# Closed and Open Sets in R^d

• Sep 5th 2011, 12:54 AM
xsavedkt
Closed and Open Sets in R^d
This is the question:
Let A be an open set and B a closed set. If B ⊂ A, prove that A \ B is
open. If A ⊂ B, prove that B \ A is closed.

Right before this we have a theorem stated as below:
In R^d,
(a) the union of an arbitrary collection of open sets is open;
(b) the intersection of any finite collection of open sets is open;
(c) the intersection of an arbitrary collection of closed sets is closed;
(d) the union of any finite collection of closed sets is closed.

So in each case, I think we can just say A\B=A∩B^c (B^c means B complement), and since both A and B^c are open (by assumption and then by definition of a closed set where if B is closed, B-complement is open) we use part (b) of the previous theorem, A\B is open. But in this proof, I didn't use the assumption that B⊂A, so I know it isn't right. Same goes for the second part of the question. PLEASE HELP!!!!!
• Sep 5th 2011, 04:11 AM
Plato
Re: Closed and Open Sets in R^d
Quote:

Originally Posted by xsavedkt
This is the question:
Let A be an open set and B a closed set. If B ⊂ A, prove that A \ B is open. If A ⊂ B, prove that B \ A is closed.
Right before this we have a theorem stated as below:
In R^d,
(a) the union of an arbitrary collection of open sets is open;
(b) the intersection of any finite collection of open sets is open;
(c) the intersection of an arbitrary collection of closed sets is closed;
(d) the union of any finite collection of closed sets is closed.

So in each case, I think we can just say A\B=A∩B^c (B^c means B complement), and since both A and B^c are open (by assumption and then by definition of a closed set where if B is closed, B-complement is open) we use part (b) of the previous theorem, A\B is open. But in this proof, I didn't use the assumption that B⊂A, so I know it isn't right. Same goes for the second part of the question.

Actually, what you did is correct!
The subset condition is irrelevant.
This true because \$\displaystyle R^n \;\& \,\emptyset \$ are both open and closed.
• Sep 5th 2011, 12:54 PM
xsavedkt
Re: Closed and Open Sets in R^d
Quote:

Originally Posted by Plato
Actually, what you did is correct!
The subset condition is irrelevant.
This true because \$\displaystyle R^n \;\& \,\emptyset \$ are both open and closed.

Ah, This makes sense!!! Thanks for your help! I guess I thought since the question indicated the "if"s that I've got to be doing something wrong, haha! Thank you so much for your help though!