Results 1 to 5 of 5

Math Help - togology-set proof

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    35

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: togology-set proof

    Quote Originally Posted by gbux512 View Post
    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
    Sorry to say this, but your post is completely unreadable.
    Please, please review what you posted, correct it, shorten it, and clean it up.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,279
    Thanks
    671

    Re: togology-set proof

    i gather what you are trying to prove is this:

    for A,B ⊆ S:

    A ∩ (S - B) = iff A ⊆ B.


    suppose A ∩ (S - B) = . 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 = ). 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) = . thus A ∩ (S - B) = .

    (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 ⊆ , S =

    (the empty set has no other possible subsets)).
    Last edited by Deveno; August 29th 2012 at 04:49 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: togology-set proof

    Quote Originally Posted by Deveno View Post
    A,B ⊆ S:
    A ∩ (S - B) = 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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,279
    Thanks
    671

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: June 8th 2011, 11:13 AM
  2. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  3. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  4. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum