MHB Distance between Compact Subsets

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In a metric space \((X, d)\), the discussion focuses on proving that for two nonempty compact subsets \(F_1\) and \(F_2\), there exist points \(x_1 \in F_1\) and \(x_2 \in F_2\) such that the distance \(d(x_1, x_2)\) equals the infimum of distances between points in \(F_1\) and \(F_2\). The compactness of these sets ensures that sequences drawn from them have convergent subsequences, which is crucial for establishing the existence of such points. The conversation also highlights the importance of using specific epsilon values to refine the distances and ensure convergence. Ultimately, the compactness property is key to demonstrating that the infimum is attained. The participants express appreciation for clarifying the role of epsilon in the proof process.
Sudharaka
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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 \(S\).
 
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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? :)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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