Heine-Borel Theorem: Bounded & Closed Set Covered by Finite Open Subsets

[pivoxa15]

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]

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:

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

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

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

 

[AKG]

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]

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]

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

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]

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]

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]

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]

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]

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]

"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.

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]

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]

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]

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 realized this.

 

[pivoxa15]

You could- but that's a different theorem.

Just out of interest, what is the theorem called?

 

[kaosAD]

Just out of interest, what is the theorem 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 possesses such cover with no finite (closed) subcovers. For example:

Define closed interval $I_n = [1/n, 2]$ for all $n \in \mathbb{N}$. Then the cover

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

covers a closed and bounded set $A = [0,1]$. But there is no finite closed subcovers of $C$ that can cover $A$.

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

 

[HallsofIvy]

Just out of interest, what is the theorem called?

Not every theorem has a name!

 

[pivoxa15]

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 possesses such cover with no finite (closed) subcovers. For example:

Define closed interval $I_n = [1/n, 2]$ for all $n \in \mathbb{N}$. Then the cover

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

covers a closed and bounded set $A = [0,1]$. But there is no finite closed subcovers of $C$ that can cover $A$.

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

I accepted that every subcover must be open and even indicated that it was essential in the proof of Heine-Borel.

What I observed was that Some closed cover of S can also have a finite number of open subcovers that cover S. Although this observation is quite useless and irrelevant to the Heine-Borel.

 

[HallsofIvy]

Sorry, I didn't mean to imply that there was a direct analog to Heine-Borel using "closed" covers.

In fact, since (in metric spaces at least) all singleton sets are closed, one possible "closed" cover for a set would be the collection of singleton sets for all points in the set. In other words:

Every "closed" cover for A contains a finite sub-cover if and only if A is a finite set!

And, of course, since every finite set is both closed and bounded, "closed and bounded" is necessary but not sufficient.

 

[kaosAD]

I accepted that every subcover must be open and even indicated that it was essential in the proof of Heine-Borel.

What I observed was that Some closed cover of S can also have a finite number of open subcovers that cover S. Although this observation is quite useless and irrelevant to the Heine-Borel.

I am not surprise (in fact, none of us is!) with your observation -- it is trivially true! You must have missed the most crucial point about Heine-Borel theorem that many had tried to get to you. The theorem does not argue anything about there does not exist a cover that has finite (open/closed) subcovering for the set in question. In fact, the statement is trivially not true at all, fullstop. But, it does state that every open cover must have a finite subcover for the set in question for it to be compact. You should read carefully https://www.physicsforums.com/showpost.php?p=960197&postcount=12".

Also it is useless to talk about a cover being closed or open. What is more important is its subcovers are all open. The union of arbitrary open sets is open, so it follows directly that the cover must be open. As a side note, the converse is not true.

Hope that helps.

 

[HallsofIvy]

What I observed was that Some closed cover of S can also have a finite number of open subcovers that cover S. Although this observation is quite useless and irrelevant to the Heine-Borel.

kaosAD said this is trivially true. I'm wondering what you MEAN by it!

A "closed cover", A, of S means (in analogy with "open cover") a collection of closed sets such that every point of S is in at least one of the closed sets in A. What could you possibly mean by "open subcover"? There are no open sets in in A so there cannot be any collection of open sets in A that cover S.

 

[kaosAD]

kaosAD said this is trivially true. I'm wondering what you MEAN by it!

A "closed cover", A, of S means (in analogy with "open cover") a collection of closed sets such that every point of S is in at least one of the closed sets in A. What could you possibly mean by "open subcover"? There are no open sets in in A so there cannot be any collection of open sets in A that cover S.

It was my fault, HallsofIvy, for being ignorant. I could have pointed out his mistake about "open subcover". I took that as a typo, as I did not wish to argue much on that matter (I wished to go straight to the main point). Anyway I did mention the union of arbitrary open sets is open in passing (so I was not at all that ignorant) ;)

 

[pivoxa15]

kaosAD said this is trivially true. I'm wondering what you MEAN by it!

