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Real Analysis (Set Theory) Proof

  1. Jan 16, 2006 #1
    Let A and B be subsets if a universal set U. Prove the following.
    a) A\B = (U\B)\(U\A)

    To do this, show it both ways.
    1) A\B contains (U\B)\(U\A)
    2) (U\B)\(U\A) contains A\B

    I'll start with 2)

    if x is in (U\B)\(U\A),
    then x is in (U\B) and x is NOT in (U\A).
    then (x is in U and x is NOT in B) and (x is NOT in U and x is in A)
    so, x is in U and x is NOT in U and x is NOT in B and x is in A
    *I think i made a mistake "x is in U and x is NOT in U" doesnt make sense? Any ideas?

    Back at 1)

    if x is in A/B,
    then x is in A and x is NOT in B
    then either x is in B or x is NOT in A
    *this is where im lost :S

    Anyone have any ideas on how to solve this?

    Thanks in advance
     
  2. jcsd
  3. Jan 16, 2006 #2

    AKG

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    This is not a correct inference. In particular, "x is NOT in (U\A)" does not give "x is NOT in U and x is in A".
    This is not a correct inference.
     
    Last edited: Jan 16, 2006
  4. Jan 17, 2006 #3

    HallsofIvy

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    If you are doing 2) (U\B)\(U\A) contains A\B then you are going the wrong way: you must prove "if x is in U\A then x is in (U\B)\(U\A)".
    Saying "if x is in (U\A)\(U\A)" would be correct for proving 1).

    Obviously x is in U since U is the universal set. What you are saying now is "x is A and x is NOT in B". That gives 1) directly.

    NOW you are proving 2)!

    You know that x is NOT in B, therefore it is in U\B. You know x is in A therefore it is NOT in U\A.
     
  5. Jan 17, 2006 #4
    HallsofIvy: Oh i see that i did it backwards.

    I'll start all over again since i am getting confused.

    Let A and B be subsets if a universal set U. Prove the following.
    a) A\B = (U\B)\(U\A)

    To do this, show it both ways.
    1) A\B contains (U\B)\(U\A). show: if x is in A\B then x is in (U\B)\(U\A).
    2) (U\B)\(U\A) contains A\B. show: if x is in (U\B)\(U\A) then x is in A\B.

    1) A\B contains (U\B)\(U\A)
    Let x be in A\B. Then x is in A and x is NOT in B. then x is in (U\B) and x is NOT in (U\A). Therefore, x is in (U\B)\(U\A).

    2) (U\B)\(U\A) contains A\B
    Let x be in (U\B)\(U\A), then x is in (U\A) and x is NOT in (U\A).
    "then x is in U and x is NOT in B and x is NOT in U and x is in A." <--- that makes no sense to me. Could somebody comment if that is correct or what?
    --obviously x is in U.
    --x is NOT in B and x in A means that x is in (A\B)

    Can somebody please check this.

    Thanks
     
  6. Jan 17, 2006 #5

    matt grime

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    YOu can never say x is not in U, U is a universal set, since everything in your model is in that set, so what you ask about in quotes is obviously wrong. The reason for your confusion seems to be that you think not(X and Y) is "not(X) and not(Y)", which it isn't. It is not(X) or not(Y) and since "x is not in U" is always false, then x is not in (U\A) is true if and only if "x is in A", or more succinctly, (A^c)^c=A.


    Remember that U\A is simply A^c and X\Y is just X intersected Y^c, and that makes all of these questions so much more readable, not to mention it makes them so easy to prove.

    I mean the first one is simply saying that you need to show

    [tex]B^c \cap (A^c)^c = A \cap B^c[/tex]

    which is obviously true since

    [tex](A^c)^c=A[/tex]

    as I mentioned at the start.
     
    Last edited: Jan 17, 2006
  7. Jan 21, 2006 #6
    matt grimm...

    so what you are saying is...

    if X is in (U\B)\(U\A), then X is in (U\A) and X is NOT in (U\A). Further, X is in U and X is NOT in A, and X is NOT in U or X is NOT in A.

    See how i bolded that OR? Is that what you were trying to say? infact, it makes sense if it were and OR and not an AND like i thought it was previously.
     
  8. Jan 22, 2006 #7

    matt grime

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    It is impossible to tell what you mean since you're mixing up ands and ors at will and there are at least two ways of reading what you've read.

    A and (B or C)

    is very different from

    (A and B) or C

    I still don't see why you're making this so hard but there you go.

    Note that there is s typo in you first line where you assert

    X is in U\A and X is not in U\A

    which is not possible. But I think you jsut mistyped the second part.

    If you insist on doing it with 'is in' can you at least use latex?

    click to see the code

    [tex]\in \notin \cap \cup[/tex]


    for in not in and and or respectively. can't recall what disjunction and conjunction are but we can abuse notation and have interesection for and, and union for or, since they are the same thing.
     
  9. Jan 22, 2006 #8
    Yes, Latex would make things nicer, however, i don't know how to use it.
    Do i have to install something on my computer? or do i just type the commands into the reply box?
     
  10. Jan 22, 2006 #9

    matt grime

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    Read the latex tutorial in these forums; it is widely linked to, and can be found by searching the forums if you ever forget where it is.

    however there are much easier ways to typset maths, especially for your subject. compare

    {(x in U) and (x not in A)} or {(x in U) and (x not in B)}

    with what you wrote. this is unambiguous, becuase of the bracketing. you wouldn't write three times four plus one, would you?

    you can even go further and suggestively use u and n for union and intersection.
     
    Last edited: Jan 22, 2006
  11. Jan 22, 2006 #10
    im just testing this:
    [tex]\notin[/tex]

    okay!!! it works!!
     
  12. Jan 22, 2006 #11
    if x[tex]\in[/tex](U\B)\(U\A), means that x[tex]\in[/tex](U\B) and x[tex]\notin[/tex](U\A).

    x[tex]\in[/tex](U\B) means that x[tex]\in[/tex]U and x[tex]\notin[/tex]B

    x[tex]\notin[/tex](U\A) means that x[tex]\notin[/tex]U and x[tex]\in[/tex]A

    so all together, we have:

    x[tex]\in[/tex]U and x[tex]\notin[/tex]B and x[tex]\notin[/tex]U and x[tex]\in[/tex]A.

    and this is where im still confused:

    how can x[tex]\in[/tex]U and x[tex]\notin[/tex]U.

    So obviously i made a mistake somewhere and i think i made a mistake with the and and or's
     
    Last edited by a moderator: Jan 22, 2006
  13. Jan 22, 2006 #12

    JasonRox

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    As matt grime pointed out...

    You can not have x[tex]\notin[/tex]U. This is the underlying set. It's impossible for this to happen.

    I think you're misunderstanding the definition of B/A.
     
  14. Jan 22, 2006 #13

    matt grime

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    One of the first things I said was that you were getting negations wrong.

    X not in (U\A) ***does not mean**** (X is not in U) and (X is in A).

    X is in U\A means (X is in U) and (X is not in A).

    The negation of that statement is not what you claim it still is as i told you in post 5.

    The negation of UandV is (notU) or (notV)

    do you get the idea of complements? Something is NOT in the complement of A exactly when it is in A, double negatives make a positive, minus one times minus 1 is plus one etc.
     
    Last edited: Jan 22, 2006
  15. Jan 22, 2006 #14
    so x[tex]\notin[/tex](U\A) means that "either x[tex]\notin[/tex]U OR x[tex]\in[/tex]A

    and since "x[tex]\notin[/tex]U is always false, then x[tex]\notin[/tex](U\A) is true if and only if "x[tex]\in[/tex]A"
     
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