Re: The set of necessary FISONs

Am Mon, 03 Mar 2025 17:46:43 +0100 schrieb WM:

On 03.03.2025 12:58, joes wrote:

Am Mon, 03 Mar 2025 09:48:16 +0100 schrieb WM:

Ah no, you're shifting the quantifier there.

That is allowed if Zermelo can do it.

He can't.

But it has been accepted that with all defined elements the set is
defined.

No, the set is not contained in itself.

By induction Zermelo produces the set. Quantifier shift, obviously
allowed by Zermelo. There is no other way to construct an infinite
set.

A quantifier shift is never a valid deduction (even though the
resultant sentence may be true otherwise).

Zermelo has no other deduction. The set is constructed by its elements.

Zermelo doesn't even need to shift quantifiers.

The set of sets of FISONs that can be removed together does not
contain the set of all FISONs (although it does contain the infinite
sets of the odd or even FISONs).

In exactly the same way as Z₀ is constructed by its elements, the set
of removable FISONs is constructed by its elements.

And in the same way Z_0 doesn't contain the set of all elements.

Of course not. It is the set of all elements.

And the set of removable sets of FISONs is just that and doesn't contain
the set of all FISONs (it contains every finite set of them).

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.