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Math Help - Cardinal exponentiation

  1. #1
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    Cardinal exponentiation

    I need to prove the following rules on cardinals:

    Let κ, μ, ν be cardinal numbers. Prove:

    1. κ^(μ + ν) = κ^μκ^ν.

    2. κ^μν = (κ^μ)^ν.

    --

    How do I start proving such thing? It looks so obvious, but I'm sure there has to be a formal way to prove it.

    Please help me!

    Thank you
    Last edited by adam63; August 23rd 2009 at 10:30 AM.
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  2. #2
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    Are k, μ, v infinite cardinal numbers? Because otherwise, 1 is obviously not true, unless there is a typo there.
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  3. #3
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    I'm sorry, I fixed it now. (a typo)
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  4. #4
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    Hi again.
    Cardinal power \kappa^{\lambda} is defined as \kappa^{\lambda}=|^{\lambda}\kappa| , that is cardinality of the set of mappings from \lambda to \kappa.
    So to prove for example (2) it suffices to show that ^{(\nu \times \mu)}\kappa has the same cardinality as ^{\nu}(^{\mu}\kappa).
    For an arbitrary map f:\nu \times \mu \rightarrow \kappa and an element a \in \nu let us define map f_a : \mu \rightarrow \kappa like this: f_a(t) = f(a,t) for t \in \mu.
    Then, relation r = \{<f,\{<a,f_a>:a \in \nu \}>:f \in  ^{(\nu \times \mu)}\kappa\} is an injective mapping from ^{(\nu \times \mu)}\kappa to ^{\nu}(^{\mu}\kappa). If g \in ^{\nu}(^{\mu}\kappa), then f=\{<a,b,c>:a \in \nu \& <b,c> \in g(a)\} is preimage of g under mapping r. So r is injective mapping from ^{(\nu \times \mu)}\kappa onto ^{\nu}(^{\mu}\kappa).
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  5. #5
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    I'm sorry, I lost you in the middle... What do I mainly need to do?
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  6. #6
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    Hello,
    main idea is to construct a bijection between two sets. Firstly make sure you're ok with these series of equalities:

    \kappa^{\nu \mu}=|^{\nu \mu}\kappa|= \left|^{|\nu \times \mu|}\kappa \right| = |^{(\nu \times \mu)}\kappa|
    and

    (\kappa^{\mu})^{\nu} = |^{\mu}\kappa|^{\nu} = \left|^{\nu}\left|^{\mu}\kappa \right|\right| =\left|^{\nu}(^{\mu}\kappa)\right|

    They justify that all you need to do is to show that |^{(\nu \times \mu)}\kappa| = \left|^{\nu}(^{\mu}\kappa)\right|. And that is done by the construction of the relation r in my previous post and contemplating that r is really a bijection.
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  7. #7
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    ooh, now I understand.

    Thank you, I'll do it now...
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