Why can no one in sci.math understand my simple point?
in [Logic]
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From: |-|ercules on 20 Jun 2010 21:51 "Virgil" <Virgil(a)home.esc> wrote >> Virgil's anti-diagonal construction >> BEGIN ... > Then the anti-diagonal function g:N --> {0,1} is defined by > g(n) = 1-F(n,n) in {0,1}, > so that g(n) <> F(n,n) for all n, > and thus g(--) <> F(n,--) for all n.. BRAVO! There it is! CONSTRUCT A NEW NUMBER g(n) = 1 - F(n,n) PROOF ITS A NEW NUMBER g(n) <> F(n,n) He didn't finish his HIGHER INFINITIES bit, let's add that... ( g(n)=1-F(n,n) -> g(n)<>F(n,n) ) -> HIGHER INFINITY Notice the lack of claims about new_digit_sequences as predicted, since they don't actually prove anything of the sort! Herc
From: |-|ercules on 20 Jun 2010 21:55 "Transfer Principle" <lwalke3(a)lausd.net> wrote > On Jun 20, 12:09 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "Sylvia Else" <syl...(a)not.here.invalid> wrote >> > I've told you which step I have reservations about. >> What? What a [expletive] womans answer that is. > > And so Herc plays the gender card here. > > I don't like it when anyone appeals to sexism to make an argument, > regardless of which side of the debate he is on. It's as wrong for > Cooper to make fun of Else because she's female as it is for > others to make fun of galathaea because she's female. > > I applaud Herc for standing up to those whom he considers to be > "religionists," but it hurts his cause when he makes sexist posts. But I thought it was a typical woman's copout. You know you aren't allowed to call colored people black any more? Watch "The Hangover" how the guy gets insulted over being called "black"! Herc
From: Tim Little on 20 Jun 2010 22:02 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Tim Little" <tim(a)little-possums.net> wrote in message >> 1) Recursive enumerability has nothing to do with it. > > So you assert. Personally, I think that the fact that the computable Reals > are not recursively enumerable is intimately related, as a set being > recursively enumerable in this context basically means the same thing as > there being a mapping from N to exactly that set, ie a list of elements. Look at the first theorem on http://en.wikipedia.org/wiki/Countable_set . I use Wikipedia since if I quoted from a textbook you might not be able to verify it. What you have just said as the definition for "recursively enumerable" is *exactly* equivalent to being a countable set! I suggest you look up the definition for a recursively enumerable set, and see how it differs from the definition of a countable set. >> 2) The word you should be using instead of "recursively enumerable" >> is "surjective". > > Well, recursively enumerable is a property of sets, and surjective is a > property of mappings, so dunno what you are talking about. Did you forget what you wrote? Look again at what you claimed: "What Cantor proved (in more modern parlance) is that there is no recursively enumerable function from N -> R". You were the one applying the term "recursively enumerable" to a function. I was merely correcting your mistake. > Well, that's very rude of you. Yes, it was. Polite correction of your errors wasn't getting anywhere, and you did not appear to be aware of how bad it looks to be declaiming "Cantor didn't prove that the reals were uncountable!" while obviously not grasping some of the most basic concepts in set theory, proof, and computability. - Tim
From: Jesse F. Hughes on 20 Jun 2010 22:02 Transfer Principle <lwalke3(a)lausd.net> writes: > On Jun 20, 8:51 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Newberry <newberr...(a)gmail.com> writes: >> > That's right. There is no formula or algorithm to construct the list. >> > It means that you would have to flip each and every digit one by one. >> > And that is impossible. >> Of course, it is perfectly reasonable to accept that there is no such >> list, because Newberry says so. > > Ah, so it's nice to see that Newberry has joined Herc and WM in > this ever-growing thread. Yes, how nice! Er, how come? Why do you think this is a nice thing? Honestly? It's not that I think it's a bad thing --- though, to be honest, I think Newberry's better than the argument he gives here --- but why do you think it's a positive development? > >> Cantor was this utterly insane freak who chose not to accept Newberry's >> word for it, and instead *prove* that there was no list of all real >> numbers. > > But this proof that there is no list of all real numbers must use > the axioms of some _theory_ (such as ZFC) -- a theory which > Newberry isn't required to accept just because Hughes says so. Of course he's not required to accept it. So? Nonetheless, it is a *theorem*, namely a theorem of ZF. > > So Hughes continues to be a Herc-religionist (i.e., a religionist > according to Herc), appealing to ZFC to prove Newberry wrong even > though there's no reason to assume that Newberry is working in ZFC (or > any other theory proving Cantor's Theorem). If Newberry means that Cantor's Theorem is not a Theorem in some other (specified) theory, then, of course, I'd have no opinion at all, until I saw the theory. So, let's ask Newberry! Newberry, what theory did you have in mind? -- Jesse F. Hughes "I'm not going to forget what I've seen. I understand the devastation requires more than one day's attention." -- G. W. Bush reassures Hurricane Katrina victims. Two days, minimum.
From: Tim Little on 20 Jun 2010 22:10
On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > I can't see how the move from addition of "computable" in front of "Reals" > requires any additional substitution in my "definition of a list". It's not just "definition of a list" in Cantor's proof. It's definition of a "list of reals". If you change that to "list of computable reals", then of course the definition changes. > I still can't accept that you are providing a list of computable > Reals when you aren't even telling me what the first one actually > is. That's cheating. I provide a mathematical definition for it. That is more than required, as Cantor's proof applies even to *undefinable* lists of reals. And yes, it is easily proven that there exist undefinable lists of computable reals. So if it is "cheating", it is only according to rules that never existed in the first place. - Tim |