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Set Theory

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that:

    [itex] A \cap B \subset (A \cap C) \cup (B \cap C') [/itex]

    2. Relevant equations


    3. The attempt at a solution

    I tryed distribute [itex] (A \cap C) over (B \cap C') [/itex] but i'm always walking in circles and i don't came to a satisfactory answer. This exercise was in a section "some easy exercies on complementation" but i don't see how to use complements here.

    Thanks
     
  2. jcsd
  3. Sep 6, 2012 #2

    micromass

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    Is [itex]C^\prime[/itex] supposed to be the complement of C??
     
  4. Sep 6, 2012 #3

    HallsofIvy

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    The standard way to show that "[itex]X\subseteq Y[/itex]" is to start "if [itex]x\in X[/itex]" and then use the definitions of X and Y to conclude "[itex]x\in Y[/itex]".

    Here, if [itex]x\in A\cap B[/itex], what can you say about x?
     
  5. Sep 7, 2012 #4
    Yes, [itex]C'[/itex] is the complement of [itex] C [/itex]

    if [itex] x \in A \cap B [/itex],

    by the definition of intersection:

    [itex] A \cap B = \{x \in A : x \in B\} [/itex]

    and we can conclude that x is simultaneously in A and B.

    But my doubt, is how to reduced the [itex] (A \cap C) \cup (B \cap C') [/itex] to a expression that i can readly see that [itex] A \cap B \subset (A \cap C) \cup (B \cap C') [/itex].

    I tryed ...

    [itex] (A \cap C) \cup (B \cap C') = [/itex]
    [itex] (A' \cup C')' \cup (B' \cup C)' = [/itex]
    [itex] [(A' \cup C') \cap (B' \cup C)]' = [/itex]
    [itex] (...) [/itex]
    [itex] A \cup B \cup C' [/itex]

    But it takes me a lot of work, i'm not sure if this result is correct and i think that exists a better way of doing this...

    Thanks
     
  6. Sep 7, 2012 #5

    ehild

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    You certainly know that [itex]X\subset Y[/itex] if and only if [itex]X \bigcap Y=X[/itex].
    [itex]X=A\bigcap B [/itex] and [itex]Y=(A \cap C) \cup (B \cap C')[/itex]

    ehild
     
  7. Sep 7, 2012 #6

    HallsofIvy

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    Good, we know x is in both A and B. And we know that x is either in C or it is NOT! That means x is C or it is in C'
    Case 1: Suppose x is in C. We know it is in A therefore ....
    Case 2: Suppose x is in C'. We know it is in B therefore ....

    In my opinion, to much focus on "formulas", not enough on basic "definitions".
     
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