# Intro Topology - Proving these statements

• Sep 30th 2010, 03:05 AM
AKTilted
Intro Topology - Proving these statements
Hi, I'm in an intro topology course and need help proving these statements using balls.

1) Prove for any set S that Interior(S) = Complement of (Closure of S Complement)

2) Prove that if A ⊆ B then Interior(A) ⊆ Interior(B)

3) Prove that if A ⊆ B then Closure(B Complement) ⊆ Closure(A Complement)

Thanks a lot!
• Sep 30th 2010, 03:25 AM
HappyJoe
It sounds like you're working with metric spaces, right?

Here's a quick answer, let me know, if anything seems strange.

Let me postpone part 1).

As for part 2), take a point x in Interior(A). By definition of the interior, there exists an r>0, such that the open ball B_r(x) of radius r and center x is contained in A. But A is assumed to be inside of B, hence B_r(x) is contained in B. But this means that x is an interior point of B - hence Interior(A) is a subset of Interior(B).

Part 3) follows from part 1) and 2). Indeed, we have from part 1) that Complement(Interior(A)) = Closure(Complement(A)), and similarly for B. From 2), it follows that Complement(Interior(A)) contains Complement(Interior(B)), and hence that Closure(Complement(A)) contains Closure(Complement(B)), as wanted.

As for part 1), take a point in Interior(S), and take an r>0, such that the ball B_r(x) is entirely contained in S. Then the point x is _not_ an element of Closure(Complement(S)), because otherwise the ball B_r(x) would intersect Complement(S) (by definition of the closure), but this ball is entirely contained in S. Hence x is an element of Complement(Closure(Complement(S))).

For the other direction, let x be a point of Complement(Closure(Complement(S))). Then x is not contained in Closure(Complement(S)), and hence there exists an r>0, such that the ball B_r(x) does not intersect Complement(S). But then this ball is entirely contained in S, whereas x is an interior point of S, that is x is an element of Interior(S).