Virgil
Posts:
311
Registered:
5/26/09


Re: Answer to Dik T. Winter
Posted:
Jun 15, 2009 11:00 PM


In article <30b4af790a18464aa2aca47933732167@x29g2000prf.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 15 Jun., 18:09, Owen Jacobson <angrybald...@gmail.com> wrote: > > On 20090614 14:41:40 0400, WM <mueck...@rz.fhaugsburg.de> said: > > > > > > > > > > > > > On 14 Jun., 17:31, Owen Jacobson <angrybald...@gmail.com> wrote: > > >> On 20090613 14:29:05 0400, WM <mueck...@rz.fhaugsburg.de> said: > > > > >>> We agree that in Cantor's diagonal argument, applied to real numbers, > > >>> the numbers are represented and identified solely by their digits. No > > >>> further information is available. > > > > >>> We assume that a real number p can be distinguished from a set Q of > > >>> real numbers q by general considerations, for instance, if p is a > > >>> transcendental number and Q consists of rational numbers q only. > > > > >>> Of course it would be impossible to distinguish p from all q, because > > >>> for every digit d_n of p, there is a number q that shares all digits > > >>> up to d_n with p. > > > > >> Wait, so you're arguing that you can't distinguish 1/pi (roughly > > >> 0.31830988...) from a set containing all rationals in [0, 1] because, > > >> for example, 3/10 is equal to 1/pi to the first decimal place, 31/100 > > >> is equal to the second, and so on? What about the nonzero difference > > >> between any one rational number and 1/pi? > > > > >> 1/pi  3/10 is greater than the rational 1/50. 1/pi  31/100 is > > >> greater than the rational 1/125. In fact, for each rational q in [0, > > >> 1], 1/pi differs from q by an amount greater than some other rational r > > >> in [0, 1], so we can distinguish 1/pi from every rational in [0, 1]. > > > > > If 1/pi exists as an actually infinite digit sequence. > > > > Since there is a function g from N to {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} > > that, for *every* element n of N, produces the n'th decimal place of > > 1/pi, it's safe to say that 1/pi has an infinite decimal > > representation. There's also a function g' from N to {0, 1} that > > produces, for *every* element n of N, the n'th bit of the binary > > representation of 1/pi. > > That does not help, because *every* element of N is at a finite place, > and for every element of N the sequence 1, ..., n is finite.
but for every n in N, there is a susequent finite place. > > > > > > > There are two statements: > > > 1) Path p can be distinguished from every path of the countable set P > > > that is used to construct the tree. > > > 2) Path p cannot be distinguished from every path of P. > > > > Reading upstream a bit, I think you've misapprehended the point people > > are making. If P is a set of paths, and the union of all elements of P > > is the maximal complete binary tree, then there can be paths in the > > union of all elements of P that are not in P. > > Magic?
Only to those without the wit to understand it. > > In fact there is no union. There is in ZF.
> We simply write the paths p_n of P in > slightly different form: The beginning > zeros of all paths are written only once, the following 1 or 0 also > are written only once each and so on. Note, we have not done anything > else but writing the list in slightly different form, saving some > ink. But it is now impossible to recover the original forms from their abbreviations, so it is no longer possible from those abbreviations to declare that a given path is or is not a member.
> In particular we have not added any new path.
But you have carefully obscured which ones are there. > > This yields the complete binary tree.
But not any maximal infinite binary tree.
> You are unable to distinguish > any binary sequence representing a real of the unit interval from > every path of that tree.
If the tree is a maximal infinite binary tree then every path is in it, but if it is any of your less than maximal trees, here may be paths missing. > > >If P is a countable set, > > then there are necessarily paths in the union of all elements of P that > > are not in P. > > Why should they? "Necessarily" would be correct if actual infinity > existed, but it necessarily doesn't.
