Proving compact iff bicompact in metric spaces

[phoenixthoth]

I'd like hints only please. I have an analysis book and I could look up the proof myself but I'm trying to prove it myself as an exercise; so giving the full proof would be redundant as well as counterproductive to my own learning.

X is a metric space.

In this other book, K is compact iff every sequence in K has a convergent subsequence. I know that topologically, this is not the standard definition of compact; some authors would call this sequentially compact.

K is bicompact if every open cover of K admits a finite refinement that also covers K. This is how compactness is usually defined as far as I knew.

Hints on how compact implies bicompact and bicompact implies compact would be very much appreciated.

Right now, I'm working on compact implies bicompact so that is the current priority. I don't need hints on the other direction at this time since I haven't tried it.

A general hint like "prove the contrapositive" or "use contradiction" might be helpful though I've tried the first of the two already. I suppose as I await a reply, I'll try contradiction...

Thanks!

 

[plover]

Did you mean "K is a metric space"?

 

[phoenixthoth]

Sorry for the oversight. I meant that X is a metric space and that K is a subset of X.

 

[arildno]

I'd like hints only please. I have an analysis book and I could look up the proof myself but I'm trying to prove it myself as an exercise; so giving the full proof would be redundant as well as counterproductive to my own learning.

X is a metric space.

In this other book, K is compact iff every sequence in K has a convergent subsequence. I know that topologically, this is not the standard definition of compact; some authors would call this sequentially compact.

K is bicompact if every open cover of K admits a finite refinement that also covers K. This is how compactness is usually defined as far as I knew.

Hints on how compact implies bicompact and bicompact implies compact would be very much appreciated.

Right now, I'm working on compact implies bicompact so that is the current priority. I don't need hints on the other direction at this time since I haven't tried it.

A general hint like "prove the contrapositive" or "use contradiction" might be helpful though I've tried the first of the two already. I suppose as I await a reply, I'll try contradiction...

Thanks!

I'll give you 2 lemmas to prove first:

1. If K is (sequentially) compact, then K is totally bounded.
(A set is called totally bounded, if for any e>0, there exist a finite set $$\{x_{k}\}\in{K}$$ such that K is covered by the union of disks $$D(x_{k},e)$$)
2.Let $$\{U_{i}\}$$ be an open cover of K.
Then there exist r>0, such that for any $$y\in{K}$$, the open disk
D(y,r) is contained in some $$U_{i}$$

(I thought 2. was shockingly strong the first time I saw it..)

Proof by contradiction..

 

[phoenixthoth]

Thank you so much. I went ahead and proved that compact --> bicompact using the two lemmas; now I'm working on proving the first lemma. I'm sort of stuck on it though I haven't tried everything yet. In the next day or two, I may ask for a hint on how to prove lemma 1. I'm now resorting to proving it with contradiction. I want to exhibit a sequence with no convergent subsequence... Still working on it. Thanks again.

 

[arildno]

Good luck!
I guess you've already figured out how the two lemmas together will prove your main objective..

 

[phoenixthoth]

How does this look for a proof of the first lemma?

We will show that if $$K$$ is not totally bounded then $$K$$ is not compact by constructing a sequence in $$K$$ that has no convergent subsequence. Let $$x_{1}$$ be any element of $$K$$. Note that since $$K$$ is not totally bounded, $$K$$ is not covered by $$B_{1}\left( x_{1}\right)$$. So let $$x_{2}\in K\backslash B_{1}\left( x_{1}\right)$$. Likewise, $$K$$ is not covered by $$B_{1}\left( x_{1}\right) \cup B_{2}\left( x_{2}\right)$$. For $$n>2$$ let $$x_{n}\in K\backslash \bigcup_{i=1}^{n-1}B_{1}\left( x_{i}\right)$$. $$x_{n}$$ exists because if $$K$$ is covered by $$\bigcup_{i=1}^{n-1}B_{1}\left( x_{i}\right)$$, it would be totally bounded. The sequence $$\left\{ x_{i}\right\}$$ has no convergent subsequence because for all (unequal) $$i,j$$, $$d\left( x_{i},x_{j}\right) >1$$ and if there were an increasing function $$\tau :Z^{+}\rightarrow Z^{+}$$ such that there is an $$I\in Z^{+}$$ such that if $$i\geq I$$, then $$d\left( x_{\tau \left( i\right) },x\right) <1/2$$ for some $$x\in X$$. That would imply that $$d\left( x_{\tau \left( I\right) },x_{\tau \left( I+1\right) }\right) \leq d\left( x_{\tau \left( I\right) },x\right) +d\left( x,x_{\tau \left( I+1\right) }\right) <1$$ when we know that $$d\left( x_{\tau \left( I\right) },x_{\tau \left( I+1\right) }\right) >1$$. Therefore $$K$$ is not compact as we have exhibited a sequence with no convergent subsequence.

 

[arildno]

Is $$B_{1}$$ a ball with radius 1?
If so, your argument is (slightly) flawed:
The negation is:
There exist some e>0 such that K cannot be covered by finitely many balls of radius e.
(As far as I can see, you can merely substitute e for 1 in your argument to get the proof)

 

[phoenixthoth]

Cool deal. Now I'm working on the second lemma...

 

[phoenixthoth]

Could I have a hint for the second lemma? Thanks in advance.

 

[arildno]

The negation is:
For every r>0, there exist some y in K, such that the disk D(y,r) is not (fully) contained in any Ui