Cantor Confusion
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From: mueckenh on 1 Feb 2007 11:50 On 1 Feb., 14:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170328870.612961.224...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: Continued: > > > > Why do you accept then that the union of sets of nodes (finite trees) > > establishes an infinite set of nodes (an infinite tree) without such > > an infinite set (infinite tree) being contained in the union? > > Because that is something different. The union of sets of paths > establishes an infinite set of paths. But that infinite set of > paths does not contain an infinite path, because none of the > constituent sets contains one. So the union of finite chains of nodes like the following 0 12 6543 does not contain an infinite chain of nodes of the form 0 12 6543 789... ??? Frankly, I do not see much difference to: The union of finite paths contains an infinite path. > > To translate to sets of nodes: the union of the sets of nodes establishes > an infinite set of nodes that does not contain an infinite node. > > > > More formal, define > > > a path p_k as {n_0, n_1, n_2, n_k-1}, where the n_i are nodes. The sets > > > of paths from the finite trees contain p_k for all k. The union of the > > > trees contains a path {n_0, n_1, n_2, ...}, but the union of the sets of > > > paths does consist of: > > > { {n_0, n_1, n_2, ..., n_k-1} | n_0 is a node, k in N} > > > and so does not contain {n_0, n_1, n_2, ...}. So there is *no* infinite > > > path in the union of the sets of paths from finite trees. > > > > How then does this infinite path manage to enter the union of the > > finite trees? > > Because the set of paths of the union of finite trees is *not* contained in > the union of the sets of finite paths. And it is not contained in any of the finite trees. Why is it said to be contained in the union of all finite trees? > > > > > Because I need the identity of path-lengths and numbers for my > > > > argument. > > > > > > Well, it does not hold. If you measure the lengths as you did define, the > > > union of all lengths is not a length, because it is not a natural number. > > > > If all natural numbers are finite, their union as defined for > > pathlengths cannot be infinite. > > Why not? Because there cannot be more than infinitely many natural numbers. All those infinitely many numbers, however, are allegedly finite. > > > Except you assert that an infinite union requires a number of infinite > > size. > > It is not a natural number. So it cannot be in the set of natural numbers. > > > > > > > If T(oo) is constructed as the union of all finite trees T(k), > > > > > > then every path in P is a path which is in the union > > > > > > P(1) U P(2) U ... of the elements of P_C. > > > > > > > > > > Prove it. The union of P(i) contains finite paths only, *by the > > > > > definition*. > > > > > > > > The infinite tree contains finite trees only. Is it finite on that > > > > behalf? > > > > > > No. The union of finite sets can be infinite. But the union of sets > > > with finite elements does not contain an infinite element. > > > > Finite trees are sets with finite elements, namely the finite number > > of nodes. > > That number is not an element of a tree. The elements of the tree are > *nodes* (by your own definition). The union of the finite trees contains > finite elements only (i.e. nodes at a finite distance from the root). > But there are infinite sequences of such elements. Don't you accept that a finite tree simutaneously represents a finite chain of nodes? And that an infinite tree represents an infinite chain of nodes? > > > > And, as > > > your P(i) are sets with finite elements, their union does not contain > > > an infinite element. On the other hand, their union *is* an infinite > > > set. And so it is an infinite set of finite elements. > > > > The cardinal number is infinite, each pathlength is finite. > > Right. But you said that there is an infinite pathlength corresponding to an infinite set of paths. > > > The same you can require for the tree. A finite tree is nothing but a > > sequence of nodes (if you count them as Cantor did count the rational > > numbers: > > > > 0 > > 12 > > 6543 > > 789...) > > > > What was your argument distinguishing infinite unions of finite trees > > and finite paths, respectively? > > None. I distinguish unions of finite trees and unions of *sets* of > finite paths. Above you are using P(1) U P(2) U ..., where each > P(i) is a *set* of finite paths. What about unions of sets of finite trees? Would they work completely different from unions of sets of paths? In particular, would the union of sets of finite trees not contain or establish the inifnite tree? > > > > Wrong. Infinite paths do not have a path-length that is a natural number. > > > > Pathlengths ARE natural numbers. > > For finite paths. In all cases, by definition. Regards, WM
