\displaystyle \delta\le d(x_n,y_n)<\delta+\epsilon_n$$.

Now use compactness to get limits points of the two sequences.
Thank you so much. The thing I missed was to use the right hand side of the inequalities,
$$d(x_1,\, y_1)< \delta \,\cdots ,\, d(x_n,\,y_n)<\delta+\epsilon_n$$ and so on. Thanks again for your help, I really appreciate it.

But can you please explain why you specifically wanted to use the epsilons? I mean, we know that there exist $$x_1$$ and $$y_1$$ such that,

$\delta\leq d(x_1,\,y_1)<\delta+1$

and $$x_2$$ and $$y_2$$ such that,

$\delta\leq d(x_2,\,y_2)<\delta+\frac{1}{2}$

and generally, $$x_n$$ and $$y_n$$ such that,

$\delta\leq d(x_n,\,y_n)<\delta+\frac{1}{n}$

Using the compactness we can still arrive at the result isn't?