Re: R^R = R^2 = R

On Wed, 28 Nov 2007 15:31:00 EST, tommy1729 wrote:
On Wed, 28 Nov 2007 14:16:57 EST, tommy1729 wrote:
David wrote :

( and DAVE seaman basicly wrote the same )


tommy1729 wrote:
set theory :

N = aleph_0

2^aleph_0 = 2^N = R = C = aleph_1

R = aleph_1

How do you know that?

***

what is aleph_2 ?

The smallest ordinal of cardinality greater than
alsph_1.

it has been said aleph_2 IS : R -> R or R ^ R.

Who said that?

LOL , the cantorians did !

dave seaman just asked me the same question !

yet he himself said this many times here on the
forum :)

That's a lie.

example :


http://mathforum.org/kb/message.jspa?messageID=6004726
&tstart=0

I did not use the word "aleph" at all in that
posting.

quote :

R^R is the set of all
mappings from R to R, and I gave you an example of
such a mapping (the
sin function).

end quote .

As I said. No aleph in sight.



--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>

but you said R -> R = R ^ R !!

I don't use "R -> R" when I mean the set of all functions from R to R.
That expression by itself has no meaning. It makes sense in a larger
context such as "f : R -> R", which means that f is a (single) function
from R to R.

and others said R ^ R > R

Yes, and so did I. That is not what I objected to.

( cantor's theorem implies this !! )

thus if R = aleph_1 then R ^ R > = aleph_2

The "if" clause is not satisfied in ZF, which is the set theory that I
and most others are using.

and thus it follows.

No, it doesn't.

--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
.

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