From: mueckenh on 1 Feb 2007 11:53 On 1 Feb., 13:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 1, 7:02 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 30 Jan., 17:43, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > On Jan 30, 8:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > You can try to utter again and agan this nonsense, but after a while > > > > aI will cease to reply. Every such thing including only natural > > > > numbers is covered by induction. All natural numbers are subject to > > > > induction. > > > > True, you can use induction to prove something about > > > any natural number. However, the question is: "Can > > > you use induction to prove something about a set > > > of natural numbers?" > > > This question has a trivial answer: > > If the set of all natural numbers is nothing than all natural numbers, > > then yes. > > If the set of all natural numbers is more than all natural numbers, > > then no. > > No there are two questions. > > i: Is X true for every element of the set A? > ii: Is X true for the set A. > > The two questions are different. > The answer to question ii can be No, even when > the answer to question i is yes. That difference concerning quantity is correct. But this concerns finite sets as well as infinite sets. > Your trivial answer only applies to question i. And only this question is of interest. For instance: Is every segment of the set of finite natural numbers a finite set? > > So you believe that type II sets have some esoteric supplement? > > No. The answer to question ii can be No, even if > there is no "esoteric supplement" (i.e. the answer to > question i is Yes). Yes. That is not in question, but the difference of answers for finite and infinite sets. > > We can prove that every number is a number while a set of several > > numbers is not a number. > > As you point out there is a difference between > the elements of a set and the set. > Yes. That is not in question. What I dispute is that the difference depends on the cardinal number of the set. > > There is no reason to distinguish between > > finite and infinite sets. > > > > By running induction longer and longer you can get more and more > > > sets of type I. By running induction for an infinite time you can > > > get all > > > sets of type I. > > > No. By two steps of induction, namely > > P(1) > > P(n) ==> P(n+1) > > you can prove P for all natural numbers and for all sets of natural > > number (except quantitative statements as I mentioned above). > > No you can prove things for all elements of all sets > of natural numbers. There is a difference between > proving that something is true of every element in a set A > and proving that something is true of a set A. Yes. That is not in question. What I dispute is that the difference (induction is possible or is not) depends on the cardinal number of the set. Regards, WM
From: William Hughes on 1 Feb 2007 12:44 On Feb 1, 11:53 am, mueck...(a)rz.fh-augsburg.de wrote: > On 1 Feb., 13:38, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 1, 7:02 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 30 Jan., 17:43, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Jan 30, 8:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > You can try to utter again and agan this nonsense, but after a while > > > > > aI will cease to reply. Every such thing including only natural > > > > > numbers is covered by induction. All natural numbers are subject to > > > > > induction. > > > > > True, you can use induction to prove something about > > > > any natural number. However, the question is: "Can > > > > you use induction to prove something about a set > > > > of natural numbers?" > > > > This question has a trivial answer: > > > If the set of all natural numbers is nothing than all natural numbers, > > > then yes. > > > If the set of all natural numbers is more than all natural numbers, > > > then no. > > > No there are two questions. > > > i: Is X true for every element of the set A? > > ii: Is X true for the set A. > > > The two questions are different. > > The answer to question ii can be No, even when > > the answer to question i is yes. > > That difference concerning quantity is correct. But this concerns > finite sets as well as infinite sets. > > > Your trivial answer only applies to question i. > > And only this question is of interest. For instance: Is every segment > of the set of finite natural numbers a finite set? > > > > So you believe that type II sets have some esoteric supplement? > > > No. The answer to question ii can be No, even if > > there is no "esoteric supplement" (i.e. the answer to > > question i is Yes). > > Yes. That is not in question, but the difference of answers for finite > and infinite sets. > > > > We can prove that every number is a number while a set of several > > > numbers is not a number. > > > As you point out there is a difference between > > the elements of a set and the set. > > Yes. That is not in question. What I dispute is that the difference > depends on the cardinal number of the set. It may or may not. For instance the question "Is the cardinal number of the set a (finite) natural number (if the set does not have a cardinal number then