A "closed cover", A, of S means (in analogy with "open cover") a collection of closed sets such that every point of S is in at least one of the closed sets in A. What could you possibly mean by "open subcover"? There are no open sets in in A so there cannot be any collection of open sets in A that cover S.

a) If you can argue that the set {(-1,2)} covers the closed and bounded set S=[0,1]. Moreover, there is a finite number of open sets in {(-1,2)} that cover S, namely {(-1,2)}.

b) Than I could state that the set {[-1,2], (-1,2)} covers the closed and bounded set S=[0,1]. Moreover, there is a finite number of open (sub)sets in {[-1,2], (-1,2)}, namely {(-1,2)} that also cover S. Would you call the set {[-1,2], (-1,2)} a closed or open cover of S?
If closed (because [-1,2] is larger and covers (-1,2) hence their union is closed) than it is the case that some closed covers of S has a finite number of open subcover(s).

I hope what I have stated is correct. If it is correct than I agree it is a trivial point. I understand that the main point of the Heine-Borel theorem is that Every open cover of S has a finite subcover <=> S is a closed and bounded set.

 

[kaosAD]

Let me clarify. An open cover for a set $$A$$ is a collection of open sets whose union contains $$A$$. Similar definition goes for closed cover, as noted by HallsofIvy. If you stick to this definition, then it resolves your question b).

With that in mind, and to avoid confusion, don't use "open/closed cover" to mean "cover being open/closed (in union sense)". The following shows they are not the same.

To find out if an open cover is open or closed (in union sense), we use the following argument. Since open cover is a collection of open sets and that the union of arbitrary open sets is open. Then open cover is open [in union sense]. A closed cover, on the other hand, is not necessarily closed [in union sense].

In summary, you cannot take a cover being open (in union sense) to imply open cover and closed cover to imply cover being closed (in union sense).

Note:
1) "Finite number of open subcover(s)" and "Finite open subcover(s)" mean different things.
2) I spot a typo in https://www.physicsforums.com/showpost.php?p=961214&postcount=20". What I meant to write was - What is more important is its elements are all open.

 

[pivoxa15]

Note:
1) "Finite number of open subcover(s)" and "Finite open subcover(s)" mean different things.
2) I spot a typo in my previous post. What I meant to write was - What is more important is its elements are all open.

I take it that the elements you were referring to are open subsets of R.

Does the difference lie in the fact that "Finite number of open subcover(s)" mean a finite number of subsets of the original set that can cover S with each subset not necessilary containing finitely number of open subsets of R.

And "Finite open subcover(s)" mean one or more subset of the original set that can cover S with a finite number of elements or open subsets of R.

 

[loopgrav]

I just noticed this thread, and it seems really silly to me that it has gone on so long. So as a new observer, let me summarize what I think the essential points are from a fresh point of view.

First of all, in any metric space (and hence in an arbitrary topological space) the union of an arbitrary collection of open sets is open, and the intersection of a finite collection of open sets is open. By deMorgan's theorems (the complement of a union is the intersection of the complements etc) and the fact that a set is defined to be closed if its complement is open, it follows that the intersection of an arbitrary collection of closed sets is closed, and the union of a finite collection of closed sets is closed. Note that the union of the (infinite) collection of closed subsets of R given by [-1+1/n, 1-1/n] is equal to the open set (-1, 1), but no finite subcollection will cover this open interval. Similarly, the intersection of the infinite collection of open sets (-1/n, 1/n) is the closed set {0}.

As pointed out by someone else above, a set A is defined to be compact if EVERY open cover of A has a finite subcover. In other words, for ANY collection {A_i} of open sets that covers a set A (ie, for every x in A there exists an A_j such that x is in A_j) there must be a finite number of the A_i that also covers A. Thus we see that (-1, 1) is not compact.

One last needed definition. First, let a, b be points in R^n (ie, a and b are each n-tuples of numbers a^i and b^i for i=1, ..., n) and assume that a^i < b^i for each i. Then the set of all points x in R^n such that a^i is less than or equal to x^i is less than or equal to b^i is called an n-cell (or an n-dimensional rectangle). We say that a subset A of R^n is bounded if it can be enclosed in an n-cell.

Finally, the Heine-Borel theorem says that a subset A of R^n is compact if and only if it is closed and bounded.

I think this is a finite subcover of all of the important things covered in the posts above.

 

[pivoxa15]

Let me clarify. An open cover for a set $$A$$ is a collection of open sets whose union contains $$A$$. Similar definition goes for closed cover, as noted by HallsofIvy. If you stick to this definition, then it resolves your question b).

