# Relatively open, closed, disconnection

1. Jun 13, 2007

### redyelloworange

The problem has 2 parts. For the first part, I've got a solution that I'd like to confirm. I'm stuck on the second part.

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

PART 1
Show that if (A,B) is a disconnection of S, then A and B are both relatively open and relatively closed in S.

PART 2
Show that if A is a proper subset of S and A is both relatively open and relatively closed, then (A, A-S) is a disconnection of S.

In essence, show that A ∩ closure(S-A) = Ø and closure(A) ∩ S-A = Ø
and that A, A-S ≠ Ø

2. Relevant equations

Relatively open: A is relatively open in S if there exists an open set U in R^n s.t.
U ∩ Rn = A

(A,B) disconnection implies A ∩ closure(B) = Ø and closure(A) ∩ B = Ø

3. The attempt at a solution

PART 1
1) Show that A is relatively open in S:
closure(B) is closed, so closure(B)c is open.
Choose U = closure(B)c
Show this set intersected with S equals A.
Well, S = A U B and closure(B) ∩ A = Ø
Thus all the points that are not in closure(B) are in A, so closure(B)c = A
And S ∩ U = S ∩ closure(B)c = S ∩ A = A
Thus, A is relatively open in B.
2) by the same procedure, show that B is relatively open in S
3) A, B are both relatively open in S. But A ∩ closure(B) = Ø so if A is relatively open, B is relatively closed. And since closure(A) ∩ B = Ø A is relatively closed. Therefore A and B are both relatively open and relatively closed in S.

PART 2

A ≠ S and A ≠ Ø, so S-A ≠ S and S-A ≠ Ø
We know A U S- A = S.

A is relatively open in S.
This implies that there exists some open set U in Rn s.t. S ∩ U = A

Let this set U be closure(S-A)c. We know this is open because closure(S-A) is closed by definition.

We want to show that U = A. We see this because

I'm not sure how to bring up the intersection of the closure of A with S-A.

That's everything for now. Thanks for your help! =)

Last edited: Jun 13, 2007