Cantor Confusion
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From: Michael Stemper on 25 Jan 2007 13:22 In article <virgil-5263D4.14184524012007(a)comcast.dca.giganews.com>, Virgil writes: >In article <ep7k43$79j$1(a)mailhub227.itcs.purdue.edu>, Dave Seaman <dseaman(a)no.such.host> wrote: >> On Wed, 24 Jan 2007 09:19:22 +0100, G Frege wrote: >> > "In the beginning God created the heavens and the earth. Now the earth >> > was formless and empty, darkness was over the surface of the deep, and >> > the Spirit of God was hovering over the waters." >> >> > This happened about 6000 years before Christ's birth, or so. >> >> It was supposed to be 4004 BC, according to Bishop Ussher. > >What o'clock? Not specified. It was either October 23, or the night before, based on what I can find. -- Michael F. Stemper #include <Standard_Disclaimer> You can lead a horse to water, but you can't make him talk like Mr. Ed by rubbing peanut butter on his gums.
From: imaginatorium on 25 Jan 2007 13:41 Andy Smith wrote: > G. Frege <nomail(a)invalid.?.invalid> writes > >On Thu, 25 Jan 2007 08:38:08 GMT, Andy Smith > ><Andy(a)phoenixsystems.co.uk> wrote: <snip banter> > Cantor says, you say, almost everybody says (apart from a few denizens > of this NG) that the reals are uncountable. I think that is clearly > right, but am not wholly convinced by Cantor's diagonalisation proof, > not least because I need to take on faith the validity of some of the > reasoning and defer my judgement to people more expert in these matters. You shouldn't need to. The argument is so simple you can check it yourself. > In any event, while Cantor's proof may show that the reals are > uncountable, it doesn't explain why - other than in terms of the proof. Oof! Well, supposing we prove that the angles of a triangle add up to 180 degrees. Now, *why* do the angles of a triangle add up to 180 degrees? It's extremely hard to attach meaning to such questions (and attempts to do so are not maths but philosophy... you know, the department without wastepaper baskets). > So I thought that it would be useful to consider a systematic scheme > that would count all the reals if they were countable i.e. for a given > finite integer n one can read out the corresponding real, and for a > given real determine its integer index. I suggested such a systematic > counting scheme, which, of course, fails - it fails because the reals > are intrinsically infinite, whereas the natural numbers just have no > greatest number. What does "intrinsically infinite" mean? In maths we define terms like "infinite" [of a set], and investigate the consequences of our definition. How do you define "infinite" so that the naturals are somehow only quasi-infinite?? Incidentally, having no largest member absolutely does not imply that a set is infinite: consider a platonic solid to be less than another if it has fewer faces, sides, and vertices. Then a tetrahedron is less than a dodecahedron, and so on. But this set has no largest member. (Yes, yes, yes, yes, yes, yes, yes, I know. This is "a different case". But mathematics is all about being formal: we do not cut slack, actually.) Brian Chandler http://imaginatorium.org
From: Virgil on 25 Jan 2007 13:55 In article <1169713155.574340.275560(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 14:41, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Jan 24, 6:58 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > I guess you have a different definition for the union of all finite > > > > trees. > > > > > > Let R be a set of finite trees with the property that: > > > > there is a (fixed) tree t_D in R, such that: > > > > if s is in R, then > > > > s is a subtree of t_D > > > > > > Definition i:Here you snipped both definitions that I gave without > > noting this fact. Naughty! I will restore this section. > > Sorry. Both definitions are not my definition and are of no relevance > in further discussion. > > > > Definition i: > > > > The union of all finite trees in R is > > the tree t_D. (Note that using this definition, > > the union of all finite trees in R is a finite tree.) > > > > Defintion i': > > > > The union of all finite trees in R > > is a set of paths, S, where S contains any > > path that is in a tree in R. (Note that i and i' > > are almost equivalent.) > > > > Now let W be the set of all finite trees. > > We know that there does not exist a (fixed) t_D > > in W such that every tree in W (that is > > every finite tree) is a subtree of t_D. So > > we cannot use definition i. Instead I use > > definition i'. By definition i', T1 contains > > every finite path. > > I do not use your definitions. Further you introduced new symbols. > which might confuse lurkers. Therefore I snipped them, now with notice. It seems much more likely that WM is protecting himself from the confusion of having to deal with what he