With that in mind, and to avoid confusion, don't use "open/closed cover" to mean "cover being open/closed (in union sense)". The following shows they are not the same.

To find out if an open cover is open or closed (in union sense), we use the following argument. Since open cover is a collection of open sets and that the union of arbitrary open sets is open. Then open cover is open [in union sense]. A closed cover, on the other hand, is not necessarily closed [in union sense].

In summary, you cannot take a cover being open (in union sense) to imply open cover and closed cover to imply cover being closed (in union sense).

Note:
1) "Finite number of open subcover(s)" and "Finite open subcover(s)" mean different things.
2) I spot a typo in https://www.physicsforums.com/showpost.php?p=961214&postcount=20". What I meant to write was - What is more important is its elements are all open.

Are you saying
'open cover <=> collection of only open sets'
'closed cover <=> collection of only closed sets'

What happens if you have a combination of open and closed sets in your cover? What would the cover be called? Would it be called 'neither open nor closed cover'?

You said 'the union of arbitrary open sets is open'
What happens if you have an infinite number of open sets?

 

[pivoxa15]

I just noticed this thread, and it seems really silly to me that it has gone on so long. So as a new observer, let me summarize what I think the essential points are from a fresh point of view.

First of all, in any metric space (and hence in an arbitrary topological space) the union of an arbitrary collection of open sets is open, and the intersection of a finite collection of open sets is open. By deMorgan's theorems (the complement of a union is the intersection of the complements etc) and the fact that a set is defined to be closed if its complement is open, it follows that the intersection of an arbitrary collection of closed sets is closed, and the union of a finite collection of closed sets is closed. Note that the union of the (infinite) collection of closed subsets of R given by [-1+1/n, 1-1/n] is equal to the open set (-1, 1), but no finite subcollection will cover this open interval. Similarly, the intersection of the infinite collection of open sets (-1/n, 1/n) is the closed set {0}.

I take it that whenever you write arbitrary, it includes an infinite quantity.

How can you justify that the union of an infinite number of open sets is open only? i.e. {(-n,n)} where n an element of the integers. The complement of this set is the empty set which is clopen or open and closed. So {(-n,n)} can be closed as well. Therefore it is clopen.

What about your claim that the intersection of an arbitrary number of closed sets is closed only? i.e. {[-n,n]} where n an element of the integes. The intesection of this set is again the empty set which is clopen. So the intersection is not just closed but open as well so it is clopen.

I like to clear these questions first before talking about the Heine-Borel theorem.

 

[loopgrav]

An arbitrary (ie possibly infinite) union is easily seen to be open, and it may very well also be closed (for example, if it is equal to the whole space under consideration), but that is irrelevant to this discussion.

As to the closed sets [-n, n] for n=0, 1, 2, ... the intersection is {0} which is not empty-- it is closed.

I think you need to do some thinking on your own, study something good and readable like Simmons' "Introduction to Topology and Modern Analysis," Gemignani's "Elementary Topology" or the appendices in Broida & Williamson's "A Comprehensive Introduction to Linear Algebra", or maybe find a new subject to interest you.

 

[matt grime]

I take it that whenever you write arbitrary, it includes an infinite quantity.

correct.

How can you justify that the union of an infinite number of open sets is open only?

because it is easy to prove from the definitions.

i.e. {(-n,n)} where n an element of the integers. The complement of this set is the empty set which is clopen or open and closed. So {(-n,n)} can be closed as well. Therefore it is clopen.

what has this to do with anything?

What about your claim that the intersection of an arbitrary number of closed sets is closed only? i.e. {[-n,n]} where n an element of the integes. The intesection of this set is again the empty set which is clopen. So the intersection is not just closed but open as well so it is clopen.

The intersection of those set is the set {0} which is closed and is not the empty set, if we allow 0 te be a natural number. If we do not then the intersection is [-1,1], which is also closed. Even if it were the empty set, which it is not, that does not make the assertion that the intersection is closed false, does it? If something is both closed and open it is certainly closed. This is not an 'exclusive or' property.
The union of any collection of open sets WILL be open, where as the union of some collection of closed sets might be open, closed, both or neither.

The intersection of a finite number of open sets WILL be open, the intersection of an infinite number of closed sets WILL be closed.

The intersection of an infinite number of open sets might be open, closed, neither or both.