Cantor Confusion
in [Math]
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From: Virgil on 11 Apr 2007 14:30 In article <1176301248.833221.170050(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 10 Apr., 20:52, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Apr 10, 2:32 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > Do all paths exist separated from one another in the infinite binary > > > tree or not? > > > > Yes, but no level separates an uncountable number of paths. > > Why then do you answer "yes" while the answer is "no"? If no level of > infinitely many levels separates all paths, then not all separated > paths are separated in any level of the tree. Not being all separated at the same level does not mean that any path cannot be separated from any other at some level. What is this garbage about "separations" all having to be at the same level? Your argument is no more sensible than demanding that all decimal expansions must be cut off at some fixed number of decimal places, the same for all decimals. > > > > i. No level of the infinite binary tree separates an > > uncountable number of paths. But > > > > ii. The union of all countable levels is the union of > > countable levels. > > > > And we get the usual logical jump As the jump arrives at a falsehood, either it is not logical or some premise is false. It is the premise that separation's must all have occured at some level. In finite trees, paths separate at all sorts of levels. In a finite tree they must all be separated by the last level, as there is nowhere beyond that where they can separate, but in an infinite tree, there is no last level so that there are at any level always separations no yet made but which will be made in later levels. > > > > iii. The union of all countable levels does not separate > > an uncountable number of paths > > > > Which bit of "i and ii do not imply iii" do you fail to > > understand? > > I do not understand the following. If, as you seem to suggest, iii is > correctly spelled out: The union of all countably many levels > separates an uncountable number of paths. Where are the uncountably > many separation points = nodes? Why are you insisting on all those unneccessary nodes? At each node, the infinite set of paths through that node splits into two equally infinite sets of paths. This is equally true whether one claims the path sets are countably infinite or uncountably so. > The union of all levels contains only countably many nodes? Precisely! 2^(n-1) nodes at level n, counting the root node as level 1. But for each ordered partition of N, of form (L,R), where the union of L and R is N and their intersection is empty, there is a path which branches left or right at each level n e N according to whether n e L or n e R. Which make the set of paths equinumerous with P(N), the power set of N.
From: MoeBlee on 11 Apr 2007 14:49 On Apr 11, 11:05 am, Virgil <vir...(a)comcast.net> wrote: > Note that in mathematics one defines finiteness (of sets) in terms of > naturals, not the other way round. That is common but not universal. There are also treatments in which 'finite' is defined first and then 'natural number' is defined later. > A set is finite if and only if it can be put into bijection with the set > if naturals less than some fixed natural. And there are also definitions of 'finite' that do not require having first defined 'natural number'. MoeBlee
From: Virgil on 11 Apr 2007 15:26 In article <1176302166.249971.221780(a)q75g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 10 Apr., 22:02, Virgil <vir...(a)comcast.net> wrote: > > In article <1176230527.931928.12...(a)y5g2000hsa.googlegroups.com>, > > > > > All mathematics is equally virtual. Finite trees and their finite paths > > > > are as virtual as infinite trees and their infinite paths, as they exist > > > > only in imaginations. > > > > > Wrong. The finite tree > > > 0. > > > /\ > > > 0 1 > > > > > does exist here, on your screen. > > > > That is only a picture of a tree, and is no more an actual tree than a > > picture of a triangle is an actual triangle. > > > > 'Actual' trees, like 'actual' triangles are only actual in one's > > imagination. > > Without pictures or words, there would be no imagination of > mathematics. There have been blind geometers, who certainly did not rely on pictures. Without words there would certainly be no communication of mathematics, but whether that wold prevent imagination is not so obvious. The finite tree shown above is the best image one can > give, mental or electronical. Then that one is severely limited. > > > > Cantor's proof does not cover the number behind the last one proved. > > > > Since the proof covers all members of N simultaneously, there is no > > "last one" > > There is a last position to be recognized so far. Not when one has done ALL of them simultaneoulsy. > To imagine all > numbers simultaneously can only mean that your are incapable of > imagine anything at all. That your imagination is so limited, I agree, but your limitation is not universal. > > > > > Same as: > > > The binary tree does not contain all paths, because otherwise it must > > > contain a level with uncountably many nodes. > > > > Not in my world: > > (1) A tree without all paths is not actually a tree at all. > > > > (2) each partition of WM's set of countably many node levels into an > > ordered pair of two sets determines a path in which the first set of the > > pair contains those nodes levels from which the path branches left and > > the other contains the set of node levels from which the path branches > > right. This works equally well for finite or infinite trees. > > > > For a finite tree with n levels, excluding leaves, > > this produces 2^n paths. > > > > For an infinite tree with aleph_0 levels, there being no leaves, > > this produces 2^aleph_0 paths. > > How does the production produce separated paths without separation > points? WM seems to think that only one path get separated from all others at a separation point. That only happens at the parent of a leaf node, which can only happen in finite trees, never in infinite trees, which have no leaf nodes. Otherwise each node is the separation point of a set of several paths from another set of several paths. In finite trees such sets of paths are necessarily finite, but in infinite trees such sets are necessarily infinite. So that even if the set of all paths in an infinite tree were countable, his demand for "separation" of any one path from the collection of all others would be impossible. Thus WM's "separation" arguments are irrelevant to the cardinality of the set of all paths in a complete infinite binary tree. > > > > That's how it works in mathematics, including ZF and NBG and most other > > systems. > > Deplorable mathematics. It is the critic who is deplorable in his misunderstanding and misrepresenting of mathematics. > > > > If that is not how it works in WM's system, then WM's system is no part > > of any standard mathematics.- > > No these are unavoidable considerations the logic of which forces upon > us. As mathematics follows from and is entirely consistent with logic, but WM's presumptions are not, he speaks only for himself. For everyone else, avoiding such irrelevant considerateness is quite easy.
