A problem in Real Analysis/Topology

[mtayab1994]

Homework Statement

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

Homework 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?

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.

 

[Dick]

Homework Statement

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

Homework 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?

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.

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 $sup(A)$?

 

[mtayab1994]

Part 1 isn't going well. 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 $sup(A)$?

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?

 

[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.

 

[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.

 

[Dick]

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?

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?

 

[Dick]

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.

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

 

[Dick]

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

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).

 

[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.

 

[Dick]

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:

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.

\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.

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?

 

[mtayab1994]

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?

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).

 

[Dick]

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).

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.

 

[pasmith]

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).

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?

 

[mtayab1994]

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.

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

 

[Dick]

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

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?

 

[mtayab1994]

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?

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

 

[Dick]

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

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

 

[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.

 

[Dick]

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.

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?

 

[mtayab1994]

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?

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?

 

[Dick]

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?

And why must it be that if $a$ is not in $A$ implies that $a$ must be in the closure of A (I think 'closure' is a common term for that than 'adherent')?

 

[mtayab1994]

And why must it be that if $a$ is not in $A$ implies that $a$ must be in the closure of A (I think 'closure' is a common term for that than 'adherent')?

Because we said that a is a limit point of A.

 

[Dick]

Because we said that a is a limit point of A.

Saying it doesn't prove it. There is a simple argument just based on the definition of 'limit point' and 'sup'.

 

[mtayab1994]

For the second question can I prove that B is nonempty by saying that B is a subset of A?

 

[vela]

No. $\emptyset \subset A$ but $\emptyset$ is certainly not non-empty.

 

[mtayab1994]

No. $\emptyset \subset A$ but $\emptyset$ is certainly not non-empty.

I can say that if B were empty than A will be bounded above and that's a contradiction because in the first question we've proved that A is not bounded above so we can conclude that B is nonempty.

 

[vela]

If B is empty, that means A contains no elements less than or equal to $t$. Why does that imply A is bounded from above?

How exactly is $t$ chosen? It seems to me you can't say much about B without knowing this.

 

[mtayab1994]

If B is empty, that means A contains no elements less than or equal to $t$. Why does that imply A is bounded from above?

How exactly is $t$ chosen? It seems to me you can't say much about B without knowing this.

I'm sorry I made a mistake in the original thread post the set B is : B={tεA : x≤t}

 

[mtayab1994]

For the second question knowing that B is nonempty and bounded below by x inf(B) exists in B. So now to prove that m=Inf(B). So we suppose that there exists another lower bound m' and we have to prove that m>m'.
Using a proof by contradiction I supposed that m'>m and I let ε=m'-m>0 .

So by definition if the infimum: There exists a t in B such that t<m+ε=m+m'-m=m'. And we reached a contradiction with m' being a lower bound of B so therefore m=inf(B).

 

[vela]

Why are you saying $m$ is inf(B)? The problem statement only says to show that $m$ bounds B from below and is greater than or equal to $x$. It's not necessarily the greatest lower bound.

 

[mtayab1994]

Why are you saying $m$ is inf(B)? The problem statement only says to show that $m$ bounds B from below and is greater than or equal to $x$. It's not necessarily the greatest lower bound.

Wow I'm really sorry that's another typo the problem statement says to show that B has a and inferior bound not just a lower bound. Hence why I proceeded like that. Is my reasoning correct?