MHB Distance between Compact Subsets

  • Thread starter Thread starter Sudharaka
  • Start date Start date
  • Tags Tags
    Compact Subsets
Sudharaka
Gold Member
MHB
Messages
1,558
Reaction score
1
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 \(S\).
 
Physics news on Phys.org
Sudharaka said:
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 $$F_1\cap F_2=\emptyset$$, otherwise there is nothing to prove.
Let $$D(F_1,F_2)=\delta=\inf\{d(x,y):x\in F_1~\&~y\in F_2\}$$
$$\exists x_1\in F_1~\&~y_1\in F_2$$ such that $$\delta\le d(x_1,y_1)<\delta+1$$

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

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

Now use compactness to get limits points of the two sequences.
 
Plato said:
Assume that $$F_1\cap F_2=\emptyset$$, otherwise there is nothing to prove.
Let $$D(F_1,F_2)=\delta=\inf\{d(x,y):x\in F_1~\&~y\in F_2\}$$
$$\exists x_1\in F_1~\&~y_1\in F_2$$ such that $$\delta\le d(x_1,y_1)<\delta+1$$

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

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

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? :)
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top