Re: R^R = R^2 = R
- From: tommy1729 <tommy1729@xxxxxxxxx>
- Date: Wed, 28 Nov 2007 15:31:00 EST
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 theforum :)
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 allsuch a mapping (the
mappings from R to R, and I gave you an example of
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 !!
and others said R ^ R > R
( cantor's theorem implies this !! )
thus if R = aleph_1 then R ^ R > = aleph_2
and thus it follows.
.
- Follow-Ups:
- Re: R^R = R^2 = R
- From: Dave Seaman
- Re: R^R = R^2 = R
- References:
- Re: R^R = R^2 = R
- From: Dave Seaman
- Re: R^R = R^2 = R
- Prev by Date: Re: Finding the remainder from very large division
- Next by Date: Any recommendations for affordable SPSS equivalents.
- Previous by thread: Re: R^R = R^2 = R
- Next by thread: Re: R^R = R^2 = R
- Index(es):
Relevant Pages
|
