# Thread: Compactness; show inf is achieved

1. ## Compactness; show inf is achieved

===================================

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]

2. Originally Posted by kingwinner

===================================

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]
What have you tried mon capitain?

3. 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?

4. This is a lot harder than I initially thought. The idea behind the following is that this is obvious if $Y$ is bounded, and if it isn't we can "erase" the parts that are far enough away from $X$ to be ignored.

So, we know that $X$ is compact and thus totally bounded (specifically bounded). Thus, $X\subseteq \left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqsla nt M\right\}$ for some constant $M$. Let $N=\left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqs lant M+d(X,Y)+1\right\}$. This is a compact subspace of $\mathbb{R}^n$ and so $G\cap N=K$ is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that $d(E,G)=d(E,K)$, but I leave this for you to prove. Thus, we must merely show that $d(E,G)=d(E,K)=d(e,k)$ for some $e\in E,k\in K$. To do this we note that since $G,K$ are compact that $G\times K$ is compact by Tychonoff's theorem. Also, it is clear that $d:G\times K\mapsto\mathbb{R}$ is continuous and so it assumes it's minimum on $G\times K$ and the conclusion follows.

5. Originally Posted by Drexel28
This is a lot harder than I initially thought. The idea behind the following is that this is obvious if $Y$ is bounded, and if it isn't we can "erase" the parts that are far enough away from $X$ to be ignored.

So, we know that $X$ is compact and thus totally bounded (specifically bounded). Thus, $X\subseteq \left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqsla nt M\right\}$ for some constant $M$. Let $N=\left\{\bold{x}\in\mathbb{R}^n:\|\bold{x}\|\leqs lant M+d(X,Y)+1\right\}$. This is a compact subspace of $\mathbb{R}^n$ and so $G\cap N=K$ is compact (since the intersection of a compact set and a closed set is compact). Now, we next claim that $d(E,G)=d(E,K)$, but I leave this for you to prove. Thus, we must merely show that $d(E,G)=d(E,K)=d(e,k)$ for some $e\in E,k\in K$. To do this we note that since $G,K$ are compact that $G\times K$ is compact by Tychonoff's theorem. Also, it is clear that $d:G\times K\mapsto\mathbb{R}$ is continuous and so it assumes it's minimum on $G\times K$ 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...