Re: Can you prove that 2^omega < 2^omega_1?
- From: rkam2001@xxxxxxxxxxx
- Date: 14 Apr 2005 16:32:55 -0700
Okay, I did some more searching on the internet, and found another old
post from this newsgroup about Luzin's conjecture and Easton's theorem
which says that 2^aleph_0 = 2^aleph_1 (cardinal exponentiation) is
independent of ZFC.
Thanks to everyone who posted and Stan Liou for the ordinal
exponentiation solution.
.
- References:
- Can you prove that 2^omega < 2^omega_1?
- From: rkam2001
- Re: Can you prove that 2^omega < 2^omega_1?
- From: Stan Liou
- Re: Can you prove that 2^omega < 2^omega_1?
- From: A N Niel
- Re: Can you prove that 2^omega < 2^omega_1?
- From: rkam2001
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