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Topic: Tim Chow in Forcing for dummies
Replies: 2   Last Post: Nov 9, 2017 4:45 AM

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Alan Smaill

Posts: 1,099
Registered: 1/29/05
Re: Tim Chow in Forcing for dummies
Posted: Nov 9, 2017 4:23 AM
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WM <wolfgang.mueckenheim@hs-augsburg.de> writes:

> Am Mittwoch, 8. November 2017 20:15:08 UTC+1 schrieb Alan Smaill:
>> 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).

>
> It talks about all subsets of Z. A subset is a combination or
> collection of elements of Z.


Yes; but as ever you miss the point;
it is the *converse* of that claim that you wrongly assume needs no proof.

>> If you want to say that is equivalent, *you* need to
>> justify that equivalence.

>
> On the contrary. Everybody who denies that the power set is
> uncountable has to show at least one subset of Z that is not a set.


More diversionary tactics.

Try and justify your own claim: the burden of proof is with you.

Try again.
Fail again.

"There is no need for proof": that was his catchphrase.


>
> Regards, WM


--
Alan Smaill



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