Re: Non-aleph cardinals in set theory without axiom of choice?

On Dec 23, 1:10 am, mike3 <mike4...@xxxxxxxxx> wrote:
On Dec 22, 10:04 pm, Butch Malahide <fred.gal...@xxxxxxxxx> wrote:





On Dec 22, 9:40 pm, mike3 <mike4...@xxxxxxxxx> wrote:

On Dec 22, 8:35 pm, Dave Seaman <dsea...@xxxxxxxxxxxx> wrote:

On Sat, 22 Dec 2007 18:16:30 -0800 (PST), mike3 wrote:
Hi.
I saw this:
http://en.wikipedia.org/wiki/Cardinal_number
It says that "if the axiom of choice fails there are additional
infinite cardinals that are not alephs".
So, could one provide an example of a set that has such a non-aleph
cardinality, in set
theory where the axiom of choice does fail? If so, what is it?

If AC fails, then there are sets that cannot be well ordered.  No such
set can have an aleph cardinality, because each aleph is an ordinal and
therefore is a well ordered set.

Ah. What would be an example of such a set then?

The set of all real numbers?

But this doesn't seem to work. Wouldn't this imply the continuum
hypothesis is false, as then beth_1, the cardinal number of the
continuum, could not equal aleph_1, since it's not an aleph!
But that is independent of the Zermelo-Fraenkel axioms!

This would suggest at best we could say that whether or not beth_1 is
an aleph in ZF (note: no AC) is undecidable.

This suggests that you must be using some set theory other
than ZF, or ZF with some axiom that's not AC but you have
not specified, or your "example" simply doesn't cut it. I was
presuming that ZF set theory was under consideration, perhaps
I should've stated that up front...
.

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