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!!!!!

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 **are both open and closed**.

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

**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!