An uncountable countable set
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From: Mike Kelly on 4 Oct 2006 08:19 Tony Orlow wrote: > Virgil wrote: > > In article <45215d2f(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Han de Bruijn wrote: > >>> stephen(a)nomail.com wrote: > > > >>>> how many balls are in the vase at noon? > >>>> > >>>> What does your "mathematics" say the answer to this > >>>> question is, in the "limit" as n approaches infinity? > >>> My mathematics says that it is an ill-posed question. And it doesn't > >>> give an answer to ill-posed questions. > >>> > >>> Han de Bruijn > >>> > >> Actually, that question is not ill-posed, and has a clear answer. The > >> vase will be empty, if there is any limit on the number of balls, and > >> balls can be removed before more balls are added, but it is not the > >> original problem, which states clearly that ten balls are inserted, > >> before each one that is removed. That's the salient property of the > >> gedanken. Any other scheme, such as labeling the balls and applying > >> transfinitology, violates this basic sequential property, and so is a ruse. > > > > One can pose any gedanken one likes. > > > > If TO does not like to be able to tell one ball from another, he does > > not have to play the game, but he should not ever try to pull that in > > games of pool or billiards. > > If distinguishing balls gives a less exact answer, Less exact how? >and a nonsensical one to boot It makes sense to me that if you put a ball into a vase and later remove it then it isn't there. It also makes sense to me that if you put a ball in a vase and don't remove it then it is still there. What *doesn't* make sense to me is that if you put some number of balls in a vase and remove them all then there are still some left. That seems to be what you are claiming. 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. >then that attention can be judged to be ill spent, and not > contributing to a solution at all. It is clear that sum(x=1->oo: 9) > diverges, is infinite, not 0. It's ridiculous to think otherwise. But the number of balls in the vase at noon *isn't* the limit of that sum, Tony. Nobody disagrees that that sum diverges (of course, we might disagree that it diverges to a "specific actually infinite value", but I digress...), people disagree that the limit of that sum is the same thing as the number of balls in the vase at noon. It seems a little barmy to spend a year arguing something that nobody disagrees with - that the sum 9, 18, 27... diverges. You should instead try to argue that the limit of that sum is equal to the number of balls in the vase at noon - that is what people are disagreeing with! Just saying "clearly" doesn't quite cut it. Here's my take on things... Problem : at one minute to noon, balls 1 thru 10 are added to the vase and ball 1 is removed. At half a minute to noon balls 11 thru 20 are added and ball 2 is removed. etc. Let noon = 0 and "one minute to noon" = -1. Let A(n,t) be 1 if the ball n is in the vase at time t, 0 if it is not in the vase at time t. Let B(n) be the time that the nth ball is added to the vase and C(n) be the time that it is removed. B(n) = -1/(2^(floor((n-1)/10))) C(n) = -1/(2^(n-1)) Note that B(n) and C(n) are strictly less than 0. Now A(n,t) = { 1 if B(n) <= t < C(n) 0 otherwise } Note that A(n,0) = 0. Let S(t) be the number of balls in the vase at time t. Then S(t) = { sum(n=1..) A(n,t) } Then S(0) = { sum (n=1..) A(n,0) } = { sum (n=1..) 0 } = 0 QED. Now I'm going to look a little at your argument, based on limits. Let's look at the sequence of S_ns where S_n = S(-1/(2^(n-1)) That is, S_n is the number of balls in the vase "just after" the nth iteration. Then, you say, the limit of the sequence of S_ns must equal S(0) (ie, the number of balls at noon) and the limit of the sequences of S_ns (9, 18, 27,...) is "positive infinity" and thus the number of balls at noon is infinite. But this is fallacious reasoning, Tony! What justification do you have for asserting that the limit of the S_ns is equal to S(0)? Presumably something along the lines of... + oo = lim((n->oo), S_n) = lim((n->oo), S(-1/(2^(n-1))) = S(lim(n->oo), -1/(2^(n-1))) = S(0) But this is not a valid manipulation of limits. lim(n->oo,S(T(n))) = S(lim(n->oo),T(n)) is true when S is continuous. But S is not everywhere continuous in our problem. So you have no justification for your assertion that S(0) is equal to the limit of the S_ns. -- mike.
From: Mike Kelly on 4 Oct 2006 08:24 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? > > Brian Chandler > http://imaginatorium.org Lunch doesn't exist. It's a wrong-headed attempt to extend what we know of actual eating to something that doesn't exist in reality. In fact, the whole concept of "meals" is ludicrous. You can approach a meal by individual bites, or even snacks, but you never eat an actual meal. -- mike.
