Give me ONE digit position of any real that isn't computable up to that digit position!
in [Math]
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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 17:09 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
From: |-|ercules on 5 Jun 2010 21:07 "George Greene" <greeneg(a)email.unc.edu> wrote > 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. You could always state whether you agree with the proposition or not, and THEN argue WHY it holds or it doesn't. (NoBoxOfIndexes -> NoBijectionToPowerset -> HigherInfinity) -> (NoBoxOfIndexes -> HigherInfinity) You explicitly agree with what's inside the first set of parenthesis but not the second set of parenthesis. Herc
From: Transfer Principle on 5 Jun 2010 23:00
On Jun 5, 3:13 am, Ostap Bender <ostap_bender_1...(a)hotmail.com> wrote: > On Jun 5, 3:06 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > You're a demented idiot Barb. I disproved Halt, Turing, and Godel, and I have > > Cantor's proof down to.. > You think that "Halt" is the name of a scientist? It should be evident that by "Halt," Herc/Cooper means the Halting _Theorem_, especially since the word "Halt" is immediately followed by the name "Turing." Of course, this doesn't necessarily mean that Herc has actually disproved Turing's Halting Theorem in _ZFC_. > > Maybe you wouldn't all knock me if some of you ACKNOWLEDGED my theories.. Ah, _theories_. This seems to indicate that Herc/Cooper really is discussing more than one theory -- possibly a theory other than ZFC in which he has actually proved the negations of statements that are theorems in ZFC. This would put us in Case 1 of my four-case list. If this is the case, then Bender and the others are no longer justified in criticizing Herc just because he contradicts standard theory. And so let me do what Cooper requests of us, and at least _acknowledge_ his theories. So I ask of Cooper, how do your theories differ from ZFC? Do ZFC and Herc's theories have any axioms in common? Which axioms of ZFC aren't included in Herc's theories? There have always been posts criticizing Cantor and ZFC here on sci.math, but for some reason, ever since we turned the calendar to June, the number of such posts and threads has exploded! So let me try to keep up with all of these threads. |