Heine-Borel Theorem

**pivoxa15** · Apr8-06, 10:22 PM

Why does the set have to be (bounded) and closed in order for there being finitely many open subsets that can completely cover it?

My question is concerned with the closed aspect (I know why it has to be bounded). So why can't a bounded and open set be able to be covered by finitely many open subsets?

**Rach3** · Apr8-06, 10:40 PM

Keep in mind - you're talking about a finite subcover of any arbitrary open cover!

An example, of a non-closed bounded set in R which is not compact:

$LaTeX Code: S=\\{1,\\frac{1}{2},\\frac{1}{4},\\frac{1}{8}...\\}$

and this particular open cover:
$LaTeX Code: O=\\{(1-\\frac{1}{4},1+\\frac{1}{4}),(\\frac{1}{2}-\\frac{1}{8},\\frac{1}{2}+\\frac{1}{8}),...\\}$

(So that the ith open interval contains only the ith element in S, and no other element of S)

**AKG** · Apr9-06, 01:58 AM

Every set can be covered by finitely many open sets. In fact, every set can be covered by a single open set, that open set being the entire space. Compactness says that given any open covering, there is a finite subcovering. Both (0,1) and [0,1] have {(-1,2)} as an open covering. And in both cases, both of them can take a finite subcovering from this covering. But I could give you a much "nastier" open covering of (0,1) for which you won't be able to pick a finite covering. On the other hand, no matter what nasty open covering I try to think up for [0,1], you can always pick a finite number of sets from that covering that will cover [0,1].

**pivoxa15** · Apr9-06, 02:27 AM

The version of the Heine-Borel Theorem as I understand it is:
Let S be closed and bounded subset of R. Then S is a subset of (or can be covered by) finitely many open subsets of R.

Could you relate the above to 'every open cover of S has a finite subcover'.
What is 'a finite subcover'? One cover is always finite.

**Euclid** · Apr9-06, 02:46 AM

A cover of a set E by open sets is a collection $LaTeX Code: \\mathcal{C} = \\{O_\\alpha\\}_{\\alpha \\in J\\}$ such that $LaTeX Code: E \\subset \\bigcup\\limits_{\\alpha \\in J} O_\\alpha$ .
A subcover of $LaTeX Code: \\mathcal{C}$ is any subset of $LaTeX Code: \\mathcal{C}$ . A finite subcover is simply a finite subset of $LaTeX Code: \\mathcal{C}$ .

Edit: The subcover must of course still cover E (i.e., E is contained in the union).

As AKG points out, any set can be covered by finitely many open sets, so the version of the Heine-Borel theorem you are talking of is clearly not the same one as "every open cover has a finite subcover".

**matt grime** · Apr9-06, 05:58 AM

Originally Posted by pivoxa15

The version of the Heine-Borel Theorem as I understand it is:
Let S be closed and bounded subset of R. Then S is a subset of (or can be covered by) finitely many open subsets of R.

Then you do not understand the Heine-Borel theorem. Indeed that 'theorem' you state is meaningless. Any subset of any topological space can be coverered by a finite number of open sets, namely 1: the whole space.

**pivoxa15** · Apr9-06, 08:16 AM

I did not fully understand the Heine-Borel theorem. Now I understand that it is, For any open cover of S there exist a finite number of open subcovers that also completely cover S.

I have also understood why S must be closed and bounded in order to satisfy the Heine-Borel property.

But why must the (first) cover of S be open? I understand why each subcover must be open.

**matt grime** · Apr9-06, 08:21 AM

That question doesn't make any sense. It doesn't make any sense to state you understand why one is open and the other not. They are after all just hypotheses.

**pivoxa15** · Apr9-06, 07:36 PM

Originally Posted by matt grime

That question doesn't make any sense. It doesn't make any sense to state you understand why one is open and the other not. They are after all just hypotheses.

In the version of the proof I have, it used the fact that all the elements in the subcover are open. But the fact that the cover was open was not mentioned or used.

I could have a set S=[0,1] than I could cover it by the set [-1,2]. I could than take out a finite number of open subcovers of [-1,2] that completely cover S.

But than with the same set S=[0,1], I could cover it with [0,1] but I can't create a finite number of open subcovers of [0,1] that also cover S. So in this sense, one can't claim that Every closed cover of S has a finite subcover. Instead Every open cover of S has a finite subcover. But its not false that Some closed cover of S has a finite subcover.

**Muzza** · Apr10-06, 03:39 AM

In the version of the proof I have, it used the fact that all the elements in the subcover are open. But the fact that the cover was open was not mentioned or used.

"The covering is open" <=> "all elements in the covering are open".

