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Math Help - Another ordinal problem

  1. #1
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    Another ordinal problem

    I have this kind of problem:
    Show, that for every ordinal number \alpha there exists a cardinal number  \kappa so that \alpha \in \kappa.

    I have no idea how to prove that, so little help would be nice. Thanks!
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  2. #2
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    Quote Originally Posted by Ester View Post
    I have this kind of problem:
    Show, that for every ordinal number \alpha there exists a cardinal number  \kappa so that \alpha \in \kappa.

    I have no idea how to prove that, so little help would be nice. Thanks!
    What about \kappa=cardinality of powerset of \alpha? (This is the smallest ordinal equinumerous to the powerset of \alpha.)

    You know that cardinality of \kappa is greater than the cardinality of \alpha. You also should know that any two ordinals are comparable. The cardinality argument implies that \kappa\le\alpha is not possible, hence you get \alpha<\kappa. (This is equivalent to \alpha\in\kappa.)
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  3. #3
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    The above solution relies on the axiom of choice for the result that every every set x does have a cardinality that is indeed equinumerous with x. That is fine, but also we have a solution that does not require the axiom of choice.

    By a result of Hartogs, let k be the least ordinal not dominated by a. So k is not less than nor equal to a. Then by comparability of ordinals, we have a is less than k. And k is a cardinal, for if k were equinumerous with an ordinal less than k, then k would not be the least ordinal not dominated by a.

    (Granted, this requires that Hartogs's result has already been proven in the course.)
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