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Stupid question

  1. Jan 9, 2005 #1
    Let R be the set of real numbers.

    What is card(R)^card(R)? I mean, is that expression an aleph (assuming CH, if necessary) or is it not an aleph? Any name given to card(R)^card(R)?

    (I know that card(R)^card(R) is the card(real valued functions defined on R).
  2. jcsd
  3. Jan 9, 2005 #2
    This can't be:
    can it?

    Here's what I came up with and I suspect I'm wrong:
    1. 2^|R|=|P(R)|
    2. It suffices to show that |P(R)|=|R|^|R|.
    3. To show P(R)<=R^R, let A be a subset of R. Map it to the characteristic function on A (ie this function is 1 for x in A and 0 for x in R\A). Clearly, this mapping of A to Char(A) is 1-1 for if Char(A)=Char(B) then A=B.
    4. To show R^R<=P(R), let f be in R^R. Invoke the (set) isomorphism that maps R^2 to R. Call this map S (for squish). f is a subset of R^2. Then S(f) is a subset of R. As S is 1-1, S(f)=S(g) implies f=g. Thus R^R<=P(R).

    Poke away!
  4. Jan 10, 2005 #3

    matt grime

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    4. doesn't seem to be true at all. I mean, what has the map from R^2 got to do with a map from R^R?

    wait, i get it now.

    If we are assuming the CH, then it is an aleph, though I don't know which one.
    Last edited: Jan 10, 2005
  5. Jan 10, 2005 #4
    This site has info on the exponentiation of cardinals assuming GCH, which is something I feel comfortable assuming.

    http://planetmath.org/encyclopedia/CardinalExponentiationUnderGCH.html [Broken]

    Then if lambda=kappa=Aleph1=|R|, then

    L^K=K+, using the top row of their formula, where K+ is the cardinal successor of K.

    Then since K=Aleph1, Isn't K+=Aleph2?

    And isn't Aleph2=|P(R)|=2^K, as stated above? (Assuming GCH, of course.)

    This kind of messes with my intuition of the Alephs if it's true.
    Last edited by a moderator: May 1, 2017
  6. Jan 10, 2005 #5


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    Yep, it can.

    Here's another proof using cardinal arithmetic:

    |R|^|R| = (2^|N|)^|R| = 2^(|NxR|) =2^|R|
  7. Jan 10, 2005 #6
    Cool. I never see the most elegant way to do things, do I???

    Well, so it is decided. That's rather strange to me. THere are as many subsets of R as there are functions from R to R?

    Hmmm.... I guess so.

    But the GCH nor CH is needed, huh? Rats. I thought this would be used in my study of Awareness and I'd have a use for the CH.

    Thanks, you two.
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