hints only please: compact iff bicompact

**phoenixthoth** · Aug2-04, 07:47 PM

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 definiton 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** · Aug2-04, 07:59 PM

Did you mean "K is a metric space"?

**phoenixthoth** · Aug2-04, 08:31 PM

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

**arildno** · Aug3-04, 01:54 PM

Originally Posted by 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 definiton 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 $LaTeX Code: \\{x_{k}\\}\\in{K}$ such that K is covered by the union of disks $LaTeX Code: D(x_{k},e)$ )
2.Let $LaTeX Code: \\{U_{i}\\}$ be an open cover of K.
Then there exist r>0, such that for any $LaTeX Code: y\\in{K}$ , the open disk
D(y,r) is contained in some $LaTeX Code: U_{i}$

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

Proof by contradiction..

**phoenixthoth** · Aug3-04, 09:40 PM

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** · Aug4-04, 03:51 AM

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

**phoenixthoth** · Aug4-04, 04:57 PM

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

We will show that if

is not totally bounded then

is not compact by constructing a sequence in

that has no convergent subsequence. Let $LaTeX Code: x_{1}$ be any element of

. Note that since

is not totally bounded,

is not covered by $LaTeX Code: B_{1}\\left( x_{1}\\right)$ . So let $LaTeX Code: x_{2}\\in K\\backslash B_{1}\\left( x_{1}\\right)$ . Likewise,

is not covered by $LaTeX Code: B_{1}\\left( x_{1}\\right) \\cup B_{2}\\left( x_{2}\\right)$ . For

let $LaTeX Code: x_{n}\\in K\\backslash \\bigcup_{i=1}^{n-1}B_{1}\\left( x_{i}\\right)$ . $LaTeX Code: x_{n}$ exists because if

is covered by $LaTeX Code: \\bigcup_{i=1}^{n-1}B_{1}\\left( x_{i}\\right)$ , it would be totally bounded. The sequence $LaTeX Code: \\left\\{ x_{i}\\right\\}$ has no convergent subsequence because for all (unequal)

, $LaTeX Code: d\\left( x_{i},x_{j}\\right) >1$ and if there were an increasing function $LaTeX Code: \\tau :Z^{+}\\rightarrow Z^{+}$ such that there is an $LaTeX Code: I\\in Z^{+}$ such that if $LaTeX Code: i\\geq I$ , then $LaTeX Code: d\\left( x_{\\tau \\left( i\\right) },x\\right) <1/2$ for some $LaTeX Code: x\\in X$ . That would imply that $LaTeX Code: d\\left( x_{\\tau \\left( I\\right) },x_{\\tau \\left( I+1\\right) }\\right) \\leq d\\left( x_{\\tau \\left( I\\right) },x\\right) +d\\left( x,x_{\\tau \\left( I+1\\right) }\\right) <1$ when we know that $LaTeX Code: d\\left( x_{\\tau \\left( I\\right) },x_{\\tau \\left( I+1\\right) }\\right) >1$ . Therefore

is not compact as we have exhibited a sequence with no convergent subsequence.

**arildno** · Aug4-04, 05:14 PM

Is $LaTeX Code: B_{1}$ 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** · Aug4-04, 07:36 PM

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

**phoenixthoth** · Aug4-04, 11:14 PM

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

**arildno** · Aug5-04, 06:37 AM

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