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Topic:
? 533 Proof
Replies:
8
Last Post:
Aug 4, 2014 12:48 PM




Re: ? 533 Proof
Posted:
Aug 2, 2014 8:32 PM


mueckenh@rz.fhaugsburg.de writes:
> On Saturday, 2 August 2014 00:34:22 UTC+2, Ben Bacarisse wrote: > > >> >> Ah. Can I just clarify: you don't, in fact, have an proposition of set >> >> theory, P, where you can prove both P and not(P)? >> >> > You think that the negation can be proved, don't you?. You think that >> > all rational numbers can be enumerated. You do not comprehend that >> > only every can be proved and that acceptance of "all" would require a >> > last one. >> >> Another question to add to the list that you won't answer. Are you >> embarrassed at having answered my question ("What actually *is* the >> provable statement whose negation is also provable?) with an example >> that you can't prove? >> > Look, I will it make as easy as possible for you. But it is not really > simple. So you might fail to understand. > > Consider the sequence a_n = {n}. It describes the super task, also > visualized in my lesson GU12 > http://www.hsaugsburg.de/~mueckenh/GU/GU12.PPT#394 that every natural > number is put into a box and then removed. What is the limit? > According to set theory, the limit is the empty set. But this is not > true for any finite step n. Therefore the limit must describe the > state of the box after every finite step. This implies that all finite > steps exis  actual infinity. > > Assuming this kind of infinity, you get from set theory the result > that all rationals can be enumerated. My proof shows that this is > wrong. My proof is independent of any assumption on infinity. It is > true for potential infinity (not all steps exist) and for actual > infinity (all steps exist). The result is then the limit of the > sequence s_n, which is also a set with infinite cardinality. > > So we have the result: Assuming actual infinity, then set theory is > right and there is a contradiction. Assuming not actual infinity, then > the settheoretic limit is wrong. My proof is valid in both cases.
So, no. No probable statement of set theory whose negation is also provable. Just withdraw the claim. Stop digging a deeper hole.
What you can prove (and have done repeatedly) is that you are puzzled by s_n and it's limit. You can't prove P and not(P) for any settheoretic proposition P.
 Ben.



