Proving Finite Subcovering with Compactness for A and B: Homework Solution"

[afkguy]

Homework Statement

A is compact and B is an open covering of A. Each a in A is contained in at least 2 subsets of B. Show that B has a finite sub-covering where A is still contained in at least 2 members of this finite sub-covering.

Homework Equations

I just posted the general idea of my solution. If you could let me if that's the right idea, I'm not really looking for a solution, just to know if what I posted is right/correct.

The Attempt at a Solution

We can break up B into 2 sets B1 and B2 that still cover A. We'll break them up such that every a is contained in both B1 and B2 at least once.

These B1 and B2 can be beaten down to finite subcovers because of compactness condition.

The union of these finite subcovers is a finite subcovering of A that satisfies this condition

 

[Dick]

Well, yes. That's kind of a strange question but sure, I think that works.

 

[afkguy]

Ok, thank you. I wasn't sure if it was that simple or not.

 

[LeonhardEuler]

It doesn't quite work because B1 and B2 may contain some of the same elements, so when you union them, you won't necessarily end up with two distinct sets covering every element.

 

[Dick]

It doesn't quite work because B1 and B2 may contain some of the same elements, so when you union them, you won't necessarily end up with two distinct sets covering every element.

You make an extremely good point. I didn't give this problem the thought it deserves. Sorry.

 

[LeonhardEuler]

The basic idea is right. You just need to start with one finite subcover, and then construct another that avoids copying the wrong elements. Have you tried this afkguy?

 

[LeonhardEuler]

I just posted the general idea of my solution. If you could let me if that's the right idea, I'm not really looking for a solution, just to know if what I posted is right/correct.

I'm a little confused because I don't see any solution posted by you. The solution posted by afkguy had a mistake.

 

[afkguy]

I'm a little confused. we'll break up B into B1 and B2 so that the intersection of B1 and B2 is empty, aka they have no subsets in common. If we reduce B1 and B2 into finite sets, wouldn't the intersection of the finite subcoverings of B1 and B2 still be empty?

 

[arestes]

Ricky Doherty is a troll, he's been posting copied and pasted excerpts randomly taken from posts inside a thread... weird guy...

 

[LeonhardEuler]

I'm a little confused. we'll break up B into B1 and B2 so that the intersection of B1 and B2 is empty, aka they have no subsets in common. If we reduce B1 and B2 into finite sets, wouldn't the intersection of the finite subcoverings of B1 and B2 still be empty?

If you just arbitrarily break B into B1 and B2 so that the intersection is empty, then there is no guarantee that either covers A.

 

[LeonhardEuler]

Ricky Doherty is a troll, he's been posting copied and pasted excerpts randomly taken from posts inside a thread... weird guy...

That is weird.

 

[afkguy]

Blah, I feel like I'm farther away now than closer!

So somewhere in B there are two subsets that both contain some a in A. This holds for any a.

So let's just take out just enough subsets that contain every a in A only once, call this B1.

This is still an open cover, and hence there exists some finite subcovering of B1.

If we define B2 = B - B1, every B2 still must contain a at least one more time, and they share no sets in common.

if B2 is an open cover, it has a finite subcovering as well.

so the union of these two finite subcovers contains no sets in common, but contains every A at least twice.

 

[LeonhardEuler]

So let's just take out just enough subsets that contain every a in A only once, call this B1.

This is still an open cover, and hence there exists some finite subcovering of B1.

You need to be more careful in defining new sets. You didn't prove that a set like B1 exists, and actually, it might not.

Suppose a, b, and c are elements of A. Suppose X, Y and Z are sets in the open cover B. Suppose a is in X and Y, b is in Y and Z, and c in in Z and X. How could you remove sets from B so that each of these elements is in only one set? It's impossible.

 

[LeonhardEuler]

Notice, by the way, that if a,b and c are the only elements of A, and X,Y and Z the only sets in B, then {X,Y,Z} is a finite sub-cover that meets all of the requirements, yet it is impossible to come up with sub-covers that covers each element only once like the B1 and B2 you wanted to define.

This suggests that trying to come up with a sub-cover that covers each element only once is impossible, but also unnecessary.

How about starting by generating a finite sub-cover, and then changing only the parts it is necessary to change to get your second finite sub-cover?

 

[afkguy]

Hmmm, so B is an open cover that has every a in A twice, but it doesn't mean that we can make finite subcover that contains every A only once.

So let's just take any subset of B contains every element of A, this can be ensured because B is an open cover. We'll call this B1 and it has a finite subcover.

This finite subcover of B1 may already satisify our condition, but we don't know.

If it does ok. If not, it means there were some sets we missed when we created B1 from B.

If we say B2 = B - B1, this may not be an open cover. It might even be empty. So we'll add just enough sets from B1 to B2 so that it forms another open cover of A. So we'll say B2 = (B-B1) + enough subsets of B1 to cover A. By this I mean, For the elements in A that we missed, we'll take the rest from B1. This has a finite subcovering.

If we now consider the union of these. The finite subcovering of B1 covers A at least 1 or more times. The finite subcovering of B2 covers A also but it also contains some of the same sets from B1. This is ok, because our condition still holds because this is accounted for in the subsets (B-B1).

