|
|
David C. Ullrich wrote:
> >So how did you get card(T) < card(P(S)) without the axiom of choice?
>
> Come now.
>
> Without AC I'm not sure what card even means. So forget that.
Without AC, I take card comparisons to be quick jotted notation for the
existence and non-existence of injections. In my original problem I
didn't mention card:
Show (in Z set theory, without choice):
AxEy(x equinumerous with y & x disjoint from y).
Then, in my response to Rupert, I did use card, but I could have
written it out without card by asserting the existence or non-existence
of injections instead (that proof used the axiom of choice anyway, but,
if I'm not mistaken, it did not use the axiom schema of replacement so
I was still in Z not ZF). What I'm interested in is whether this is a
theorem of just Z - without choice, regularity, or replacement.
I'm not trying to give you a hard time. I just wanted to understand
whether you meant for your prove to rely upon choice.
> But come on now - all we need here is that there exists y such
> that y is not in T; the fact that card(T) < card(P(S)) was just
> supposed to be a convenient way to see there is such a thing,
> but in fact P(S) is irrelevant.
Then I don't understand the proof. I understand this:
> Theorem: For every T there exists y such that y is not an element of
> T.
>
> Proof: Surely you've seen the proof of this? QED.
Of course, and I used it, though unstated, in my proof of the finite
case. And I understand your y not in T. And, by filling in the step I
mentioned, I understand your original proof, but using the axiom of
choice. I just wanted to know if we can get the theorem without choice.
Thanks,
MoeBlee
|
|