# Thread: Prove that E is bounded and that sup E and inf E both belong to E.

1. ## Prove that E is bounded and that sup E and inf E both belong to E.

Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E.

For the first part can we use the fact that if E is compact iff it is closed and bounded? So since its compact than its closed and bounded..or is do I have to actually prove that its bounded? And also for the sup and inf, i'm not sure but would we set E=the interval and show that they are the inf and sup?

2. Originally Posted by tn11631
Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E.

For the first part can we use the fact that if E is compact iff it is closed and bounded? So since its compact than its closed and bounded..or is do I have to actually prove that its bounded? And also for the sup and inf, i'm not sure but would we set E=the interval and show that they are the inf and sup?
The first part follows from the open cover $\Omega=\left\{B_1(e)\right\}_{e\in E}$ and the second comes from the fact that those two points are limit points and $E$ is closed.

3. Hey this is what i've been working on. Can you let me know if i'm on the right track?

If E is compact then E is closed and bounded. Since E is nonempty, supE and infE both exist. To show supE∈ E, suppose not. Then supE ∈ Ec(compliment) which is an open set. Hence there exists δ > 0 such that
(supE − δ, supE + δ) ⊆ Ec(compliment). This is a contradiction because there must exist an element of E greater than
supE − δ....Then I just have to show it for inf..

4. Originally Posted by tn11631
Hey this is what i've been working on. Can you let me know if i'm on the right track?

If E is compact then E is closed and bounded. Since E is nonempty, supE and infE both exist. To show supE∈ E, suppose not. Then supE ∈ Ec(compliment) which is an open set. Hence there exists δ > 0 such that
(supE − δ, supE + δ) ⊆ Ec(compliment). This is a contradiction because there must exist an element of E greater than
supE − δ....Then I just have to show it for inf..
That's great! The only thing that worries me is that of course a compact subspace of a metric space is closed and bounded, but you have yet to prove that...I think.

5. Originally Posted by Drexel28
That's great! The only thing that worries me is that of course a compact subspace of a metric space is closed and bounded, but you have yet to prove that...I think.
well I was refering to the Heine-Borel Theorem which states Let E C R. then E is compact iff it is closed and bounded. So I thought since that was already proved as a theorem I could just state that it was closed and bounded b/c its compact. So instead of proving it I could refer to the theorem (at least my prof let us do somthing like that on one of our workshops)..

6. Originally Posted by tn11631
well I was refering to the Heine-Borel Theorem which states Let E C R. then E is compact iff it is closed and bounded. So I thought since that was already proved as a theorem I could just state that it was closed and bounded b/c its compact. So instead of proving it I could refer to the theorem (at least my prof let us do somthing like that on one of our workshops)..
Well, if you know the Heine-Borel theorem then I guess you're good.

7. Originally Posted by Drexel28
Well, if you know the Heine-Borel theorem then I guess you're good.
Sweet thanks! sorry for so many questions lol I'm just trying to organize and understand the problems I don't know on my sheet thats why I'm being so specific! thank you!

8. You don't need "Heine-Borel". Heine-Borel says that any closed and bounded subset of the real numbers is compact. You want the other way- in any metric space (so that "bounded" is defined) all compact sets are both closed and bounded.

9. Originally Posted by HallsofIvy
You don't need "Heine-Borel". Heine-Borel says that any closed and bounded subset of the real numbers is compact. You want the other way- in any metric space (so that "bounded" is defined) all compact sets are both closed and bounded.
That is true. As you well know any compact subspace of a Hausdorff space is closed and any compact subspace of a metric space is bounded.

That said, these ideas in most analysis courses are not pursued and so the Heine-Borel theorem (in some of my books) is an iff statement.

10. Originally Posted by Drexel28
That is true. As you well know any compact subspace of a Hausdorff space is closed and any compact subspace of a metric space is bounded.

That said, these ideas in most analysis courses are not pursued and so the Heine-Borel theorem (in some of my books) is an iff statement.
I hope those books note that this is only true for subsets of the real numbers with the usual topology! For example, the set of all rational numbers, whose square is less than 2, is both closed and bounded in the rational numbers but is not compact.

11. Originally Posted by HallsofIvy
I hope those books note that this is only true for subsets of the real numbers with the usual topology! For example, the set of all rational numbers, whose square is less than 2, is both closed and bounded in the rational numbers but is not compact.
I agree, you could have equally well defined a compact metric space to be one that is complete and totally bounded, but that doesn't change the fact that in a beginning analysis course there is only one "topology". They don't consider anything but the usual topology on $\mathbb{R}$