Cantor Confusion
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From: Lester Zick on 23 Oct 2006 10:43 On Mon, 23 Oct 2006 02:00:06 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >In article <1161518242.756958.103660(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: [. . .] > Virgils definition, although not wrong, definitions >can not be wrong, does not lead to desirable results. Since when can definitions not be wrong? ~v~~
From: Tony Orlow on 23 Oct 2006 10:51 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. > >> Topology and analysis define >> limits. > > They are based upon set theory. >>> 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. >>> 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. >>> lim [t-->oo] X(t) = X(omega). >>> >> In what topology? Without specification, your statement is useless. > > Like that of Euler? >> lim n->oo f(n) is /not/ in general defined as f(omega). If it were, >> then your first limit would not be N; it would be N union {N}. > > That observation shows but an inconsistency of set theory. Cantor, in > his first proof, used the notation a_oo. Later he chnaged oo to omega. >> For example, if we restrict ourself to the rationals only, then the >> sequence you mention above regarding pi^2/6 "converges" in some sense, > > It is called Cauchy-convergence. > >> but not to a rational number - the rational numbers are not complete >> with respect to the usual metric. > > Therefore the irrational numbers are not a subset of the rational > numbers. >>>> ZFC excluding AoI does /not/ say "you can get to omega" in the sense >>>> you are using "can get to" here. >>> Therefore ZFC excluding INF is not sufficient to prove the existence of >>> numerical representations of irrational numbers, which in fact do not >>> exist. Don't misunderstand me: The ratio of circumference to diameter >>> of a circle is pi. But pi is not representable as a number. >>> >> 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. >>>> Instead, ZFC /with/ AoI says, "since, >>>> if we are honest, we have to admit that you /can't/ 'get to' omega >>>> using the other axioms, we must therefore /assume/ omega's existence, >>>> in order to talk logically about arguments that assume omega is a set >>>> in the first place". >>> In order to assume that an irrational number does exist with its omega >>> digits in decimal representation. >> First of all, there are other ways to describe irrationals such as >> sqrt(2) which don't rely on describing it as the limit of a sequence of >> approximations. For example, we can describe it as "that number which, >> when squared, equals 2" or "the number which is the ratio of the >> diagonal of a unit square to the number 1". > > Yes. That, however, does not yield a decimal representation as needed > in Cantor's argument. >> 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. > > >>>> Of course that assumes that you have /some/ axioms in mind: otherwise >>>> it is simply not a mathematical question whether your statements >>>> actually follow, one from the other, in your argument. It is instead an >>>> argument of philosophy. >>> True mathematical questions can be answered by physical experiments. >> I find this philosophical stance quite puzzling, and extremely >> unsatisfying in its circularity; since generally physical experiments >> rely on deductions made using some pre-existing mathematical model. > > Simple physical experiments like counting III and II together yield the > foundations of mathematics. Advanced physical experiments may use > advanced mathematics. >>> 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. >> 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? >> 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. Otherwise you can > never be sure that all elements are included in a bijection with a > subset, because you can never be sure that all do exist. >>> 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? > They all stay in the finite domain if you start with existence of the > empty (or any other finite) set. Comprehension or replacement cannot go > further. So, how would you like to achieve it? > > Regards, WM > Inductive subdivision of the unit continuum? We certainly seem to be able to specify, or approximate arbitrarily closely, some values with infinite strings of digits. It seems obvious that any finite interval in the continuum has more than any finite number of points within
From: Randy Poe on 23 Oct 2006 11:48 Lester Zick wrote: > On Mon, 23 Oct 2006 02:00:06 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > wrote: > > >In article <1161518242.756958.103660(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > [. . .] > > > Virgils definition, although not wrong, definitions > >can not be wrong, does not lead to desirable results. > > Since when can definitions not be wrong? A definition is a statement by a person that a certain symbol will be used by them to stand for some concept. What is there in such a statement that can be wrong? Example: I will use the term "gleeb" to refer to an integer which is divisible by 2. How can that statement be wrong? How would you define "wrong" for such a statement? - Randy
From: David Marcus on 23 Oct 2006 11:54 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. Don't you mean Newton? -- David Marcus
From: David Marcus on 23 Oct 2006 12:05
Sebastian Holzmann wrote: > mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > > How would you construct an actually infinite set? Pair, power, union? > > They all stay in the finite domain if you start with existence of the > > empty (or any other finite) set. Comprehension or replacement cannot go > > further. So, how would you like to achieve it? > > 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. -- David Marcus |