Results 1 to 12 of 12

Math Help - Real Analysis - Sets and Proofs

  1. #1
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318

    Real Analysis - Sets and Proofs

    I have 2 problems that are annoying the heck out of me...

    #1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.

    #2. Under what conditions does A\(A\B)=B. Again, how do I start this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    I point you towards a website I've been contributing to for some few weeks now ... here's the page containing some proofs that might help.

    Subset Equivalences - Proofwiki.org

    explore and enjoy.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Oi, miss Green !
    Quote Originally Posted by kathrynmath View Post
    I have 2 problems that are annoying the heck out of me...

    #1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.
    To prove a set P is contained in another one, Q, prove :
    if x is in P, then x is in Q.

    If you want to know whether it's true or false, draw a diagram. Just to give you the correct direction !

    Note, A \backslash B=A \cap \overline{B}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    Quote Originally Posted by Moo View Post
    Oi, miss Green !

    To prove a set P is contained in another one, Q, prove :
    if x is in P, then x is in Q.
    Does this involve assuming P=Q?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    Quote Originally Posted by Moo View Post
    Oi, miss Green !

    To prove a set P is contained in another one, Q, prove :
    if x is in P, then x is in Q.

    If you want to know whether it's true or false, draw a diagram. Just to give you the correct direction !

    Note, A \backslash B=A \cap \overline{B}
    So, I figured out #1 now.
    For #2, is it something with A equaling B?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Not quite.

    A - (A - B) \Longrightarrow A \cap B (sorry, I use - for set difference)

    The above may need some thought, it's not trivial.

    But that means A \cap B = A which is one of the classic subset equivalences:

    A \cap B  = A \iff A \subseteq B

    That's the direction you want to go in, but proving them needs thought.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    Hmmm, this question is still frustrating me.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    Is B a null set?
    Because say A={1, 2, 3, 4, 5} and B={1, 2,3}.
    We get:
    A\({4,5}
    ={1,2, 3}

    If B is a null set, and A is the same, we get:
    A\{1, 2, 3, 4, 5}.
    ={}
    Therefore, we have B.

    Also, what about A=B? A = {1, 2 , 3, 4, 5} and B={1,2, 3, 4, 5}
    A\({}
    {1,2,3,4,5}
    Therefore, we get B.

    So, these 2 cases appear to work. these are my thoughts so far.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Sorry, I thought I just told you: A has to be a subset of B.

    Note that if A=B then it is also true that A \subseteq B so yes, the fact that A=B does make your equation work, it's just not the full answer.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    So, B has to be contained in A? For example A={1,2,3,4,5}, B={1,2,3}
    A\(A\B)
    A\{4,5}
    {1,2,3,4,5}\{4,5}
    {1,2,3}
    B

    That works....
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Did you actually check out the link I gave you a few postings back in the thread?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Arrow

    Quote Originally Posted by kathrynmath View Post
    I have 2 problems that are annoying the heck out of me...

    #1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.

    #2. Under what conditions does A\(A\B)=B. Again, how do I start this?
    I just looked at #1.
    Clearly, we have the following identities

    1. if A\subset B, then B^{c}\subset A^{c}, where ~^{c} stands for the compliment of the sets.
    2. if A\subset B, then A\cap C\subset B\cap C
    3. A\backslash B=A\cap B^{c}

    Hence, from the properties above, we get A\subset B implies B^{c}\subset A^{c}. Intersection both sides with C, we get C\cap B^{c}\subset C\cap A^{c}, which is equivalent to C\backslash B\subset C\backslash A.

    I hope you may easily find #2 by using 3. for twice and thinking on the resultant situation.

    Note. The easiest way to work on such problems may be using Venn diagrams.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Real analysis: Compact Sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 7th 2010, 07:27 PM
  2. Real Analysis - Compact Sets #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 16th 2008, 02:02 PM
  3. Real analysis - limit proofs
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 20th 2008, 06:29 PM
  4. more Proofs (real analysis)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2007, 03:27 PM
  5. Proofs Questions (Real Analysis)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2007, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum