An uncountable countable set
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From: imaginatorium on 4 Oct 2006 14:33 Ross A. Finlayson wrote: > Han de Bruijn wrote: > > imaginatorium(a)despammed.com wrote: > > > > > Han de Bruijn wrote: > > > > > >>The question is: how many balls are there in the vase at noon. > > >>This question is meaningless, because noon is never reached. > > > > > > Really? When's lunch, then? > > > > Time is _suggested_, but not present, in the Balls in a Vase problem. > > > > Han de Bruijn > > It certainly is, time. Uh, Ross, his name's "Han", not "time". I think you may have the "From" and "Subj" lines muddled up. Brian Chandler http://imaginatorium.org
From: imaginatorium on 4 Oct 2006 14:36 Lester Zick wrote: > On 3 Oct 2006 23:15:04 -0700, imaginatorium(a)despammed.com wrote: > > > > >David R Tribble wrote: > >> Tony Orlow wrote: <snip> > >> in fact. So how do you choose two points in the real number line > >> that "touch"? > > > >Me! Me!! I can do this one!! > > > >Er, well, here's how. You choose a first point (P). Easy, yes? <snop> Very disappointed with your answer, Lester. You didn't use the word "technically" even once. Brian Chandler http://imaginatorium.org
From: Virgil on 4 Oct 2006 14:43 In article <4523c1ac(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45231438(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > > > > > >>> The property of not being an infinite natural holds for the first > >>> natural, and holds for the successor of each non-infinite natural, so > >>> that it must hold for ALL naturals. > >> It holds for all finite naturals > > > > It holds for ALL naturals, as that " inductiveness" is an essential part > > of the definition of the naturals. > > > > Anything without that property, whatever it may be, is not the set of > > naturals, though it may contain the naturals as a proper subset. > > > > That's because the current view of the naturals is as containing only > finite values. And it is the view that will rule what is allowed to be call a natural number for a good deal longer than TO will live to dispute it. > > It does not follow that having an endless supply of something means that > > any of them are infinite. > > A contradiction follows from having an infinite number of elements each > separated from its closest neighbors by a unit of difference, without > there being two elements with an infinite difference between them. TO has often claimed this but never proven it. In mathematics, such unproven claims carry no weight. > If > every pair of naturals are within a finite difference of each other, > Then there is no infinite count in either direction, quantitatively. But every pair of (finite) reals is within a finite distance of each other, even though the set of reals is of more than any finite length. So TO is trying to say that the standard image of the naturals in the reals is more spread out that the set of reals which contains it. Just one more of TO's many impossible notions. > The > reason it fits the Dedekind definition is that there is no largest > finite, but with the identity relation between element count and value > in the reals, the set cannot be said to be quantitatively infinite, but > merely unbounded and denumerable. Quantitative infiniteness is a chimera. > Countable infinity is potential, not actual. TO's comprehension of mathematics and logic is potential, not actual. > > > > > TO has repeated that falsehood, many times but it remains false, and > > will continue to remain false however many times TO repeats it in future. > > > > According to every standard definition of natural numbers, the first one > > is finite, and the successor of any finite natural is also a finite > > natural, so that by induction (by definition, a necessary property of > > naturals) EVERY natural is finite. > > > > TO may choose to ignore this irrefutable proof that all naturals are > > finite, but he cannot refute it. > > That ignores sum(n=1->oo: 1)=oo There is no such sum. > and it rests on the notion that all > naturals are finite, when proving its own premise. It rests on the undeniable facts that (1) the first natural is finite and (2) the successor of any finite natural is also a finite natural, and those two facts, together with the inductive principle required of naturals, prove that every natural is a finite natural. Nothing that TO has said, or can say, can refute that fact. > > >>>>> ERGO: If the first natural is finite and the successor of a finite > >>>>> natural is a finite then every natural is finite. > >>>>> > >>>> Only for a finite number of successive increments. > >>> The inductive property says FOR EVERY NATURAL, idiot. The induction > >>> property does not exempt any naturals. > >>> > >>> If a property is true