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Topic: Tim Chow in Forcing for dummies
Replies: 3   Last Post: Nov 9, 2017 3:07 AM

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 Alan Smaill Posts: 1,103 Registered: 1/29/05
Re: Tim Chow in Forcing for dummies
Posted: Nov 8, 2017 2:12 PM

WM <wolfgang.mueckenheim@hs-augsburg.de> writes:

> Am Mittwoch, 8. November 2017 18:30:07 UTC+1 schrieb Alan Smaill:
>> WM <wolfgang.mueckenheim@hs-augsburg.de> writes:
>>

>> > Am Mittwoch, 8. November 2017 17:35:11 UTC+1 schrieb Alan Smaill:
>> >
>> >

>> >> Of course the axiom does not depend on the model.
>> >>
>> >> But the models simply have to respect the axiom, in that
>> >> all the collections *that happen to be sets* and
>> >> where every element is in X will be in the power set of X.

>> >
>> > The models have to respect these axioms (among others):
>> >
>> > Axiom VII. The domain contains at least a set Z which contains the
>> > null-set as an element and is such that each of its elements a is
>> > related to another element of the form {a}, or which with each of its
>> > elements a contains also the related set {a} as an element.
>> >
>> > Axiom IV. Every set T is related to a second set U(T) (the 'power
>> > set' of T), which contains all subsets of T and only those as
>> > elements.
>> >
>> > Therefore { }, {{ }}, {{{ }}}, ... happen to be sets in every
>> > model. And all combinations of these sets happen to be sets,

>>
>> Non sequitur.

>
> No sequentur required. Simply read Axiom IV.

Axiom IV does not talk about "all combinations of these sets";
it talks about sets (ie those combinations which happen to be sets).

If you want to say that is equivalent, *you* need to
justify that equivalence.
Hand-waving and reiteration do not constitute proof.

>>
>> Try and prove that in ZF --
>> you claim it holds in all models;

>
> *By definition* a model of a theory is a system that satisfies all
> axioms of the theory.

When in doubt repeat a banality.

You claim: "And all combinations of these sets happen to be sets,"

that is what you need to prove.

Try again.
Fail again.

>> it follows that it has an FOL
>> proof. Find that proof, and you have inconsistency of ZF.

>
> You mean I should prove that Zermelo's "all" means all?

No.

I mean that you should adopt the reasoning of ZF to show its
inconsistency. Your argument is so simple, what could the problem
possibly be in doing this?

> It is hardly
> possible to convince a gang of fools that they are fools.

Frege was convinced by Russell when his set theory was shown
to be inconsistent.

You claim not to understand why Russell would choose to produce
a proof in Frege's theory.

Another joke?

"No proof is necessary": that was his catchphrase.

> Regards, WM

--
Alan Smaill