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Math Help - Proof involving functions

  1. #1
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    Proof involving functions

    Prove the following:

    If  f : A \rightarrow B and  g : C \rightarrow D , then  f \cap g : A \cap C \rightarrow B \cap D.

    Here's my thoughts/attempt:

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



    This is all I have so far.

    Would I have to break it into cases where  a = c and  a \not= c ? If  a = c ,  A \cap C contains an element, but if  a \not= c ,  A \cap C is empty since a and c were arbitrary. The same argument holds for  B \cap D . So, taking these things into account,  f \cap g 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.
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  2. #2
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    Quote Originally Posted by Jacobpm64 View Post
    Prove the following:
    If  f : A \rightarrow B and  g : C \rightarrow D , then  f \cap g : A \cap C \rightarrow B \cap D.
    As stated the proposition is false.
    It may be that f \cap g = \emptyset.
    Then what?
    Please check the exact wording of the question
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  3. #3
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    my homework just says Prove the following theorem.

    Theorem 1
    If  f : A \rightarrow B and  g : C \rightarrow D , then  f \cap g : A \cap C \rightarrow B \cap D .

    Isn't  f \cap g still a function even if it is the empty set? In this case the definition of function would be vacuously true, right?
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  4. #4
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    As stated the proposition is still false.
    Consider this example A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}
    f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}
    g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}

    But f \cap g = \left\{ {(2,r)} \right\} while A \cap C = \left\{ {2,4} \right\} clearly f \cap g:A \cap C \not{\mapsto} B \cap D
    There is no mapping for the term 4.
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  5. #5
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    I'm convinced now.

    Thanks.
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