Cardinality of Real-Valued Functions on R

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Discussion Overview

The discussion revolves around the cardinality of the set of real-valued functions defined on the real numbers, specifically examining the expression card(R)^card(R) and its implications under various assumptions such as the Continuum Hypothesis (CH) and the Generalized Continuum Hypothesis (GCH).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether card(R)^card(R) can be equal to 2^|R| and proposes a series of steps to explore the relationship between these cardinalities.
  • Another participant challenges the validity of a mapping argument related to R^R and P(R), expressing confusion about its relevance.
  • A later reply suggests that if the CH is assumed, then card(R)^card(R) is indeed an aleph, though the specific aleph is uncertain.
  • One participant references a source discussing cardinal exponentiation under GCH, suggesting that if |R|=Aleph1, then the successor cardinal K+ could be Aleph2, raising questions about the implications for their understanding of the Alephs.
  • Another participant provides an alternative proof using cardinal arithmetic to show that |R|^|R| equals 2^|R|, indicating a potential resolution to the earlier question.
  • One participant expresses surprise at the conclusion that there are as many subsets of R as there are functions from R to R, noting that the assumptions of GCH or CH may not be necessary.

Areas of Agreement / Disagreement

Participants express differing views on the relationships between the cardinalities discussed, with some proposing proofs and others challenging those proofs. The discussion does not reach a consensus on the implications of the CH or GCH for the cardinalities in question.

Contextual Notes

Some arguments depend on the assumptions of CH or GCH, and there are unresolved questions regarding the mappings and relationships between the sets involved. The discussion includes various mathematical steps that may not be fully resolved.

phoenixthoth
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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).
 
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This can't be:
2^|R|=|R|^|R|,
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. 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.
 
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This site has info on the exponentiation of cardinals assuming GCH, which is something I feel comfortable assuming.

http://planetmath.org/encyclopedia/CardinalExponentiationUnderGCH.html

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.
 
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Yep, it can.

Here's another proof using cardinal arithmetic:

|R|^|R| = (2^|N|)^|R| = 2^(|NxR|) =2^|R|
 
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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