Real Analysis/Topology problem

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

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 is not in A nor in A complement.

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?

My attempt:

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-ε,a+ε[nA is nonempty so therefore there exist another number a+ε/2>a so therefore A is unbounded.

Thank you for any help before hand.

Re: Real Analysis/Topology problem

Quote:

Originally Posted by

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

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 is not in A nor in A complement.

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?

**This is a trick question** (or else whoever wrote it knows really little).

The only non-empty open subset of $\displaystyle \mathbb{R}$ having its complement open is $\displaystyle \mathbb{R}$ itself.

This easily seen by noting that $\displaystyle \mathbb{R}$ is a connected set.

Re: Real Analysis/Topology problem

Yes I believe the goal of this exercise is to get to A=R. I forgot to add to the third question that you have to conclude that A=R.