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




Re: Tim Chow in Forcing for dummies
Posted:
Nov 9, 2017 4:23 AM


WM <wolfgang.mueckenheim@hsaugsburg.de> writes:
> Am Mittwoch, 8. November 2017 20:15:08 UTC+1 schrieb Alan Smaill: >> WM <wolfgang.mueckenheim@hsaugsburg.de> writes: >> >> > Am Mittwoch, 8. November 2017 18:30:07 UTC+1 schrieb Alan Smaill: >> >> WM <wolfgang.mueckenheim@hsaugsburg.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 >> >> > nullset 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



