Re: proving not(c = aleph_aleph_null) diagonally without choice

David Hartley wrote :

In message
<hYqdnYHlz-vqfp7UnZ2dnUVZ_u-dnZ2d@xxxxxxxxxxxx>,
galathaea
<galathaea@xxxxxxxxxx> writes
i am questioning your proof
because i have had poor choice-spotting
capabilities
and want them to improve

i don't trust my abilities yet
so i am blowing every step up
like an antonioni flic
and trying to get some kind of better understanding

The problem is that in blowing these steps up, you're
throwing away
information. aleph_0 x aleph_0 can be seen as an
example of a countable
union of countable sets but it's not just that. It's
equivalent to the
union of a specific collection of copies of a
specific set with all
necessary choice functions "built in". I'm sure all
these weird and
wonderful models you've been finding still have |N x
N| = |N|.
--
David Hartley

no that is is incorrect.

they have dimension build in.

its only a choice function if you assume axiom of countable choice or stronger.

aleph_2 * aleph_2 is square with side aleph_2 and area aleph_2.

no AC or ACC neccessary.

regards

tommy1729
.