From: Han de Bruijn on 4 Oct 2006 09:47 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
From: Tony Orlow on 4 Oct 2006 10:14 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. > > , but if there are an infinite number of >> naturals generating using increment, then there are naturals which are >> the result of infinite increments, > > 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. If every pair of naturals are within a finite difference of each other, Then there is no infinite count in either direction, quantitatively. 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. Countable infinity 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, and it rests on the notion that all naturals are finite, when proving its own premise. >>>>> 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. > >> 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. You can prove 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". Clearly lim(n->oo: n) is not less than oo. > >> lim(n->oo: n)=oo, 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. So, limits are nonsense? > > 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 misrepresentations. > > The inductive proof is an inherent part of the definition of naturals, > and has been so ever since Peano. It will not go away simply because > TO's intuition balks at it. Who's trying to make it go away? I'm trying to expand it a little. Why so against any innovation? >>>>>>> Von Neumann's N is such that >>>>>>> (1) {} is a member of N >>>>>> {} is the empty set, not a natural number. >>>>>> >>>>>>> (2) If x is a member of N then so is x \union {x} >>>>>>> (3) If S is an subset of N such that >>>>>>> (3.1) {} is a member of S, and >>>>>>> (3.2) if x is a member of S then so is x \union {x} >>>>>>> then S = N. >>>>>>> >>>>>>> And that is, by general consent, a satisfactory model for N. >>>>>> You may be satisfied with it. >>>>> Von Neumann was, and the vast majority of those who followed have been. >>>>> >>>>> That TO is not only bolsters my confidence in it. >>>> Good. >>>> >>>>>>> But that model cannot contain any of TO's allegedly infinite naturals. >>>>>> Well, it can. >>>>> Let S be the all finite naturals in vN , the set of von Neumann >>>>> naturals, then >>>>> (1) {} is a member of S >>>>> (2) if x is a member of S then so is its successor >>>>> Thus S = vN, and there is no room in vN for TO's illusionaries. >>>>>>> (1) there cannot be a first infinite natural, and, >>>>>> No, there can't, but you have a fallacious first infinite ordinal, so >>>>>> why complain? >>>>> Because TO is trying to put objects into vN that cannot be there. >>>> I reject the von Neumann ordinals as an inadequate model of the naturals. >>> The same result follows from the Peano rules, or any other set of rules >>> by which a standard set of naturals is defined. >>> >>> All versions of the standard set of naturals prohibit infinite naturals >>> by essentially the same argument. >> I reject it as circular. > > I reject TO as circular. >>> So TO will have to find some other set of rules which does not >>> explicitely and absolutely prohibit infinite naturals the way that all >>> standard systems do. >> Each of my T-riffics has a successor and predecessor. No discontinuiti
From: Tony Orlow on 4 Oct 2006 10:43
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. > >>> 2. The structure <w 0 S + *> is a model of first order PA, where w is >>> the set theoretic set of natural numbers, 0 is the set theoretic empty >>> set, S is the set theoretic successor operation on natural numbers, + >>> is the set theoretic addition operation on natural numbers, and * is >>> the set theoretic multiplication operation on natural numbers. >> Okay >> >>> 3. By Lowenheim-Skolem-Tarski, first order PA has uncountable models, >>> and Ax(x = 0 v Ey x=Sy) is a theorem of first order PA. But it is NOT >>> required that the model regard 'S' as standing for the usual set >>> theoretic successor function. So first order PA does not "define the >>> inductive set". >> It defines an inductive structure for the set, but not the meaning of >> successor. Still, there is no reason why we cannot apply the normal unit >> increment uncountable times. > > 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? Increment is a specific interpretation of successor, and the naturals a model of the Peano set based on this interpretation. > > > 3. In set theory, we define 'Peano system' in a way that captures > our >>> intuitive notion of the counting numbers starting with zero. And we >>> prove that all Peano systems are isomorphic. In particular, all Peano >>> systems have a denumerable carrier set. So even though first order PA >>> has uncountable models, such uncountable models are not part of any >>> Peano system. >> It sounds like convention more than anything. > > We define 'Peano structure'. I've already explained to you too many > times what a mathematical definition is. The definition of 'Peano > structure' does capture the intuitive notion of the counting numbers. > The rest of what I mentioned in that paragraph are theorems. > >>> 4. [I think the following is correct, and hope to be corrected if it is >>> not correct:] In Z set theory without the axiom schema of replacement, >>> it is underdermined whether there exist inductive sets that are >>> uncountable (where an inductive set is one that has 0 as a member and >>> is closed under the successor operation), but with the axiom schema of >>> replacement, there are proven to be inductive sets that are >>> uncountable. [I think this item 4. is is correct, and hope to be >>> corrected if it is not correct:] In any case, again, first order PA >>> does not define "the inductive set", since there may be many inductive >>> sets. w (which is the LEAST inductive set) is the carrier set of ONE OF >>> the models of first order PA, but w is defined in set theory, not in >>> first order PA. >> Yes. >> >>> 5. But, per item 3, no uncountable set, whether or inductive or not, is >>> a carrier set for a Peano system. That is to say, no, it is not true >>> that it is possible to have a set for a Peano system (note 'system', >>> not 'axioms') that has a successor operation "applied" an uncountable >>> number of times. And the notion of such a thing reduces to just another >>> variation on the unsupported and axiom-less confusion (suffered by many >>> cranks) that a limit ordinal, such as the first infinite ordinal, can >>> somehow (or must somehow) also be a successor. >> I am not talking about limit ordinals, but about an uncountable >> successorship. > > 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. > >> |