An uncountable countable set
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From: MoeBlee on 4 Oct 2006 11:44 Tony Orlow 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'? It has the primitive 'S' (read as 'successor'), yes. It does NOT have 'e'. The usual primitives are: 0, S, +, *, as well as = from identity theory. Sometimes '<' is primitive, but more usually it's defined. But in any case 'e' is not a primitive and and not defined. First order PA does NOT have set theory in it. > > 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. You're realizing nothing but realizations that are reflections of your own previous misrealizations of your own reflections. Why don't you just read a book on this subject already? Oh, that's right, you don't need to because you've already realized by your reflections that set theory is a religion. > The limit of finiteness on the indexes > and/or values of the naturals follows from the Axiom of Induction. The axiom schema of induction in first order PA? It applies to ANY model of the axioms of first order PA, which includes models with universes that are not just made of natural numbers. > If it > is taken to apply to all naturals, It applies to anything in a universe of a model of the axioms. > and simply mentions properties in > general as being provable, then that leads to the naturals all being > finite, No, NOT properties in general. The induction axiom schema of first order PA has only formulas OF THE LANGUAGE of first order PA. It is just hopeless for you to be contemplating, pontificating, and other 'ating' (if you know what I mean) on this subject if you continue in your determination not to read a single damn book about it. > since otherwise there are properties one can prove inductively > that aren't "true", like 0.999...<1, and you wouldn't want that. What in the world are you talking about?!!! "0.999" We were talking about first order PA! > 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. Okay, I'm harsh with you sometimes. But I have to admit you really are cute when you pretend to drive your daddy's car. > > Your "inductive structure for the set" and "apply the normal unit > > increment uncountable times" is just T-mathsoundingspeak. It's not > > something you have shown to be statable in first order PA or in any > > particular mathematical theory. > > > > So, you can't read English? We're not talking about English! We're talking about the formal language of first order PA. Sheesh! > Increment is a specific interpretation of > successor, and the naturals a model of the Peano set based on this > interpretation. It's not a "specific interpretation" of any kind, since it does not involve any specific 1-place operation on any specific set. > > 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. That's where count and measure tie together. Just shut up for a minute and read a book on mathematical logic. I can't do ALL of that job for you. > > 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. No, on the assumption of the axioms of set theory. > I have stated, again, above, what my proposed change to inductive proof > be for it to apply to the infinite case and infinite n. I don't see any > worthy counterexamples, CHas' heroic efforts notwithstanding, so moving > ahead with that notion, it's obvious that we can then prove all sorts of > inequalities between formulaically mapped infinite sets. This produces a > rich system of infinities, along with which come a rich assortment of > infinitesimals. Very nice, if I do say so myself. No one doubts that you are enamored with your own thoughts. > >> 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 countabl
From: mueckenh on 4 Oct 2006 12:15 Dik T. Winter schrieb: > > > > Of course you can set up a bijection beween the sets > > > > k + omega = {-k, -k+1, -k+2, ... , 0, 1,2,3,...,} and > > > > -k + omega = {k+1, k+2, k+3, ...}. > > > > But that does not mean that both sets have the same number of elements. > > > > > > You are utterly confused. Ordinals are concerned with order preserving > > > bijections. A set with ordinal "k + omega" is, by definition of the > > > operator, the set that is formed from the elements of a set with ordinal > > > "k" followed by the elements of a set with ordinal "omega". > > > > You can read it above. There is the order preserving union of a set of > > k negative numbers and a set of omega natural numbers including zero. > > But that has ordinal k + omega, that is what I said! >not -k + omega. A set of k negative numbers > has ordinal k; not -k. But subtraction of a set of positive numbers from the set omega is expressed, as I did, by -k + omega. > > > > Now try to > > > apply that definition with -k, where k is positive. What is a set with > > > ordinal "-k"? Do you know how ordinal addition works? > > > > You can find it in Cantor's collected works, e.g., p. 201: "Die > > Subtraktion kann nach zwei Seiten hin betrachtet werden." And you can > > read above how it works in this special case. The first k elements are > > cut off. > > And that, again, is quite different from what you did write. But I will read > and see how Cantor does define omega - 1. A left-handed subtraction is > indeed possible, but that is not the same as a left-handed addition by a > negative number. No? Who decides that? You see, I knew already that, according to Cantor, subtraction is possible. If I express this as addition of negative, what do you think did I meant? > > > > Stated without proof at all. What is erroneous about my definition? > > > Do you assert that definitions can be erroneous? If so, why? Do you > > > think the definition > > > Let a be the number such that a = 4 and a = 5 > > > is erroneous? I think not. It is a proper definition, but there is just > > > no 'a' that satisfies the definition. > > > > It is erroneous, because you say let a *be* which is false, if a cannot > > *be*. > > That is just opinion. Like everything else. Regards, WM
From: MoeBlee on 4 Oct 2006 12:28 mueckenh(a)rz.fh-augsburg.de wrote: > But subtraction of a set of positive numbers from the set omega is > expressed, as I did, by -k + omega. Did I miss something? Did you actually give a mathematical definition of a 2-place operation to be applied to ordered pairs with any negative integer as the first coordinate and omega as the second coordinate? MoeBlee
From: Lester Zick on 4 Oct 2006 12:54 On 4 Oct 2006 05:19:38 -0700, "Mike Kelly" <mk4284(a)bris.ac.uk> wrote: >Tony Orlow wrote: [. . .] >Note : I agree with those who say it makes no sense in physical terms >to have an infinite number of balls. But mathematics is an idealisation >so it can make sense to talk about the infinite, even if it is >physically impossible. So physics has to make sense but math doesn't? I think you'd find plenty of quantum theorists, hyperdimensionalists and relativists who'd disagree. ~v~~
From: Lester Zick on 4 Oct 2006 12:56
On Wed, 04 Oct 2006 10:47:02 -0400, Tony Orlow <tony(a)lightlink.com> wrote: >David R Tribble wrote: >> Tony Orlow wrote: >>>> On the other hand >>>> I don't know why I said "neither can the reals". In any case, the only >>>> way the ordinals manage to be "well ordered" is because they're defined >>>> with predecessor discontinuities at the limit ordinals, including 0. >>>> That doesn't seem "real" >> >> Virgil wrote: >>>> In what sense of "real". There are subsets of the reals which are order >>>> isomorphic to every countable ordinal, including those with limit >>>> ordinals, so until one posits uncountable ordinals there are no problems. >> >> Tony Orlow wrote: >>> In the sense that the real world is continuous, and you don't just have >>> these beginnings with nothing before them. The real line is a line, with >>> each point touching two others. >> >> That's a neat trick, considering that between any two points there is >> always another point. An infinite number of points between any two, >> in fact. So how do you choose two points in the real number line >> that "touch"? >> > >They have to be infinitely close, so actually, they have an >infinitesimal segment between them. :) There you go, Tony. Now between points you're beginning to see some light. ~v~~ |