From: Virgil on 11 Apr 2007 16:53 In article <1176317382.263867.218110(a)l77g2000hsb.googlegroups.com>, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > On Apr 11, 11:05 am, Virgil <vir...(a)comcast.net> wrote: > > Note that in mathematics one defines finiteness (of sets) in terms of > > naturals, not the other way round. > > That is common but not universal. There are also treatments in which > 'finite' is defined first and then 'natural number' is defined later. The only other common definition if finiteness that I am aware of is the Dedekind one, which can be, but need not be, invoked before defining the naturals. Given the axiom of choice, the two definitions are equivalent. > > > A set is finite if and only if it can be put into bijection with the set > > of naturals less than some fixed natural. > > And there are also definitions of 'finite' that do not require having > first defined 'natural number'. Given the axiom of choice, one can take "A set is finite if and only if it can be put into bijection with the set of naturals less than some fixed natural" as a theorem wherever it is not taken as a definition. > > MoeBlee
From: MoeBlee on 11 Apr 2007 20:45
On Apr 11, 1:53 pm, Virgil <vir...(a)comcast.net> wrote: > In article <1176317382.263867.218...(a)l77g2000hsb.googlegroups.com>, > > "MoeBlee" <jazzm...(a)hotmail.com> wrote: > > On Apr 11, 11:05 am, Virgil <vir...(a)comcast.net> wrote: > > > Note that in mathematics one defines finiteness (of sets) in terms of > > > naturals, not the other way round. > > > That is common but not universal. There are also treatments in which > > 'finite' is defined first and then 'natural number' is defined later. > > The only other common definition if finiteness that I am aware of is the > Dedekind one, which can be, but need not be, invoked before defining the > naturals. Given the axiom of choice, the two definitions are equivalent. I don't claim anything about commonness. And what I'm talking about are equivalents not of Dedekind finiteness but indeed of finite in the sense of being 1-1 with a natural number. Definition: m is a subset-minimal member of S <-> (meS & AxeS ~ x is a proper subset of m) Definition: m is a subset-maximal member of S <-> (meS & AxeS ~ m is a proper subset of x) Let 'P' be the unary power set symbol and 'u' the binary union symbol. The following five formulas can be formulated in Z set theory even before defining 'is a natural number', and then after 'is a natural number' has been defined, each of the five can be proven (using at most the axioms of extensionality, schema of separation, power set, union, and pairing) equivalent to Efn(n is a natural number & f is a bijection between n and x), thus any of the five may also serve as a definition of 'x is finite' even before 'is a natural number' is defined: AS((S subset_of Px & ~S=0) -> Em m is a subset-minimal member of S) AS((S subset_of Px & ~S=0) -> Em m is a subset-maximal member of S) AK((0eK & Ayb((yex & b subset_of x & beK) -> bu{y}eK) -> xeK) AK((0ek & Ayex {y}eK & AbeK AceK buc e K) -> xeK) AK((K subset_of Px & 0ek & Ayex {y}eK & Abek AceK buc e K) -> K = Px) > Given the axiom of choice, one can take > "A set is finite if and only if it can be put into bijection > with the set of naturals less than some fixed natural" > as a theorem wherever it is not taken as a definition. That's correct (where in the above 'finite' means Dedekind finite) but it is a different matter from what I am speaking of. By the way, the axiom of denumerable choice (more usually called 'the axiom of countable choice') is sufficient for the equivalence betweeen finite and Dedekind finite, though,, just to be clear, I emphasize that that is not of import regarding the particular point I made about 'is finite' being definable (and not in the Dedekind finite sense) without requiring having first defined 'is a natural number'. We can define 'is finite' without having first defined 'is a natural number' and in such a way that after 'is a natural number' has been defined, the previous definitions (any of the five I mentioned) are equivalent to 'is 1-1 with some natural number'. MoeBlee |