# togology-set proof

• Aug 29th 2012, 03:13 PM
gbux512
togology-set proof
Im new to proofs, and the proof class i took last semester was absolutely terrible. Im given A isa subset of S and B is a subset of S. I need to prove that a is a subset of the complement of B iff a intersection b = the empty set. i started by breaking it up into two if then statements. I said if a is a subset of the complement of B then a intersect b = the empty set. I have words that prove this, but i cannot figure out how to display this mathematically. I said the intersection of A and B is the set of all elements that are in A and B. Because A is a subset of thej compliment of B, every x in a is also in the compliment of B. because B and the compliment of B share no elements, the intersection of b and a can contain only the null set. Now as far as reversing it and proving the other iff statement I cant even get started. Im sure my other statement is unsatisfactory as well. i feel so confused and demotivated. i wrote down everything important from the book, suck as modus tollens etc as my tool box, but i feel like i have a puzzle with blank pieces, and i cant see how they fit together. any help is appreciated
• Aug 29th 2012, 03:43 PM
Plato
Re: togology-set proof
Quote:

Originally Posted by gbux512
Im new to proofs, and the proof class i took last semester was absolutely terrible. Im given A isa subset of S and B is a subset of S. I need to prove that a is a subset of the complement of B iff a intersection b = the empty set. i started by breaking it up into two if then statements. I said if a is a subset of the complement of B then a intersect b = the empty set. I have words that prove this, but i cannot figure out how to display this mathematically. I said the intersection of A and B is the set of all elements that are in A and B. Because A is a subset of thej compliment of B, every x in a is also in the compliment of B. because B and the compliment of B share no elements, the intersection of b and a can contain only the null set. Now as far as reversing it and proving the other iff statement I cant even get started. Im sure my other statement is unsatisfactory as well. i feel so confused and demotivated. i wrote down everything important from the book, suck as modus tollens etc as my tool box, but i feel like i have a puzzle with blank pieces, and i cant see how they fit together. any help is appreciated

Please, please review what you posted, correct it, shorten it, and clean it up.
• Aug 29th 2012, 04:46 PM
Deveno
Re: togology-set proof
i gather what you are trying to prove is this:

for A,B ⊆ S:

A ∩ (S - B) = Ø iff A ⊆ B.

suppose A ∩ (S - B) = Ø. we will proceed by contradiction, and show that x in A - B is not possible.

for if there were such an x, it would be in S - B, since x is not in B, and also in A (since A - B is a subset of A), and therefore in A ∩ (S - B), which is a contradiction because A ∩ (S - B) has no members.

now A = (A - B) U (A ∩ B), so every x in A must lie in A ∩ B (since A - B = Ø). in other words A ∩ B = A, which is another way of saying: A ⊆ B

(if every x in A lies in A ∩ B ⊆ B, then every x in A is also in B).

on the other hand suppose A ⊆ B. this means that (S - B) ⊆ (S - A) (draw a picture if you must, this is just the contrapositive applied to set membership).
hence A ∩ (S - B) ⊆ A ∩ (S - A) = Ø. thus A ∩ (S - B) = Ø.

(a word on the last line of reasoning: suppose B ⊆ C. then A ∩ B ⊆ A ∩ C.

to see this suppose x is in A ∩ B. then x is in A, and x is in B. since B ⊆ C, x is is also in C, so x is in A, and x is in C, so x is in A ∩ C. also note that if a set S ⊆ Ø, S = Ø

(the empty set has no other possible subsets)).
• Aug 29th 2012, 05:05 PM
Plato
Re: togology-set proof
Quote:

Originally Posted by Deveno
A,B ⊆ S:
A ∩ (S - B) = Ø iff A ⊆ B.

I must ask why you think that the OP means that?
What in the world does "a is a subset of the complement of B iff a intersection b = the empty set." mean?
Why are you guessing what the OP means?
• Aug 29th 2012, 05:50 PM
Deveno
Re: togology-set proof
oops, you are correct, i mis-read the question. all the B's and S - B's should be switched. oh dear.