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One to one functions

  1. Sep 18, 2009 #1
    Problem:
    Let A, V, and C be any nonempty sets. Prove or disprove the following:
    [tex]|A \times C| \leq |(A \times B) \times C|[/tex]


    Proof:
    Fix b [tex]\in[/tex] B.
    Set f(a,c) = ((a,b),c) for a [tex]\in[/tex] A and c [tex]\in[/tex] C.
    Now suppose [tex]f(a_{1}, c_{1}) = f(a_{2}, c_{2})[/tex].
    Thus, [tex]f(a_{1}, c_{1}) = ((a_{1}, b), c_{1}) = ((a_{2}, b), c_{2}) = f(a_{2}, c_{2})[/tex]
    Thus, [tex](a_{1}, c_{1}) = (a_{2}, c_{2})[/tex] since the corresponding ordered triples are equal.
    Thus, [tex](a_{1}) = (a_{2}),[/tex] and [tex](c_{1}) = (c_{2}),[/tex] since the ordered pairs above are equal.
    Thus, f is one to one.
    Finally, [tex]|A \times C| \leq |(A \times B) \times C|,[/tex] (by definition of one to one functions).


    Questions:
    I understand this proof and why b[tex]\in[/tex]B is fixed, but to prove [tex]|A \times C| \leq |(A \times B) \times C|,[/tex] it is not necessary to state to fix b. Since b can be any value in the triple order, it doesn't have to be fixed, in fact it can hold any value from the set B and still fulfill the condition: [tex]|A \times C| \leq |(A \times B) \times C|[/tex]. If what I am saying is true, is there an easy modification of the proof above? Personally I think by fixing b, we are limiting the scope of the proof.


    Thanks,


    JL
     
    Last edited: Sep 18, 2009
  2. jcsd
  3. Sep 18, 2009 #2

    Office_Shredder

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    You need to fix b only so you can define your function properly. You're right, the function could be any

    f: (a,c) -> (a,g(a,c),c)

    where g is any function that maps (a,c) to B. But why bother saying that mouthful when you can just say "fix b"?
     
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