Re: Aleph question
- From: "Guy L." <gihyunlee@xxxxxxxxx>
- Date: 8 May 2006 19:03:05 -0700
Thank you!
I am new to these groups things so sorry about the non-quoting.
I think I got this entire chapter messed up :-( I am going to go
re-study the text. It's tough studying because my high school teachers
have no clue... :-/
Arturo Magidin wrote:
In article <1147123641.709454.131050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Guy L. <@gmail.com> wrote:
Please learn to quote.
http://groups.google.com/support/bin/answer.py?answer=14213&topic=250
Would this solution also work:
No since, cofinality of 2^aleph_0 is uncountable by Konig's Thm but
cofinality of aleph_(w1*5+w) = aleph_(w1*5+w),
Do you know what "cofinality" means? The cofinality of aleph_{w1*5+w}
is not aleph_{w1*5+w}, it is in fact omega: the subset
{aleph_{w1*5+i} | i in omega} is cofinal in aleph_{w1*5+w}, and since
infinite cardinals are all limit ordinals, it cannot be any smaller
than omega.
therefore 2^aleph_0 cannot equal aleph_(w1*5+w)
Yes. 2^{aleph_0} cannot be a cardinal with cofinality omega. There are
other conditions that restrict possible values of 2^{aleph_0}, as I
recall (though I could be wrong), but this is the easiest one.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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