for the first and also for the such successor of > >>> every one for which it is true, then it is true for ALL naturals without > >>> exception. > >>> > >>> And being finite is such a property. > >>> > >>> So that TO's system violates the inductive property up front. > >> No, the inductive proof of an equality applies to all n, finite or > >> infinite. > > > > But in the process proves that none of them can be infinite. > > > > No, "is finite" is not an equality. Being finite is a property. And being a property is all that the inductive principle requires: if the first natural has some property and if the successor of every natural with that property also has that property, then every natural has that property. > > > > >> But "is finite" is an inequality, equivalent to "<oo". > > > > > > It is a property, which is all that is needed. > > Not when applying induction to the infinite case. Where in "If the first natural has some property and if the successor of every natural with that property also has that property, then every natural has that property." does TO find any exceptions for infinite cases? > You can prove > equalities or inequalities. The inductive property only involves properties. it does not mention either equalities or inequalities. > Equalities between expressions hold true in > the infinite case. Where inequalities are concerned, one must make sure > that the difference which establishes the inequality does not have a > limit of 0 as n->oo. Otherwise, your inequality holds only for finite n. > In this case, the inequality is "n<oo", where "oo" means "any infinite > number". As "finiteness" is a property not requiring either equalities or inequalities for its expression, none of TO's arguments are relevant. Clearly lim(n->oo: n) is not less than oo. As no such limit exists, neither is it greater than oo nor is it equal to oo. > > > >> lim(n->oo: n)=oo Wrong. No such limit exists, so it is not equal to anything. , not <oo. You can only increment a finite value a > >> finite number of times before you get infinite values out of it. > > > > I can. > > > > Nonsense is not a refutation of logic, TO. Why not? When TO uses it as his arguments so often, why should I be barred from using it, too? > > > > > ANY PROPERTY which holds for the first natural and which holds for the > > successor of every natural which has it, is a property that EVERY > > natural has. > > > > Being finite, or being less that 'oo', or not being infinite, are all > > such properties of EVERY natural, regardless of TO's
From: Lester Zick on 4 Oct 2006 14:52 On 4 Oct 2006 11:36:27 -0700, imaginatorium(a)despammed.com wrote: >Lester Zick wrote: >> On 3 Oct 2006 23:15:04 -0700, imaginatorium(a)despammed.com wrote: >> >> > >> >David R Tribble wrote: >> >> Tony Orlow wrote: > ><snip> > >> >> in fact. So how do you choose two points in the real number line >> >> that "touch"? >> > >> >Me! Me!! I can do this one!! >> > >> >Er, well, here's how. You choose a first point (P). Easy, yes? > ><snop> > >Very disappointed with your answer, Lester. You didn't use the word >"technically" even once. "Technically" is reserved for answers, Brian, not questions. And needless to say you've gotten so snippy with questions lately it's a little difficult to tell if any answers qualify for a "technically". ~v~~
From: Virgil on 4 Oct 2006 15:08
In article <4523c892(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> MoeBlee wrote: > >>> 1. Sets, countability, and uncountability, and even natural numbers are > >>> not mentioned in first order PA. The language of first PA does not > >>> include symbols - either primitive or defined - for those. > > > >> It mentions the set, as containing 0 > > > > NO IT DOES NOT. Please listen to me for a change. First order PA has no > > defined predicate for 'is a set' nor for 'containing' nor 'member' nor > > 'natural number'. > > It doesn't have primitive operators 'e'and 'succ'? > > > > >> and also the successor of any > >> member. It doesn't mention any limits on the number of iterations, nor > >> the operation that constitutes successor. It certainly doesn't mention > >> any predecessor discontinuities besides the initial one at 0. > > > > Yes, you're correct that first order PA DOES have 'S' which is > > primitive in the language of the theory and is read as 'successor of'. > > But, as to "limits on the number of iterations, nor > > the operation that constitutes successor. It certainly doesn't mention > > any predecessor discontinuities," whatever you mean by that, it is a > > theorem of first order PA that every object except [0] has a unique > > immediate predecessor (as 'immediate predecessor' is defined) and that > > there is no object between an object and its immediate predecessor nor > > between an object and its successor, and the theory doesn't have to > > address "limits on the number of iterations" for the theory to entail > > the theorems we've mentioned. > > Okay, I am realizing something. The limit of finiteness on the indexes > and/or values of the naturals follows from the Axiom of Induction. If it > is taken to apply to all naturals, and simply mentions properties in > general as being provable, then that leads to the naturals all being > finite, since otherwise there are properties one can prove inductively > that aren't "true", like 0.999...<1, and you wouldn't want that. > However, if the provable properties are taken to be either equalities or > inequalities, then the former can be said to be true in all cases, and > the latter only true in the infinite case if the difference between the > unequal expressions does not have a limit of 0 as n->oo. In this sense, > the axiom can be extended, and infinite values included in the set. Not without including TO's unnaturals in with the naturals. .... > > I understand that. But my point, even though you are blind (due to > > self-blindfolding) to this, is that, as far as set theory is concerned, > > your remarks reduce to just what I said they do. Aside from set theory, > > if your remarks are to be anything but babble, then please for > > godssakes tell us just what theory you are talking about. > > I'm trying to discuss it, but you don't seem to be able to start with > "increment is a form of successor". Peano doesn't specify what > "successor" is. When you define it as increment then you get the > naturals. Only one model of them, and anything unique to that model is an artifact of the model and not of the thing being modeled. >That's where count and measure tie together. And "measure" is an artifact of TO's model and is not inherent in the thing being modeled. > > > > I've already told you too many times that we can have inductive proof > > for MANY DIFFERENT kinds of sets. As to "what happens after an infinite > > number of increments" only you can tell us what theory you have in mind > > for which that locution makes whatever sense you would state. Anyway, > > the particular induction schema for w of course applies only to w and > > does prove that every member of w is finite. > > Based on the assumption that each is finite, sure. TO bases his theory on rejection of the inductive principle as it is stated for naturals, to be replaced by some monstrosity that allows To to decide when induction is allowed and when not. > > I have stated, again, above, what my proposed change to inductive proof > be for it to apply to the infinite case and infinite n. As it would gut the inductive principle, and allow intuition to replace proof as a methodology, no mathematician will accept it. > >> What does it lack as a Peano system? > > > > EVERYTHING that is required for a Peano system. That is, BOTH things > > required for a Peano system: > > > > A 1-1 function on the carrier set such that the function has as its > > range the entire carrier set except exactly one member of the carrier > > set. > > The carrier set being countably infinite w by definition. By proof, once the usual definition of "countably infinite" has been made and accepted. > > > > > Satisfaction of the induction schema. > > Like I said, that's what my rules about inductive proof are all about - > making room for infinite n. By eliminating the inductive schema. > > What structure of yours are you talking about? You just said in another > > post that EVERY T-riffic has a predecessor. Further, a Peano system > > also requires that is satisfy the induction schema. Your structure does > > not. > > Why not? It has a first, each has successor, and as far as proof is > concerned, it has specific rules about proof in the infinite case. But it violates an essential part of Peano structures. TO wants to build a brick house entirely out of wood, but still call it a brick house. > > > > But it is not even necessary to worry about any such particulars of > > your structure as long as YOU are correct that yours is a different > > kind of structure from <w 0 S>, since, again, if your structure > > satisfied a Peano system, then it would be isomorphic with <w 0 S>. > > Because that is how a Peano system is defined, based on w, no? Consider > it a Peano hyper-system, then. Not even close. How about calling it an Anti-Peano system? > > > > >>>>> Then it is not vN, and does not satisfy the Peano properties, and is > >>>>> irrelevant in any discussion of natural numbers. > >>>> Not really. > >>> Yes, really. Your set is not a carrier set for a Peano system. Your > >>> half-formed notion is not isomorphic with any Peano system. Your > >>> notion does not correspond to the common notion of plain ordinary > |