# Heine-Borel Theorem

1. Apr 8, 2006

### 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?

2. Apr 8, 2006

### 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)

Last edited by a moderator: Apr 8, 2006
3. Apr 9, 2006

### 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].

4. Apr 9, 2006

### 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.

5. Apr 9, 2006

### 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".

Last edited: Apr 9, 2006
6. Apr 9, 2006

### matt grime

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.

7. Apr 9, 2006

### 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.

8. Apr 9, 2006

### 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.

9. Apr 9, 2006

### pivoxa15

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.

10. Apr 10, 2006

### 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!

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...

Last edited: Apr 10, 2006
11. Apr 10, 2006

### 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 the errors in my previous post.

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.

Last edited: Apr 10, 2006
12. Apr 10, 2006

### 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.

13. Apr 10, 2006

### HallsofIvy

You could- but that's a different theorem.

14. Apr 10, 2006

### pivoxa15

15. Apr 10, 2006

### pivoxa15

Just out of interest, what is the theorm called?

16. Apr 10, 2006

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 $I_n = [1/n, 2]$ 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.

Last edited: Apr 10, 2006
17. Apr 10, 2006

### HallsofIvy

Not every theorem has a name!

18. Apr 10, 2006

### pivoxa15

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.

19. Apr 11, 2006

### 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.

Last edited by a moderator: Apr 11, 2006
20. Apr 11, 2006

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.

Last edited by a moderator: Apr 22, 2017
21. Apr 11, 2006

### HallsofIvy

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.

22. Apr 11, 2006

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) ;)

Last edited: Apr 11, 2006
23. Apr 12, 2006

### pivoxa15

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.

Last edited: Apr 12, 2006
24. Apr 14, 2006

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.

Last edited by a moderator: Apr 22, 2017
25. Apr 15, 2006

### pivoxa15

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.