Cantor Confusion
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From: Han de Bruijn on 23 Oct 2006 03:29 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. Han de Bruijn
From: jpalecek on 23 Oct 2006 10:00 mueckenh(a)rz.fh-augsburg.de wrote: > jpalecek(a)web.de schrieb: > > > mueckenh(a)rz.fh-augsburg.de napsal: > > > jpalecek(a)web.de schrieb: > > > > > > > > The set of constructible numbers is countable. Any diagonal number is a > > > > > constructed and hence constructible number. > > > > > > > > No. Definition (from MathWorld): Constructible number: A number which > > > > can be represented by a finite number of additions, subtractions, > > > > multiplications, divisions, and finite square root extractions of > > > > integers. > > > > > > > > How do you represent the diagonal number (which is sort of a limit of a > > > > series) > > > > via FINITE number of +,-,*,/,sqrt( ) ? > > > > > > Which digit should not be constructible by a finite number of > > > operations? > > > > No digit, but the number has to be constructed (according to MathWorld > > definition). So you must do something like > > > > diag=x^2+sqrt(y+z/x) > > Either such a formula or a full description like the diagonal number of > a list. The number of formulas and descriptions is countable. Or the least number which cannot be described by 20 english words? > > > > > > > > > > > Every list of reals can be shown incomplete in exactly the same way as > > > > > every list of contructible reals can be shown incomplete. > > > > > > > > No. > > > > > > A constructible number is a number which can be constructed. Definition > > > obtained from Fraenkel, Abraham A., Levy, Azriel: "Abstract Set > > > Theory" (1976), p. 54: "Why, then, the restriction to the digits 1 and > > > 2 in our proof? Just to kill the prejudice, found in some treatments of > > > the proof, as if the method were purely existential, i.e. as if the > > > proof, while showing that there exist decimals belonging to C but not > > > to C0, did not allow to construct such decimals." > > > > > > Definition (by me): A number which can be constructed like pi, sqrt(2) > > > or the diagonal of a list is that what I call constructible. If you > > > dislike that name, you may call these numbers oomflyties. Anyhow that > > > set is countable. And that set cannt be listed. Therefore the diagonal > > > proof shows that a set of countable numbers is uncountable. > > > > I like the name, but your "definition" isn't a definition. > > That depends on how you define "definition". For me it is a sufficient > definition. If not for you, then translate the quote of Fraenkel and > Levy as you like. If you do it without error, the result will be the > same. Anyhow, a diagonal number is a number which can be or has been > constructed. For me, a definition would be a predicate in the ZF language. What you have written, as well as what is written in Levy, Fraenkel, is not a definition. Moreover, if you think what I think you think, the term "constructible" cannot be defined in ZF language to mean what you think. Please note that I do not ask a precise formula, this is not a problem. The problem is, whether your term can be defined. If you don't give a definition, or predicate (if you want), then there is no set of your beasts. Therefore, asking about countability is irrelevant. Regards, JP
From: jpalecek on 23 Oct 2006 10:23 mueckenh(a)rz.fh-augsburg.de wrote: > jpalecek(a)web.de schrieb: > > > mueckenh(a)rz.fh-augsburg.de napsal: > > > jpalecek(a)web.de schrieb: > > > > > > > > > > > > > > > 0. > > > > > > > > > /a \ > > > > > > > > > 0 1 > > > > > > > > > /b \c / \ > > > > > > > > > 0 1 0 1 > > > > > > > > > ..................... > > > > > > > > > > > > > > > > > An edge is related to a set of path. If the paths, belonging to this > > > > > set, split in two different subsets, then the edge related to the > > > > > complete set is divided and half of that edge is related to each of the > > > > > two subsets. If it were important, which parts of the edges were > > > > > related, then we could denote this by "edge a splits into a_1 and a_2". > > > > > But because it is completely irrelevant which part of an edge is > > > > > related to which subset, we need not denote the fractions of the edges. > > > > > > > > Sorry, but your "proof" doesn't work. Imagine an infinite path in the > > > > tree. Which is the edge it inherits as a whole? Whenever you give me > > > > that edge, I can tell you're lying because if a path inherits an edge > > > > as a whole, it means that