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[SOLVED] Distance between Compact Subsets

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Jan 27, 2012
196
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
Feb 5, 2012
1,621
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? :)