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I don't really understand the hint in part a (actually I'm confused). HOW can we find such sequences x_n and y_n? And how can we find a subsequence x_{n_k} that converges?
Any help is appreciated!
[note: also under discussion in math link forum]
I am not sure how to begin because I don't know how to FIND x_n and y_n? Do we need to produce an actual concrete example for which d(xn,yn)->d(X,Y)? Or can we just say such sequences exist?
Also, since X is compact, and by definition we know there EXISTS a convergence subsequence x_nk whose limit lies in X., isn't it? But then the hint asks us to FIND such a subsequence, why?? What does that mean? Don't we know such a subsequence x_nk already exists by definition?
Can someone help me, please?
This is a lot harder than I initially thought. The idea behind the following is that this is obvious if is bounded, and if it isn't we can "erase" the parts that are far enough away from to be ignored.
So, we know that is compact and thus totally bounded (specifically bounded). Thus, for some constant . Let . This is a compact subspace of and so is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that , but I leave this for you to prove. Thus, we must merely show that for some . To do this we note that since are compact that is compact by Tychonoff's theorem. Also, it is clear that is continuous and so it assumes it's minimum on and the conclusion follows.
hmm...I don't have enough background to understand your proof...there are concepts and theorems that I've never seen...Thanks though...
But I have some ideas regarding the method suggested in the hint:
d(xnk,ynk) is convergent, hence bounded
Since X is bounded, d(xnk,0) is bounded
With this and the triangle inequality, I think we can show that (ynk) is bounded, and so it has a convergent subsequence.
But I can't figure out the other parts of this proof...
Any ideas about part b of this problem? I don't know how to construct a counterexample...