Re: An uncountable countable set

Virgil wrote:
In article <e8g7t4$ebo$1@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
Dave Seaman <dseaman@xxxxxxxxxxxx> wrote:

On Tue, 04 Jul 2006 21:31:16 -0600, Virgil wrote:

Also, in ZF, ZFC and NBG, any countable union of countable sets can be proved countable, so "mueckenh"'s attempts to claim otherwise must fail.
Not in ZF.

Is it the need for C?

Where i come from, Z(F)+countable choice was the standard. So if you used countable choice to prove something, that was accepted as 'not applying AC'. I've always liked this practice, and wondered (still wonder) why the rest of the world doesn't adopt it as well.
Countable choice is not implied by ZF, but of course it's much(?) weaker than full AC.

--
Cheers,
Herman Jurjus
.