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Power sets.

  1. Feb 9, 2012 #1
    If I started with a one element set and took its power set. And then I just kept taking the power set forever, would I eventually end up
    with a set that had cardinality of [itex] \aleph_0 [/itex] ?
     
  2. jcsd
  3. Feb 9, 2012 #2
    Depends what you mean by "eventually end up."

    You would have a sequence of sets of cardinalities 1, 2, 4, 16, 2^16, ...

    So what you would "end up" with is no more than a sequence of sets. You would have [itex] \aleph_0 [/itex] many sets all together, but each one would have finite cardinality.

    The sequence of cardinalities would be unbounded, meaning that there is no finite limit to how large the cardinalities get. But each cardinality would be a finite number.
     
  4. Feb 9, 2012 #3

    Deveno

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    because just taking power sets doesn't give us anything else from finite sets, except MORE finite sets, one cannot PROVE the existence of an infinite set. but we "intuitively feel" that the natural numbers ought to be a set, so an axiom is added to set theory that guarantees the existence of at least one infinite set (this is called, oddly enough, the axiom of infinity).

    this allows us to show that there is at least (and therefore several) one set with cardinality aleph-null. in other words, we feel like any worth-while system of mathematics ought to come with induction "built-in" as a standard feature. if math (in this case, "set theory") doesn't formally include arithmetic (at least), one feels that a wrong turn has been taken somewhere (sort of like, if you can't say "hello" and "good-bye" in a language, perhaps you're not using a very good language).

    of course, no one has ever checked that "+1" always gives you bigger and bigger numbers (who has that kind of patience?), but it seems illogical that it would not. this is what people mean when they say "the successor function is injective", that n+1 is always something not in the set {0,1,2,....,n}.

    to answer your original question:

    the set of cardinalities would have cardinality aleph-null, but none of the elements of that set would.
     
  5. Feb 9, 2012 #4
    interesting, thanks for the responses
     
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