Cantor Confusion
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From: Virgil on 23 Oct 2006 15:33 In article <1161613540.075660.36660(a)k70g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > Of course we cannot get to mega by counting. According to set theory we > > > get to omega by the limit: > > > > No, according to most set theories, we "get to" omega by simply > > assuming it exists. > > We get omega by assuming it exists. We get to omega by limits. > > > > Set theory doesn't define limits at all. > > Set theory defines the limits of sequences in analysis. But not merely in terms of sets, but in terms of topologies. > > > Topology and analysis define > > limits. > > They are based upon set theory. But not on any set theory prior to analysis, the way "Mueckenh" implies. > > > > > > > > lim [n-->oo] {1,2,3,...,n} = N. > > > > Here, one assumes you mean pointwise convergerence; and of course you > > are explicitly /assuming/ that N /is/ a set - otherwise, N has no > > meaning. > > Of course. That is why and how we get to omega. That may be how "Mueckenh" gets it but everyone else gets it differently. > > > > > lim [n-->oo] {1 + 1/2^2 + 1/3^2 + ... + 1/n^2} = (pi^2)/6. > > > > Here, one assumes you mean using the usual metric topology; > > No. This formula stems from Leonhard Euler. I works without any > topology. At that time topology did not exist. Then does "Mueckenh" claim that that limit does NOT exist or does not have that value in the metric topology? The fact that much of analysis was worked out before metric topology was created does not mean that any of it is necessarily invalid when one does use the metric topology. > > > > > lim [t-->oo] X(t) = X(omega). > > > > > In what topology? Without specification, your statement is useless. > > Like that of Euler? Then show it, in as much detail as Euler showed the other limit, with or without more modern tools than Euler used. So whom do you accuse of saying otherwise? > > So, pi is a number; but it is not representable as a number? > > Pi is not a number, but an idea like beauty or justice which also are > not representable as numbers. Do beauty and justice fall between 3 and 4? > > Secondly, we don't require that the set of all natural numbers exist in > > order to define the limit of a sequence of rational numbers. We merely > > require a definition of what it means to be a natural number; so that > > we can say "n is a natural". Once we have done that, we can say things > > like "there is a natural m such that, for all n, if n is a natural with > > n > m, then f(n) < f(m)"; > > That yields only rational numbers. It does not yield limits like pi. Al > representations which are not actually infinite, i.e. have not omega > digits, are representations of rational numbers. One can define Cauchy convergence for sequences of rationals, then work with such sequences, or equivalence classes of them. > Simple physical experiments like counting III and II together yield the > foundations of mathematics. Advanced physical experiments may use > advanced mathematics. But the math comes first. > > > > > For matheology you need belief or axioms. > > > > In order to get out of bed in the morning you need beliefs and > > assumptions - you need to believe, for example, that the floor will > > support your weight, and that you will not instead plunge into a fiery > > hell. > > > > In order to have a reasonable mathematical discussion, we need to > > /agree/ on what we are talking about. Those agreements are codified by > > axioms; > > That is the present, deplorable opinion of the average mathematicians. Which is far better, and more productive, than the present deplorable opinions of "Mueckenh". > > > > I wouldn't neccessarily call ZFC - AoI "finite set theory". > > It is a theory without the actual infinite, a theory without omega. One > may execute any operation. The finite domain will never be left. Which > of the remaining axioms should yield infinity? None of them require it, but none of them forbid it either. > > > > We cannot deduce "there does not exist a Dedekind-infinite set" from > > ZFC - AoI; that too would need to be asserted as an axiom. > > Dedekind infinity does assume actually existing sets. It is the absence of Dedekind infinite sets that cannot be proved in ZFC - AoI. > > > > > The axiom of choice > > > then is true too although it is no longer needed. > > > > You are assuming that ZFC - AoI implies all sets are finite; that is > > not the case. > > How would you construct an actually infinite set? Pair, power, union? How would you prove that none exist? The whole point is that without an axiom either to force some infinite set into existence or to ban all infinite sets from existence, there is no way to tell whether any infinite sets can exist or not.
