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Check my proof for cartesian product (set theory)

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]\forall[/tex] sets [tex] A, B, C [/tex], if [tex]B\subseteq C, [/tex] then [tex] A \times B \subseteq A \times C[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Haven't done set theory proofs in a while. Does this suffice in proving the statement?:

    Let [tex] x \in B [/tex] be arbitrary. Assuming [tex]B \subseteq C[/tex] is true, we know that [tex]x \in C[/tex]. Since [tex]x \in B[/tex], we know that [tex]A \times B[/tex] will produce an ordered pair (a,x) where a is an arbitrary element of A. Since [tex]x \in C[/tex], we know that [tex]A \times C[/tex] will produce the same ordered pair (a,x).

    Therefore by definition of subsets, [tex]A \times B \subseteq A \times C[/tex]

    QED


    Thanks for your help.
     
    Last edited: Apr 26, 2010
  2. jcsd
  3. Apr 26, 2010 #2

    tiny-tim

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    Hi iamsmooth! :smile:
    mmm … messy …

    and what does "[tex]A \times B[/tex] will produce an ordered pair (a,x) …" mean?

    Instead, start "for any element (a,b) of A x B … " :wink:
     
  4. Apr 26, 2010 #3
    Let [tex] x \in B [/tex] be arbitrary. Assuming [tex]B \subseteq C[/tex] is true, we know that [tex]x \in C[/tex].

    We know that for all elements [tex](a, x) \in A \times B[/tex] where a is an arbitrary element of A, there will also exist [tex](a, x) \in A \times C[/tex] since [tex]x \in A [/tex] and [tex] x \in C[/tex].

    Therefore by definition of subsets, [tex]A \times B \subseteq A \times C[/tex]

    QED

    Is this better?
     
  5. Apr 26, 2010 #4

    tiny-tim

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    Since you later say "for all elements [tex](a, x) \in A \times B[/tex]", why are you starting with "Let [tex] x \in B [/tex] be arbitrary." ?

    The proposition you are required to prove starts "for any element (a,x) of A x B",

    so your proof should start with the same words. :wink:
     
  6. Apr 26, 2010 #5
    Assuming [tex]B \subseteq C[/tex] is true, we know that if there is any element [tex]x \in B[/tex], then [tex]x \in C[/tex].

    For any element [tex](a, b) \in A \times B[/tex] where a is an arbitrary element of A and b is an arbitrary element of B, there will also exist [tex](a, b) \in A \times C[/tex] since [tex]b \in B [/tex] and [tex] b \in C[/tex] must be true (by assumption).

    Therefore by definition of subsets, [tex]A \times B \subseteq A \times C[/tex]

    QED

    Is it just my wording that's messed up?
     
  7. Apr 26, 2010 #6

    tiny-tim

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    You really don't need that opening sentence;

    also, "there will also exist" is rather a strange way of putting it, since it's the same element, (a,b), in both sets …

    you want something more like any element of A x B is of the form (a,b) with a in A and b in B, and since B is a subset of C, b is in C, and so (a,b) is in A x C. :wink:
     
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