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Proof involving functions

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the following:

    If [tex] f : A \rightarrow B [/tex] and [tex] g : C \rightarrow D [/tex], then [tex] f \cap g : A \cap C \rightarrow B \cap D[/tex].

    3. The attempt at a solution

    Here's my thoughts/attempt:

    Proof:
    Let A, B, C, and D be sets. Assume [tex] f : A \rightarrow B [/tex] and [tex] g : C \rightarrow D [/tex]. Let [tex] a \in A [/tex]. Since f is a function from A to B, there is some [tex] y \in B [/tex] such that [tex] (a, y) \in f [/tex]. Let [tex] b \in B [/tex] be such an element, that is, let [tex] b \in B [/tex] such that [tex] (a,b) \in f [/tex]. Let [tex] c \in C [/tex]. Since g is a function from C to D, there is some [tex] z \in D [/tex] such that [tex] (c, z) \in g [/tex]. Let [tex] d \in D [/tex] be such an element, that is, let [tex] d \in D [/tex] such that [tex] (c,d) \in g [/tex].

    This is all I have so far.

    Would I have to break it into cases where [tex] a = c [/tex] and [tex] a \not= c [/tex]? If [tex] a = c [/tex], [tex] A \cap C [/tex] contains an element, but if [tex] a \not= c [/tex], [tex] A \cap C [/tex] is empty since a and c were arbitrary. The same argument holds for [tex] B \cap D [/tex]. So, taking these things into account, [tex] f \cap g [/tex] is either a function from the set containing a to the set containing b, or its a function from the empty set to the empty set.

    Does this make any sense, is it necessary, and how should I write it in my proof?

    Thanks in advance.
     
  2. jcsd
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