Give me ONE digit position of any real that isn't computable up to that digit position!
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From: William Hughes on 5 Jun 2010 08:58 On Jun 5, 9:41 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? > > http://en.wikipedia.org/wiki/Cantor's_theorem Nope. There are other proofs. - William Hughes
From: George Greene on 5 Jun 2010 17:06 On Jun 5, 8:41 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? Why do you dismiss this as "verbiage"?? This is a PROOF! This "verbiage" isn't JUST "true" -- it's PROVABLE! However, you are confusing "higher infinities" with THIS proof. "This verbiage" IS a proof, so OF COURSE, IT'S provable. This proof, however, IS NOT a "proof of higher infinities". A proof of higher infinities would be an EXISTENCE proof. It would prove that higher infinities EXIST. THIS is a NON-existence proof. It proves that a bijection between a set and its powerset does NOT exist. And the set does NOT have to be N OR ANY INFINITE set, dumbass! The proof holds for ALL sets! > http://en.wikipedia.org/wiki/Cantor's_theorem > > Suppose that N is bijective with its power set P(N). Let us see a sample of what P(N) looks like: This is a poor use of citation. What you write here is not what is written at the link. But the mere fact that you are paraphrasing this in your own words PROVES you actually understand this proof. So, again, why are you dismissing it as "verbiage"?? > Given such a pairing, some natural numbers are paired with subsets that contain the very same number. And some are not, and the ones that are not ARE ALSO a subset. If EVERY subset had a number, then THIS subset would have a number that was both in AND not in this subset. Either you know a [short] contradiction when you see one, or you don't. Well, do you? > Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, > that there is a bijection between N and P(N). By deriving a contradiction from an EXISTENCE assumption (of a bijection), we have PROVED a NON-existence assertion: THERE IS NO bijection between a set and its powerset. The subset consisting of those elements not belonging to the subset- with-which-they-were-bijected is NOT in the range of the alleged bijection, so the allegation that it was a bijection is just false. FOR ALL sets AND ALL [in]jections on a set into its subsets/powerset! The fact that you keep talking about N and higher infinities really proves you don't get it. If you would just concede that you don't see how the NON-existence of a bijection winds up implying the EXISTENCE of a higher infinity, THEN we might get somewhere. But because you never make your objection specific (you just fling a bunch of poo around and then expect everybody to agree that it stinks), no argument is possible.
From: George Greene on 5 Jun 2010 17:08 On Jun 5, 8:58 am, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 5, 9:41 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > > > > Does the proof of higher infinities than 1,2,3...oo infinity rely on this verbage being true? > > >http://en.wikipedia.org/wiki/Cantor's_theorem > > Nope. There are other proofs. Well, there are not really any other CARDINAL proofs. There are ORDINAL infinities as well and THAT hierarchy can use a different proof. I.e., the collection of all countable ordinals is (obviously) not countable, since if it were, it would be a countable ordinal and therefore be a member of itself, which would violate the definition of "ordinal" and be a contradiction. But all of this is missing the point. You have get Herc to admit (at some point) that a PROOF IS a proof! until then, nothing is going to matter.
From: George Greene on 5 Jun 2010 17:09 On Jun 5, 7:34 am, William Hughes <wpihug...(a)hotmail.com> wrote: > However, y is not computable. There is no way to > combine all the finite f_k into a finite f which > produces y. Of course there is. You just take the limit.
From: William Hughes on 5 Jun 2010 17:17
On Jun 5, 6:09 pm, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 5, 7:34 am, William Hughes <wpihug...(a)hotmail.com> wrote: > > > However, y is not computable. There is no way to > > combine all the finite f_k into a finite f which > > produces y. > > Of course there is. You just take the limit. Well,in Wolkenmuekenheim a limit of finite things has to be finite, perhaps this is also true where you live (the Greene Hills ?). I live in a different universe. - William Hughes |