# E compact and nonempty

1. Nov 18, 2008

### kathrynag

1. The problem statement, all variables and given/known data
Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E.

2. Relevant equations

3. The attempt at a solution
E is compact, so for every family{$$G_{\alpha}$$}$$_{\alpha\in}A$$ of open sets such that E$$\subset$$$$\cup_{\alpha\in}A$$$$G_{\alpha}$$, there is a finite set{a1,a2...,an}$$\subset$$A such that E$$\subset$$$$\cup^{n}_{i=1}$$$$G_{\alpha_{i}}$$

2. Nov 18, 2008

### Pere Callahan

I assume you talk about E as a subset of the real numbers?
This is not an attempt at a solution but the mere definition of compactness.

Start with the easiest part of the question, that E is bounded. By the compactness of E it can be covered by a finite subset of any open cover. Can you always find an open cover which contains only bounded sets? What does it then mean for E to be a subset of the union of a finite sub-collection of this open cover?

3. Nov 18, 2008

### kathrynag

Yes, E is a subset of the real numbers.
An open cover will have bounded sets if there is a finite subcover?
If E is a subset of the finite collection, then E will be bounded and have a sup and inf?

4. Nov 18, 2008

### HallsofIvy

Staff Emeritus
Have you already proved that every compact set, in a metric space, is closed and bounded?

5. Nov 18, 2008

### kathrynag

Oh yeah I forgot about that by the Heine Borel Theorem.

6. Nov 18, 2008

### kathrynag

So that takes care of E being bounded, so now I need to show that there is sup E and inf E. Does this follow from E being bounded?

7. Nov 18, 2008

### HallsofIvy

Staff Emeritus
No, the Heine Borel theorem says any closed and bounded set is compact. The fact that every compact set is both closed and bounded is true but is not the "Heine Borel" theorem. For one thing, the Heine Borel theorem is true only in the real numbers or products of the real numbers (and so not in the rational numbers) but a compact set is both closed and bounded for all metric spaces.

Okay, E is bounded. What is the "least upper bound property" for real numbers?

8. Nov 18, 2008

### kathrynag

A set is bounded if it is bounded from both above and below.
A real number a is a least upper bound of for S if a is a upper bound for S having the property that if b is also an upper bound for S, then a<b.

9. Nov 18, 2008

### kathrynag

If there is a bounded, there will be some $$x_{1}$$<$$x_{2}$$. therefore a sup E?

10. Nov 18, 2008

### Pere Callahan

What do you mean by x1 and x2?
And what is the relationship betweeen boundedness of sets and least upper bounds?

11. Nov 18, 2008

### kathrynag

I was meaning there would be a set containing x1,x2,x3,...xn.
Well, a set is bounded from above and blow. A set must be bounded from above to have a lub, so E must have a lub?

12. Nov 18, 2008

### Pere Callahan

The least upper bound property states that every bounded set E has a least upper bound, called the supremum, denoted sup E.
You want to show that if E is compact (and thus bounded), the supremum of E actually belongs to E. How would you do that. Have you already proved that every compact set (in the reals) is closed?
If so, how does from the set E being closed follow that sup E is in E?

13. Nov 18, 2008

### kathrynag

Well, every compact set is closed and bounded. Then sup E is in E because E is a bounded set.

14. Nov 18, 2008

### Pere Callahan

No, A= (0,1) is also bounded but sup A = 1 is not an element of (0,1). $\sup E\in E$ must follow from the closedness and if you have not proved it in class yet, you will have to think of an argument yourself.

What is the definition of open and closed?

15. Nov 18, 2008

### kathrynag

Well, we did a proof or the heine Borel Theorem.
A set is closed iff every accumulation point of E belongs to E.

16. Nov 18, 2008

### Pere Callahan

So you would have to think about whether or not sup E is an accumulation point of E

17. Nov 18, 2008

### kathrynag

Ok, I would think it is because a neighborhood of sup E will contain infinitely many points of E.

18. Nov 19, 2008

### Pere Callahan

That's the definition of sup E being an accumulation point. How would you show that any neighbourhood of sup E contains infintely many elements of E?

Sorry for being picky but I want to make sure you're not just quoting lines from your notes

19. Nov 19, 2008

### HallsofIvy

Staff Emeritus
That is the definition of "least upper bound"= lub= sup. I asked if you knew the "least upper bound property" and apparently you don't. It is "every set of real numbers, having an upper bound, must have a least upper bound". And, of course the same is true of lower bound and "greatest lower bound".

But you still haven't said what x1, x2, x3, ..., xn are so that makes no sense!

Yes, but more to the point, if a set is bounded from above, it has a lub. You have the logic reversed.

No, sup E exists because E is bounded. (0, 1) has upper bound 2. What is its lub? Is that in (0,1)? The lub is in E here because of another property of E.