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Cardinal Numbers

  1. Apr 28, 2010 #1
    Let A,B,C be three cardinals. Show that...

    [tex]A^{B+C} = A^B A^C[/tex]

    I thought about using some some of distinguishing scheme where we denoted [tex]B \cup C[/tex] as [tex]B \times \{0\} \cup C \times \{1\}[/tex] so we could map thing easily but apparently that's not right and you can assume they are disjoint anyway...

    EDIT: I just noticed the sticky at the top, this was just something in the notes we had that wasn't proven, not an assignment question or anything and I wanted to know how to do it. Should I repost this in another section?
     
  2. jcsd
  3. Apr 28, 2010 #2
    Two hints:

    (1) Assuming that B and C are disjoint (which we may always assume by that construction, because it doesn't change the cardinality), any function from B + C (the disjoint union of B and C) may always be thought as the union of two functions:
    [tex]
    f_1:\B \rightarrow A
    [/tex]

    [tex]
    f_2:\C \rightarrow A
    [/tex]

    [tex]
    f=f_1\cup f_2
    [/tex]

    (2) Consider the application:

    [tex]
    \Phi:A^{B+C}\rightarrow A^{B} A^{C}
    [/tex]

    Defined by:

    [tex]
    \Phi\left(f\right)\left(a\right) = \left(f_1\left(a\right),f_2\left(a\right)\right)
    [/tex]

    For [itex]a \in A[/itex]. Is it a bijection?
     
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