From: David Marcus on 23 Oct 2006 15:51 Han de Bruijn wrote: > David Marcus wrote: > > I don't think so. Bohmian Mechanics is 100% deterministic. All of the > > uncertainty in the results of an experiment is due to uncertainty in > > setting up the initial conditions of the experiment. > > Sure. Back to the dark ages of Laplacian determinism. Are you saying that we should reject a physical theory because of our philosophy? -- David Marcus
From: David Marcus on 23 Oct 2006 16:05 Sebastian Holzmann wrote: > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > Sebastian Holzmann wrote: > >> Let's assume that ZFC - AoI is consistent, otherwise the statement is > >> void. Let \phi_n be a sentence in the language of set theory that > >> encodes the statement "There exists a set that has at least n elements" > >> (you can formulate this one as a homework problem). > >> > >> Since for every finite subset of the set of \phi_n, there exists a model > >> of ZFC - AoI where this subset is true (any model of ZFC - AoI should do > >> the trick), by compactness, there exists a model of ZFC - AoI where > >> _all_ \phi_n are simultaneously true. ("True" might not be the correct > >> English word for it, please do someone correct me in this.) A set that > >> has, for any n, at least n elements cannot be finite. Therefore it must > >> be infinite. > > > > Are you sure this argument works in ZFC - AoI? I really don't know, but > > I suspect you might have trouble proving the compactness theorem. Also, > > the fact that there is a model where the set is infinite doesn't mean > > ZFC - AoI knows the set is infinite. > > ZFC is a FO theory. Compactness for FO has been established a long time > ago. I might have enormous problems trying to prove that there is a > model of ZFC (with or without AoI) and hopefully I'll never succeed. But > apart from that I don't see a problem. I certainly don't claim to be an expert on what you can prove without AoI, but here is what I was thinking: I don't see a problem arguing that a proof of a contradiction is finite, so if each finite subset of axioms is consistent, then the entire set of axioms is consistent. But, can you prove completeness without AoI? I.e., can you prove that if a set of formulas is consistent, then it has a model? I don't know. > If "ZFC - AoI knows the set is infinite" should mean "in this model > there is a bijection of this set to a proper subset", that, indeed, > might not be true (I haven't thought about this). But certainly this set > is not finite, as that would imply that it has at most n elements, for > some n in (a naively given) IN. I agree the set is not finite in the model, but if you really want to use ZFC - AoI as your foundation for all of your mathematical work, then the question is whether ZFC - AoI can prove the set is infinite. -- David Marcus
From: David Marcus on 23 Oct 2006 16:08 Dik T. Winter wrote: > In article <1161378001.475899.279610(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1161276322.150252.120060(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > Again replying to more than one article at once without giving proper > > > > > references. That makes it very difficult to follow threads. > > > > > > > > I am sorry, I have only 15 shots per several hours, but far more opponents. > > > > > > Why do you think that is the case? > > > > I have not the slightest idea. > > I have some ideas about it. You *never* provide definitions for the terms > you use, but the terms you use are not those from standard mathematics, so > there is a huge miscommunication. I would say there is *no* communication. -- David Marcus
From: Sebastian Holzmann on 23 Oct 2006 18:14
David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Sebastian Holzmann wrote: >> David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >> > Sebastian Holzmann wrote: >> >> Let's assume that ZFC - AoI is consistent, otherwise the statement is >> >> void. Let \phi_n be a sentence in the language of set theory that >> >> encodes the statement "There exists a set that has at least n elements" >> >> (you can formulate this one as a homework problem). >> >> >> >> Since for every finite subset of the set of \phi_n, there exists a model >> >> of ZFC - AoI where this subset is true (any model of ZFC - AoI should do >> >> the trick), by compactness, there exists a model of ZFC - AoI where >> >> _all_ \phi_n are simultaneously true. ("True" might not be the correct >> >> English word for it, please do someone correct me in this.) A set that >> >> has, for any n, at least n elements cannot be finite. Therefore it must >> >> be infinite. >> > >> > Are you sure this argument works in ZFC - AoI? I really don't know, but >> > I suspect you might have trouble proving the compactness theorem. Also, >> > the fact that there is a model where the set is infinite doesn't mean >> > ZFC - AoI knows the set is infinite. >> >> ZFC is a FO theory. Compactness for FO has been established a long time >> ago. I might have enormous problems trying to prove that there is a >> model of ZFC (with or without AoI) and hopefully I'll never succeed. But >> apart from that I don't see a problem. > > I certainly don't claim to be an expert on what you can prove without > AoI, but here is what I was thinking: I don't see a problem arguing that > a proof of a contradiction is finite, so if each finite subset of axioms > is consistent, then the entire set of axioms is consistent. But, can you > prove completeness without AoI? I.e., can you prove that if a set of > formulas is consistent, then it has a model? I don't know. Oh, I think I begin to see your problem here. But before we can speak of ZFC as a theory, we must first have some sort of "background set theory" available. And if we do not allow that background theory to "have" infinite sets (in some naive way), we cannot even formulate Z, because it consist of infinitely many sentences... > I agree the set is not finite in the model, but if you really want to > use ZFC - AoI as your foundation for all of your mathematical work, then > the question is whether ZFC - AoI can prove the set is infinite. I have no idea if there is any way to avoid a naive assumption on the existence of "sets" as bases of models of a theory, but I suppose there isn't. |