the path terminates by that edge. > > > > > > How should I be able to name the last term of a sum which has no last > > > > Indeed, INDEED. But that means, that there is no term 1 in the series, > > too. > > No. That does it not mean. Unless you agree that no real numbers do > exist. Come on, real numbers do exist. > > Moreover, it means that there is no positive term in the series. > > (Because > > 1 would be counted for the last edge, which does not exist, etc.). > > > > > term? But while we cannot name any individual edge we can prove: No > > > path splits into two paths without the supply of two new edges, one > > > edge for each path. This implies there cannot be less edges than paths. > > > > No. This holds for the finite case only. > > Why? If you follow a path you can see that where ever it splits off > another path, this splitting happens by an edge. As this situation does > never change, it remains in infinity for the whole path. For every > partial sequence (i.e. for every finite sequence of the path with n > edges, starting from the top), we have the "load of edges" accumulated: > > (1 - (1/2)^n)/(1 - 1/2). > > We cannot construct a finite sequence with less edges, even if we would > like to do so. In the limit n --> oo we obtain the whole load 2. This > is the correct calculation. All other arguing is nonsense, with or > without marriage. You cannot take the limit. By taking the limit, you say "All paths with however large, but FINITE length happen to behave somehow uniformly". There is nothing about infinite paths in the statement. Or otherwise, if there is a positive sum, there should be at least one positive term. You say you can't name it - but at the same time, you say that you can name all edges by their number. > > > Or the other way round: Assume there were more paths than edges, then > > > at least two paths could no be distinguished. (A path can be > > > distinguished from every other path by at least one edge.) > > > > Which doesn't mean anything. > > Oh yes, a different edge is obviously a necessary condition to > distinguish two paths. I've written it badly, I meant "Assume there were more paths than edges, then at least two paths could no be distinguished" statement is false. > > See the marriage theorem (or Hall's > > theorem) for the conditions you would have to fulfill to have an > > injective > > mapping of paths to their edges). And note, you cannot use marriage > > theorem to prove existence of any injective mapping of paths to edges). > > > > > > This is > > > > impossible for infinite paths. > > > > > > Of course that is impossible. Therefore the sum 1 + 1/2 + 1/4 + ... is > > > > This series is irrelevant. > > This kind of arguments accumulate here. > > > > > a infinite sum. But nevertheless your argument covers only half of the > > > story. Whenever you give me two infinite paths, I can name an edge > > > which belongs to only one of them. > > > > Again, this doesn't buy you anything. See marriage theorem. > > This kind of arguments accumulate here. > > > > > > The same argument applies to other terms > > > > in the sum. (That edge is inherited by an infinite path by 1/1024! > > > > Ok, but that means that the path terminates 10 levels lower). This > > > > means that infinite path inherit zero edges in your proof. > > > > > > Then the series 1 + 1/2 + 1/4 + ... contains zeros? > > > > No, that means that series is a wrong one for your problem. The correct > > one is 0+0+0+... > > The correct answer is then: there are no paths there are no real > numbers. The difference between your "series" and real series is Series is defined by its terms. Its value is defined by limit of partial sums. So for a series sum 1/2^n, we have s_1=1 s_2=1+1/2 s_n=1+1/2+...+1/2^(n-1) and finally s_oo=1+1/2+1/2^2+... But for the edges, we have (I write the "shares" on edges) path_1=root - 1 - terminates path_2=root - 1/2 - 1 - terminates path_n=root - 1/2^(n-1) - 1/2^(n-2) - ... 1 - terminates but for an infinite path, the share of every edge in finite depth is 0 (because thepath shares it with infinitely many other paths), so we have root - 0 - 0 - 0 - ... Or, put another way, "You cannot read infinite series backwards". Regards, JP
From: mueckenh on 23 Oct 2006 10:25 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
From: Sebastian Holzmann on 23 Oct 2006 10:41
mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: >> 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? Which should prohibit it? > 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. |