# 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