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

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Ben Bacarisse

Posts: 1,972
Registered: 7/4/07
Re: ? 533 Proof
Posted: Aug 2, 2014 8:32 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply 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
> 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 set-theoretic 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 set-theoretic
proposition P.


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