I could have a set S=[0,1] than I could cover it by the set [-1,2]. I could than take out a finite number of open subcovers of [-1,2] that completely cover S.

[-1, 2] is not a covering of [0, 1].

{[-1, 2]} is a covering of [0, 1].

{[-1, 2]} is not an open covering of [0, 1] (so Heine-Borel doesn't apply anyway).

{(-1, 2)} is an open covering of [0, 1], and indeed, there is a subcovering of {(-1, 2)} which also covers [0, 1] - namely {(-1, 2)} itself!

But than with the same set S=[0,1], I could cover it with [0,1] but I can't create a finite number of open subcovers of [0,1] that also cover S.

C := {[0, 1]} covers S and there IS a finite subcover of C which also covers S (C is itself a finite covering). But still, I don't see your point - C isn't an open covering...

**pivoxa15** · Apr10-06, 04:46 AM

Originally Posted by Muzza

"The covering is open" <=> "all elements in the covering are open".

[-1, 2] is not a covering of [0, 1].

{[-1, 2]} is a covering of [0, 1].

{[-1, 2]} is not an open covering of [0, 1] (so Heine-Borel doesn't apply anyway).

{(-1, 2)} is an open covering of [0, 1], and indeed, there is a subcovering of {(-1, 2)} which also covers [0, 1] - namely {(-1, 2)} itself!

I understand that Heine-Borel applies to open coverings only. My question is why not include some closed coverings since some closed covers also has a finite subcover (that cover S)?

Thank you for pointing out the errors in my previous post.

Originally Posted by Muzza

C := {[0, 1]} covers S and there IS a finite subcover of C which also covers S (C is itself a finite covering). But still, I don't see your point - C isn't an open covering...

My point is that in this example, any open subcover of C will not cover S. The subcover {[0,1]} will cover S as you pointed out but it is not an open subcover. Therefore not every closed cover of S will have a finite (open) subcover. But some closed cover of S will have a finite (open) subcover shown from the examples I gave. As some people point out, it is a pretty irrelavant thing (concerning the Heine-Borel theorem) but an observation nonetheless.

**HallsofIvy** · Apr10-06, 06:45 AM

Originally Posted by pivoxa15

The version of the Heine-Borel Theorem as I understand it is:
Let S be closed and bounded subset of R. Then S is a subset of (or can be covered by) finitely many open subsets of R.

Could you relate the above to 'every open cover of S has a finite subcover'.
What is 'a finite subcover'? One cover is always finite.

Then your "understanding" is completely wrong. The simplest wording of the Heine-Borel theorem is that a set in R (more generally Rⁿ) is compact if and only if it is both closed and bounded.

"Compact" means: every open cover contains a finite subcover. It is NOT just that it can be covered by finitely many open subsets of R (as pointed out above, R itself is an open set that covers every subset: every subset of R can be covered by one open set). What you want is that if you have any covering of a set, A, by open sets then some finite subset of THAT covering will cover A.

**HallsofIvy** · Apr10-06, 06:45 AM

Originally Posted by pivoxa15

I understand that Heine-Borel applies to open coverings only. My question is why not include some closed coverings since some closed covers also has a finite subcover (that cover S)?

Thank you for pointing out all the errors in my previous post.

You could- but that's a different theorem.

**pivoxa15** · Apr10-06, 08:07 AM

Originally Posted by HallsofIvy

Then your "understanding" is completely wrong. The simplest wording of the Heine-Borel theorem is that a set in R (more generally Rⁿ) is compact if and only if it is both closed and bounded.

"Compact" means: every open cover contains a finite subcover. It is NOT just that it can be covered by finitely many open subsets of R (as pointed out above, R itself is an open set that covers every subset: every subset of R can be covered by one open set). What you want is that if you have any covering of a set, A, by open sets then some finite subset of THAT covering will cover A.

I have already realised this.

**pivoxa15** · Apr10-06, 08:08 AM

Originally Posted by HallsofIvy

You could- but that's a different theorem.

Just out of interest, what is the theorm called?

**kaosAD** · Apr10-06, 10:12 AM

Originally Posted by pivoxa15

Just out of interest, what is the theorm called?

No such theorem exists. If it does, it is probably useless. If we replaced open subcovers with closed subcovers, then even closed and bounded set can possess such cover with no finite (closed) subcovers. For example:

Define closed interval

for all $LaTeX Code: n \\in \\mathbb{N}$ . Then the cover

$LaTeX Code: C = \\left(\\bigcup_{i \\in \\mathbb{N}} I_i \\right) \\cup [-1,0]$

covers a closed and bounded set

. But there is no finite closed subcovers of

that can cover

.

In other words, closed subcovers just ruined what Heine-Borel Theorem suppose to prove. That is why you never see closed subcovers being discussed.