# A problem in Real Analysis/Topology

1. Dec 8, 2013

### mtayab1994

1. The problem statement, all variables and given/known data

Let A be a non empty part of ℝ, such that A and A complement are two open subsets of ℝ.

2. Relevant equations

1- Prove that A is not bounded above.
2- Assuming that A complement is non-empty and let x in A complement.
and let B={xεA such that x≤t}. Prove that B is non empty and has a lower bound m such that m≥x.
3- Prove that: $$m\notin A$$ and $$m\notin A^{c}$$ . Conclude that A=ℝ.
4- Let F be non-empty set in ℝ such that F and F complementary are closed in R. What can we say about the set F?

3. The attempt at a solution

1- A is an open subset of R so we're going to use a proof by contradiction.

Suppose that a=sup(A) so for all x in A : x≤a . But since A is open then there exists an ε>0 such that: $$]a-\epsilon,a+\varepsilon[\cap A\neq\varnothing$$ so that means there is another number say a+ε/2 that's also in A so therefore A is unbounded above.

2- B is a subset of A so there for B is nonempty. Second part I know I have to conduct a proof by contradiction but I don't know how to start . Well I can suppose that B doesn't have a lower bound.

Any help will be appreciated. Thank you very much.

2. Dec 8, 2013

### Dick

Part 1 isn't going well. Some of the parts are there but the reasoning is all muddled. The idea is to show that $a=sup(A)$ CANNOT be an element of $A$. Suppose it was? Then there's a neighborhood of $a$ that is contained in $A$. Can you see how that would create a problem with $a$ being $sup(A)$?

Last edited: Dec 8, 2013
3. Dec 8, 2013

### mtayab1994

If there is a neighborhood of a that is contained in A then there is another element other than a that's also in A so therefore a cannot be Sup(A). Is that right?

4. Dec 8, 2013

### HallsofIvy

This is only true because R is a connected set. At some point you are going to have to use the connectedness of R.

5. Dec 8, 2013

### mtayab1994

Or for number 1 can I do this:

Let N be any natural number and define V={a: lal>N} a neighborhood of a at infinity.
Thus, there exists some a in A such that A is in V.
Hence, there exists some a in A such that |x|>N.
Therefore, A is unbounded.

6. Dec 8, 2013

### Dick

There being another element besides $a$ in A does not contradict $a$ being a sup. How can it? If you could show there is an element $b$ in A such that a<b, that would contradict $a$ being a sup. Can you do that?

7. Dec 8, 2013

### Dick

No, stick to the original approach. There's no such thing as "a neighborhood of a at infinity".

8. Dec 8, 2013

### Dick

That's a good point. But I think this is a proof of the connectedness of R. Can't really use that R is connected in the proof. I think the important property of R that's not being explicitly stated is that if A is bounded above then there is such a real number as sup(A).

9. Dec 8, 2013

### mtayab1994

Okay A is open and we supposed that a=sup(A) so there exist an ε>0 such that:

$$]a-\epsilon,a+\epsilon[\subset A$$ so:

$$\exists b\neq a\in]a-\epsilon,a+\epsilon[\subset A$$

Therefor b is in the neighborhood of a and in the subset A. So for b=a+ε/2 we still have b in the neighborhood of a and in A so there for a is not the supremum of A and finally A is unbounded above.

10. Dec 8, 2013

### Dick

There is such an ε because we ASSUME that $a=sup(A)$ is an element of A when A is bounded above. That's an assumption. We may have to reconsider it.

Yes, a<b=a+ε/2 is in the neighborhood saying $a$ is not the supremum. That's a contradiction, because we defined $a=sup(A)$. When you reach a contradiction you have to go back and look at the last thing you assumed. That must be wrong. What is it?

11. Dec 8, 2013

### mtayab1994

Since we found a b>a in the neighborhood of a, and since we had assumed that a=Sup(A) we have reached a contradiction with a=Sup(A).

12. Dec 8, 2013

### Dick

You are finding the wrong contradiction at this point in the proof. I'll give you an example, take the open interval A=(0,1). a=sup(A)=1. So far, there's no contradiction. What assumption really created the contradiction. Look back. It's an assumption you keep failing to mention.

13. Dec 8, 2013

### pasmith

You've shown that if $\sup A \in A$ then $A$ cannot be open. But you're given that $A$ is open.

So either $\sup A \in A^c$ or $A$ is not bounded above, which is what you are asked to prove.

So how do you exclude the possibility that $\sup A \in A^c$, given that $A^c$ is open?

14. Dec 8, 2013

### mtayab1994

We assumed that a is in A so our contradiction is with a in A right?

15. Dec 8, 2013

### Dick

Yes, your conclusion is that $a$ is not in $A$. Therefore $a$ must be in A complement, which is also open by assumption. Now what?

16. Dec 8, 2013

### mtayab1994

If a is not in A then if it's in A complement, that means that a is a limit point of A right?

17. Dec 8, 2013

### Dick

I'll agree with that if you can give me reason.

18. Dec 8, 2013

### mtayab1994

a is a limit point of A because:

$$\exists b\neq a\in]a-\epsilon,a+\epsilon[\cap A$$ by the definition of a limit point. b is what found before to be greater than a.

19. Dec 8, 2013

### Dick

That doesn't make any sense to me as a reason. $a$ is a limit point of $A$ just because $a=sup(A)$. How would being the sup imply that it's a limit point?

20. Dec 8, 2013

### mtayab1994

A complement is open and we know that if a is not in A then it's in the adherent of A which is always a closed set because it contains all of the limit points of A?