the answer is no)?" certainly depends on the cardinal number of the set. Sets of type I have a cardinal which is a (finite) natural number. Sets of type II do not. > > > > > > There is no reason to distinguish between > > > finite and infinite sets. > > > > > By running induction longer and longer you can get more and more > > > > sets of type I. By running induction for an infinite time you can > > > > get all > > > > sets of type I. > > > > No. By two steps of induction, namely > > > P(1) > > > P(n) ==> P(n+1) > > > you can prove P for all natural numbers and for all sets of natural > > > number (except quantitative statements as I mentioned above). > > > No you can prove things for all elements of all sets > > of natural numbers. There is a difference between > > proving that something is true of every element in a set A > > and proving that something is true of a set A. > > Yes. That is not in question. What I dispute is that the difference > (induction is possible or is not) depends on the cardinal number of > the set. Induction on the elements of A is always possible given any set A of natural numbers. However, induction may not be able to prove that something is true of A For example, every one of the elements of A is finite. This can be shown by induction. However, this does not show that the set A is not (potentially) infinite. - William Hughes
From: Andy Smith on 1 Feb 2007 12:58 Dave Seaman <dseaman(a)no.such.host> writes >On Tue, 30 Jan 2007 13:45:59 GMT, Andy Smith wrote: > >> Just following up the idea of countability. > >> Am I right in thinking that any set of "finitely describable" objects is >> necessarily countable? - on the grounds that "finitely describable" >> implies that there exists some (multi-dimensional) address space which >> uniquely defines an object, and that that address is indexable by a >> finite set of natural numbers, and then that that multi-dimensional >> address (as with the "standard arrangement" for counting the rationals) >> can be ordered such that each address has a unique index in the natural >> numbers? > >Yes. The number of "descriptions" is countable. > >> On that basis I would surmise that e.g. the set of all roots of all >> polynomials with rational coefficients are countable - is that correct? > >Yes. What you have just stated is that the set of algebraic numbers is >countable. We can make a slightly stronger statement: the set of >computable numbers is also countable. That includes all of the algebraic >numbers and also some transcendentals such as pi and e, plus all the >numbers that can be derived from them. > > Thanks for that. Actually I googled it shortly after asking the question. My rather woolly phrase "finitely describable" roughly corresponds to the well defined concept of computable. Anyway, OK, I am now happy that you can't have an injection from the set of the reals to the set of the natural numbers, so finally I understand that cardinality is a meaningful concept. So I understand successive orders of cardinality to mean that there is no injection from any set with greater cardinality into any set with a lower cardinality. Two questions ( I should read a book): 1) Is there a set with cardinality greater than N but less than R ? 2) I am (dimly aware) that mathematicians talk about an infinite range of cardinalities. Is that just an abstract concept, or can you point to some set and say e.g. that has a higher cardinality than the reals? regards -- Andy Smith
From: Michael Stemper on 1 Feb 2007 13:15
In article <kAau6RggoiwFFwKO(a)phoenixsystems.demon.co.uk>, Andy Smith writes: >Anyway, OK, I am now happy that you can't have an injection from the set >of the reals to the set of the natural numbers, so finally I understand >that cardinality is a meaningful concept. Now you learn (or at least read about) all kinds of strange ideas. > So I understand successive >orders of cardinality to mean that there is no injection from any set >with greater cardinality into any set with a lower cardinality. Good enough for government work. >Two questions ( I should read a book): Or several. >1) Is there a set with cardinality greater than N but less than R ? That depends. There's an axiom called "The Continuum Hypothesis", which answers that in the negative. This axiom is sometimes accepted, other times no. It is, like the Parallel Postulate, independent from the other axioms of set theory. There's also "The Generalized Continuum Hypothesis", which states that, for *any* set B, there is no set D such that |B| < |D| < |P(B)|. This is also independent of the other axioms of set theory. >2) I am (dimly aware) that mathematicians talk about an infinite range >of cardinalities. Is that just an abstract concept, or can you point to >some set and say e.g. that has a higher cardinality than the reals? From *any* set, E, you can create its power set, P(E), which is the set of all subsets of E. It's fairly simple to show that |E| < |P(E)|. -- Michael F. Stemper #include <Standard_Disclaimer> A preposition is something that you should never end a sentence with. |