does not understand. > > > > > > > The union of all finite treesis the tree which has all nodes and > > > > edges which are in at least one > > > finite tee.As I noted, you have a different definition for the union of > > > all finite > > trees. > > > > As before, let the set of all finite trees be W. > > Let P1 be the set of all finite paths, that is if p is > > an element of P1, then there is a tree t in W, such > > that p is a path in t. > > > > We will adopt your definiton of the union of all finite > > trees. Under this definition the union of all finite trees, T1, > > is the infinite tree T2. > > Of course. Denote it better by T(oo) = T. Except that WM's T has properties that he denies. > > > However, the set of paths in T1 is > > not the union of paths in P1. Each path in T1 is the limit of a > > sequence > > of paths from P1. Union and limit are two > > different things. > > > Due to the schism "infinite set of finite numbers" there are two > answers possible (but not more). There is one case that WM carefully ignores because it proves him wrong. > 1) If we identify path-lengths with natural numbers, then there is no > infinite path in any of the finite trees, because there is no infinite > natural number. Therefore, also the union of all finite trees cannot > contain an infinite path. > 2) If we identify path-length with sets of natural numbers, then there > are infinite paths in the union tree T(oo). However, if L is a set of > limits L_k > L = { L_k | k in N } > and S is the set of corresponding sequences > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > then L cannot be larger than S, because there cannot be more limits > than sequences. Different limits for one sequence would mean a path > splitting at a level n beyond any natural number, i.e., n not in N. WM carefully ignores the one case that proves him wrong, in which we identify paths with sets of head to tail connected edges, starting at the root node, including, where appropriate, endless chains of such connections. If one can have an endless chain of natural numbers there can be nothing to prevent an endless chain of edges. Wm is so caught up with his religious doctrines that he refuses to see anything which does snot conform to them. > This is not only impossible but would also lead to undefined limits. There are all sorts of "undefined" limits in analysis. Which such limits does WM's religion declare anathema? > > Therefore, even if union and limit are two different things (due to the > schism), both things lead to countably many paths in T(oo). Only if, as usual, ones religion blinds one to all those other uncountably many paths. > > > > Let us call the union > > of all finite trees (your definition) T1. > > I prefer T(oo). If it contains the edges of all finite binary trees then, as as complete tree, it contains uncounotably many paths. As Cantor and others have repeatedly demonstrated, and which WM has not been able to disprove without circularly assuming it. > > > P1 contains > > every finite path, so P1 contains 0.11, T1 does not contain > > 0.11, so P1 is not the same as T1. > > > Of course not. The union of all finite paths contains every path which > is contained in one of the trees, including 0.1 and 0.11 and 0.111 and > so on. But even this union (which I do NOT consider) is countable. Any complete (maximal) binary tree which contains all the edges of all finite binary trees will allow uncountably many paths. And since WM has often claimed that all allowable paths must be paths, he loses. > > > 4) The set of paths in T(oo) is a subset of the countable set of finite > > > sets of all paths in the finite trees. > > > No. Each path in T(oo) is the limit of a (potentially) infintie > > sequence. So each path in T(oo) corresponds to > > a (potentially) infinite set of paths in the finite trees. > > You can say so. But see above. The schism becomes clearly visible here. Only to those blinded by their religious beliefs. > Therefore, there cannot exist an actually infinite set of finite > numbers. As "actually infinite" is not mathematically defineable, it is not mathematically relevant. There are two mathematical definitions of a set being finite: (1) A set is finite if it bijects with the set of naturals less than some natural. (2) A set is finite if there does not exist any injection from the set to any of its proper subsets. A set is infinite (not finite) relative to one of these definitions if it fails to satisfy that definition of finiteness. There is no sort of set theory extant, at least outside of religions, in which a set is anything less that actual. > It would imply the actual infinity of the union of finite > paths, i.e., the acual infinity of at least one finite path. As "actual infiniteness" is outside of every nonsectarian/agnostic form of set theory, it is only religious nuts like WM who invoke it.