 

[LeonhardEuler]

You're on to the right idea now, but you just need to be more specific about your definition of B2 to make sure you don't run into the duplicate problem.

I found it easier to think the other way around when defining B2: Instead of taking (B-B1) and then adding sets from B1 as needed, I started by thinking "which sets from B1 can we definitely not allow into B2?", and working from there.

(Hint: having a set X in both B1 and B2 is only a problem if there is an element 'a' of X which belongs to no other set in B1. Do you see why?)

 

[afkguy]

Well if we're constructing B2, we definitely don't want the same set twice and want to preserve our condition. So we'll stick with our definition of B1, that it is just any cover of A and is a subset of B.

We'll take B1 and look at (B-B1). B1 and (B-B1) might contain some of the same elements, but they won't contain the same sets.

We'll remove from B1 the sets that have the same elements in (B-B1), and add the sets from (B-B1) into B1. We'll call this B2. This is an open cover still and has a finite subcover.

If we now take the union of B1 and B2 there are 2 elements of A somewhere in there.

edit: nvm this doesn't work because B2 might not be an open cover.

 

[LeonhardEuler]

B1 and (B-B1) have no elements in common by definition.

 

[afkguy]

err I meant they might contain some of the same elements of A, sorry.so sort of continuing off that,

We'll take B1 and look at (B-B1). B1 and (B-B1) might contain some of the same elements of A, but they won't have any sets that are equal.

We'll remove from B1 the sets that have the same elements of A in (B-B1), and add those sets from (B-B1) into B1. We'll add back in whatever sets we need to so that this new collection of sets forms an open cover. We'll call this B2. This is an open cover still and has a finite subcover.

If we now take the union of B1 and B2 there are 2 elements of A somewhere in there.

 

[LeonhardEuler]

You're getting there, I had to play around for a while with this too, but there are still some confusing things to me. Is B1 a finite sub-cover of B? If it is, then that's how I started too, so that's on the right track.

But I'm unclear on all this adding and subtracting you're talking about, since you can't litterally add sets into B1 when it's already a fixed set. I know what you're getting at, but when you add and subtract repeatedly, it gets really confusing.

 

[afkguy]

Well, B1 as I have it set up is just some subset of B that covers A. It's not finite at this step yet.

if we look at B1 and (B-B1), then they may both have some elements of A, but they do contain the same sets.

So let's say there's some a in X that is in B1 that is also in Y in (B-B1). We'll define a new set that consists of B1 - X + Y, and do this for every a in A that we need to. We can add back in X ifwe need to do so to make this new collection of sets an open cover. We'll call this B2.

B1 and B2 are both open covers of A, and so have finite subcoverings. Their union should end up covering A twice.

I don't mean to cut you off, and I really, really appreciate your help; but I'm super tired and have a terrible headache and afraid I'm not being very effective anymore. I'll try and reply again when I wake up in a few hours.

 

[LeonhardEuler]

That's exactly the reasoning I used, so you're right on track. Now all you have to do is be a little more precise.

For starters, it is a problem that B1 is not finite yet. As I understand it, the ultimate goal is to take two finite sub-covers that don't necessarily contain two sets for every element, and union them so they do. If you define B1 and B2, and then take finite sub-covers, this problem can just crop up again at that point. This would happen in your example if you took X out of B1 to form B2 with Y instead, and then needed to put X back in. When you took a finite sub-cover from B2, it might contain X and not Y.

Secondly, you don't want to do this substitution for every a in A. Only the ones where 'a' occurs in only one set of B1.

Finally, it's OK to talk intuitively about a procedure for substituting elements, but since there are possibly an infinite number of a in A, when you write the final form, you should really phrase it more like constructing B2 by taking subsets, compliments, unions, etc.

 

[afkguy]

So keeping definition of B1, it has a finite subcover, say B1_i.


Could I just take the finite subcover of a subset of A? By this I mean

If we consider the subset of elements of A that are contained in B1_i only once (aka the rest lie somewhere in B-B1_i), then (B-B1_i) is still an open cover over this subset of A.

Is this subset of A compact though? It is a subset of a compact set and but I don't believe it needs to be closed, but I'm not sure.

If so, then (B-B1_i) has a finite subcover (over just this specific subset of A). We'll call this B2_j.

If this is true then I can just take B1_i and union it with B2_j. B2_j may not be a finite subcover of A, but it at least contains the extra elements that we need.

 

[LeonhardEuler]

You're getting extremely close. The first thing to notice is that B1 isn't doing anything that B1_i isn't already doing, so it would make sense to just drop the original definition of B1 and say it's a finite subcover of B.

Could I just take the finite subcover of a subset of A? By this I mean

If we consider the subset of elements of A that are contained in B1_i only once (aka the rest lie somewhere in B-B1_i), then (B-B1_i) is still an open cover over this subset of A.

This doesn't work since a subset of a compact set is not necessarily compact. But this is very close to the right way of going. Suppose A_once is the subset of A that occurs in only one element of B1. Suppose B1_once is the subset of B1 that contains at least one element of A_once. Look at (B-B1_once). Is this an open cover? Do you see where to go from there?