# Stupid question

1. Jan 9, 2005

### phoenixthoth

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. Jan 9, 2005

### phoenixthoth

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!

3. Jan 10, 2005

### matt grime

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
4. Jan 10, 2005

### phoenixthoth

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
5. Jan 10, 2005

### Hurkyl

Staff Emeritus
Yep, it can.

Here's another proof using cardinal arithmetic:

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

6. Jan 10, 2005

### phoenixthoth

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.