From: Virgil on 25 Jan 2007 13:59 In article <1169713460.685982.294480(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 15:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1169642941.554024.145...(a)k78g2000cwa.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > On 23 Jan., 19:27, Dave Seaman > > <dsea...(a)no.such.host> wrote: > > > > On Tue, 23 Jan 2007 17:15:52 GMT, Andy Smith wrote: > > > > > > > By definition, a set is "finite" if it has the size of some natural > > > > number. If a set isn't finite, then it's called "infinite". > > > > > > If a colour is not red, then it is called green by set theorists. > > > > Wrong. not-red. > > "Not red" is blue or green in the RGB axiom system. As there are all sorts of color systems, not-red can be any one of thousands of colors, or no color at all. > But you exclude > blue (potential infinity), hence you say green (green stands for > "hope", hoping on actually finishing infinity) when you can't see red. WM's religious monomania is now coloring his vision. And coloring it wrongly.
From: Andy Smith on 25 Jan 2007 14:28
IDavid Marcus <DavidMarcus(a)alumdotmit.edu> writes (snip) >> 1) Any systematic method for counting the reals is equivalent (you would >> probably say isomorphic under permutation). > >This is wrong or at least too vague to say whether it is right or wrong. >What do "systematic method" and "equivalent" mean? And, I don't know >what "isomorphic under permutation" means, so I wouldn't say that. > >Does the following qualify as a "systematic method" for enumerating the >reals? > >0 >0.1 >0.2 >... >0.9 >0.01 >0.02 >... >0.09 >0.11 >0.12 >... >0.19 >... Yes, that is exactly what I suggested - in binary that addressing is achieved by mirroring the bits about the binary point. Looks more straightforward as you have set it out. > >> 2) I proposed to try to count the reals by progressively increasing the >> degree of precision of their representation. To n bits precision, there >> are 2^n distinct numbers, requiring 2^n indices. If I wish I can >> represent a unique set of indices covering this space by e.g. reflecting >> the bit pattern of the reals to the first n bits about the binary point. >> >> 3) For all finite n, no problem. The set of indices are finite, and this >> basically is just indexing (a subset of) the rationals. However the >> reals require an infinite n and cannot be represented with indices with >> a finite number of bits. (Aristotle would say that the natural numbers >> are potentially infinite, but the reals are actually infinite). > >Is the following what you are saying? > >We can enumerate the reals by first enumerating the reals that take only >1 bit to write in binary, then the ones that 2 bits, then the ones that >take 3 bits, etc. This doesn't work because we never get to the ones >that take an infinite number of bits. > >If so, that's fine as far as it goes, but it doesn't tell us whether >some other method of enumerating the reals will work. > Well as I saw it the point is a) if you could index some permutation of the set of the reals you could necessarily index all of them - they would all be indexed, and you can permute the indices and their corresponding reals as you like, and b) if you can't index a given 1 permutation of the reals you can't index any. On reflection, maybe some finite/infinite dubious logic there... But, there is no scope for any compression of representation with the reals. If you have m bits, you have 2^m numbers, however you arrange them. Of course, if you take a subset e.g. the rationals (which are also reals with a necessarily infinite binary representation (so as to distinguish them from other members of the address space that they occupy)) then there is scope for compression, such that they do not have an infinite representation. e.g. you can represent a rational in [0,1] by <c>.<h><r> where <r> is a string of bits <c> long, repeating indefinitely, and <h> is some header string of bits. >> Incidentally, if you will excuse the cross-threading, re the >> space-filling curve, yes it fills the space, but with my variant, are >> the points in the plane painted white or black (at each stage of >> iteration by construction the white area = black area).? > >I'm sorry, but I didn't understand your variant well enough when I first >read it to answer your question. Please explain it again. > I just said, make the "generator" a thick line, such that the area of the line is the same as that of the enclosing space, with the line painted black and the external area painted white. You can consider such a generator as two lines of zero width delineating boundaries between black and white if you wish. The topology of inside and outside is preserved with each iteration of the fractal scheme, and on each iteration you have equal areas of black and white in the output. In the limit, any point on the plane is covered by both the lines defining the boundaries to the thick generator. So, is a point on the covered plane painted black or white? Regards -- Andy Smith |