Re: A question regarding to a combination between Infinity and Choice Axioms.

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 08/15/04 

Date: Sun, 15 Aug 2004 12:34:08 -0500

namducnguyen wrote:
> For a given set x, let's define {}n(x) as:
>
> {}0(x) = x
> {}1(x) = {x}
> {}2(x) = {{x}}
> {}3(x) = {{{x}}}
> .
> {}n(x) = {{...{{x}}...}} [n times for left & right brackets.]
> .
>
> Furthermore, if y = {}n(x), n is called the "order" of y.
> Let w be defined as:
>
> 1) {} \e w
> 2) if x \e w then {}n(x) \e w [for some non-zero n]
> 3) in 2), the order of {}n(x) is strictly greater than that
> of x.
>
> Question 1: in ZF, w couldn't exist because of "choice"
> being implicated in 2), right?

Your clause (3) seems to be redundant; if it isn't then
I haven't understood it. Assuming I've understood correctly,
then ZF not only doesn't refute the existence of such a w,
ZF actually *proves* that there is such a w.

BTW your wording suggests that you think ZF refutes
choice. That's not so. ZF is agnostic about choice.

> Question 2: Could w exist in (a model of) ZFC?

Anything proved by ZF is also proved by ZFC, so ZFC
also proves that such a w exists. Thus such a w
(there's more than one; you haven't actually *defined* w)
exists in *every* model of ZFC.