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Closed and Open Sets in R^d

  • Thread starter x_saved_kt
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  • #1
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!!!!!
 

Answers and Replies

  • #2
HallsofIvy
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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.
By what logic do you "know it isn't right"? Pehaps this particular part does NOT depend on B being a subset of A. For example, if A and B were disjoint, A\B= A which is, just as your proof would assert, open. The fact is that, if A is open and B is closed, the A\B is open and B\A is closed, whether A is a subset of B or B is a subset of A or not.

Same goes for the second part of the question. PLEASE HELP!!!!!
 
  • #3
By what logic do you "know it isn't right"? Pehaps this particular part does NOT depend on B being a subset of A. For example, if A and B were disjoint, A\B= A which is, just as your proof would assert, open. The fact is that, if A is open and B is closed, the A\B is open and B\A is closed, whether A is a subset of B or B is a subset of A or not.
Ah, I see. Thanks, this makes a lot of sense now! I guess all I was thinking is that if in the question they indicated the "if"s I've got to be doing something wrong! hah! Thanks a lot for your help!!
 

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