Re: 2^ aleph_x =/= aleph(x+1) !! aleph = beth !!



LOL !!!

so you all actually insist that 2^aleph_x =/=
aleph_(x+1) and i am " confused with the beth
numbers.

but APPARENTLY you guys are unaware that :

from the axiom of choice and the generalized
continuum hypothesis it follow that :

THE SEQUENCE OF BETH NUMBERS EQUALS THE SEQUENCE OF
ALEPH NUMBERS !!!!

thus 2^beth_x = beth_(x+1) and THUS 2^aleph_x =
aleph_(x+1)


aha !!

take that.


who is confused now !!!???

so its not me taking sugar every morning.

take your meds dennis, until you see AC and GCH imply
aleph = beth => 2^aleph_x = aleph_(x+1)

tommy1729

btw inaccessible cardinals like beth_beth_beth_3 cannot be proven to exist ( having the properties they should have ) in zf.

tommy1729
.



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