It does in ZF, and as WM has no axiomatically satisfactory set theory to replace it with, not any internal proof of ZF's ncinsistency, I will continue to go with ZF. > > > > In fact, one such set P is the set of finite rooted paths, which is > > isomorphic to the set of nodes in the maximal complete binary tree. > > We'll adopt a notation where each node is labelled with the set of left > > (denoted as 0) and right (denoted as 1) branchings necessary to reach > > it, prefixed by a dot (.): > > > > . is the root node. > > .0 is the left child of the root node. > > .1 is the right child of the root node. > > .00 is the left child of the left child of the root node. > > .01 is the right child of the left child of the root node. > > .10 is the left child of the right child of the root node. > > .11 is the right chind of the right child of the root node. > > (And so on.) > > > > Because every node can be reached in only finitely many branchings (as > > a consequence of the tree being representable as an infinite set of > > finite paths), every node has a finite label. > > That is similar to every bit of 1/3 = 0.010101... in binary. Evere bit > has a finite label. > > > > We'll denote paths in the standard way, as sets of nodes: > > > > {., .0} is the path from the root to the leftmost rank 1 node. > > {., .0, .01} is the path from the root to the righthand child of the > > lefthand child of the root node. > > > > Finite rooted paths can also be identified, for convenience, by their > > terminal node: there is one and only one path from the root node to the > > node .001101, so writing out the complete path is pointless. However, > > *infinite* paths do not have a terminal node, so they must be described > > in some other way  for example, by a function from N to the nodes of > > the tree. > > > > A path exists in the tree if and only if (every node on that path is in > > the tree) and (the path is a connected, linear subgraph of the tree). > > Two paths A and B are equal if and only if the set of nodes in path A > > is equal to the set of nodes on path B. > > > > There is a function f from N to the set of nodes that, for every > > natural number n, produces the node labelled by the first n binary > > places of 1/pi's binary expansion. This function describes a path. This > > path is distinct from every path in the set of finite paths used to > > describe the tree, because for every path in that set, the path > > described by this function contains more nodes. > > And for *every* bit a_n of this infinite path there is a terminating > path in the tree that contains n^n^n more nodes.
Is this supposed to be relevant to anything?
> So your argument is insufficient.
Total non sequitur. It is WM's support of that claim which is insufficient. > > > > Informally, the path q from 1/pi under f is > > {., .0, .01, .010, .0101, .01010, .010100, .0101000, .01010001, > > .010100010, ...}. (Formally, it is only finitely describable as a > > function, not as an elementbyelement list of nodes.) > > This path is distinct from each finite path: > > {.} is missing the node .0, which is in q. > > {., .0} is missing the node .01, which is in q. > > {., .0, .01} is missing the node .010, which is in q. > > > > In fact, we can continue this indefinitely; none of the finite rooted > > paths from which the tree is derived contains every node in q. However, > > every node in q is in the tree, since every one of its infinitelymany > > nodes is reachable from the root node after finitely many branchings. > > So q is a path over the maximal complete binary tree even though the > > maximal complete binary tree can be derived from a set that does not > > contain q. > > No it can't, unless magic plays a role.
Whatever is sufficiently beyond the understanding of WM, he regards as magic.
> Every path in the tree is a > finite path.
Not by the standard definition of a (maximal) path. One can, in a maximal infinite binary tree, have all kinds of finite pathlets (finite linearly ordered rooted sets of nodes), but none of them are maximal so none of them are paths according to the definition requiring that they be maximal. > > > > This is equivalent to the argument from arithmetic that there is, for > > every rational x, a rational y that is closer to 1/pi than x, presented > > above. > > And for each of these claoser rationals is far away from pi.
That does not parse in English. > > > > > One of them is false, unless it is a magic tree where something > > > happens during construction. But I do not believe in magic, least in > > > mathematics.
WM declares magical whatever he does not comprehend. > > > > That a union of infinitely many distinct finite paths can contain > > infinite paths does not surprise me. > > There is no union.
There is in ZF, and WM cannot fault ZF.
> But even in set theory a union of sets should not > contain elements that are in none of the united sets.
Nor does it. > > > In ZFC, one formulation of the > > axiom of infinity takes exactly the form of an infinite union of finite > > elements. Only a union of finitely many finite elements or a (possibly, > > infinite) union of finitely many distinct finite elements ({1} U {1} U > > {1} U {1} U {1} U ... = {1}, for example), neither of which is > > sufficient to produce the maximal complete binary tree, is a finite set. > > > > Conversely, in a set theory with no infinite sets, there is also no > > binary tree containing the path q (from above). Both kinds of theory > > can be internally consistent, which is as close to "true" as anything > > in mathematics gets. > > Actual or finished infinity as close to false as anything can get, > within and outside of mathematics.
Not as close to false as are "potentially infinite sets" in any standard set theory. And, as yet, WM has not produced any set theory in which they are not false. > > Regards, WM
 Virgil

