Cantor Confusion
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From: William Hughes on 14 Dec 2006 07:18 Han de Bruijn wrote: > Virgil wrote: > > > In article <1166011032.914983.204230(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >>Han de Bruijn schrieb: > >> > >>>William Hughes wrote: > >>> > >>>>Tony Orlow wrote: > >>>> > >>>>>Well, the proof is simple. Any finite number of subdivisions of any > >>>>>finite interval will only identify a finite number of real midpoints in > >>>>>that interval, between any two of which will remain more real midpoints. > >>>>>Therefore, there are more than any finite number of real points in the > >>>>>interval. > >>>> > >>>>This just shows that the number of real points is unbounded. > >>>>It does not show it is infinite (unless of course you use the > >>>>fact that any unbounded set of natural numbers is infinite). > >>> > >>>Isn't unbounded the same as infinite, i.e. = not finite = unlimited = > >>>without a limit? > >> > >>Unbounded is potentially infinite but it is not necessarily actually > >>infinite. > > > > AS in honest mathematics, the one implies the other, in honest math one > > has neither or one has both. In ZFC and NBG, both. > > Let's repeat the question. Does there exist more than _one_ concept of > infinity? Isn't unbounded the same as infinite = not finite = unlimited > = without a limit? Please clarify to us what your "honest" thoughts are. You are *way* in deficit on clear answers. Try answering the following question with yes or no. Is there a largest natural number? - William Hughes
From: William Hughes on 14 Dec 2006 07:30 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > (It is contained in the union of all lines, but the > > > > union of all lines is not a line) > > > > > > That is a void assertion unless you can prove it by showing that > > > element by which the union differes from all the lines. > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > that is required is: > > Given any line there is an element of the diagonal not in THAT line. > > It is not requires that: > > There is an element of the diagonal that is not in any line. > > > For linear sets you cannot help yourself by stating that the diagonal > differs form line A by element b and from line B by element a, but a is > in A and b is in B. This outcome is wrong. > > Therefore your reasoning "there is an element of the diagonal not in > THAT line. It is not required that: There is an element of the diagonal > that is not in any line." is inapplicable for linear sets. You see it > best if you try to give an example using a finite element a or b. In every finite example the line that contains the diagonal is the last line. Your claim is that there is a line which contains the diagonal. Call it L_D. Question: "Is L_D the last line?" - William Hughes
From: Dik T. Winter on 14 Dec 2006 07:45 In article <1166090594.020341.42340(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > > > > So taking cardinality and limits can not be interchanged except in some > > > > > > particular cases. But that is not unprecedented in mathematics. > > > > > > limits and integrals can also not be interchanged except in particular > > > > > > cases. And so can the interchange is not in general passoble if one > > > > > > of the things you interchange is a limit. Even interchanging limits > > > > > > is not in general possible. Consider: > > > > > > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy > > > > > > > > > > True, but is it relevant? > > > > > > > > Yes, because the same holds for cardinality. > > > > > > Could you please explain what same holds for cardinality? > > > > If you read, it is right above this. > > I do not understand why you argue that in > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy = 0 = lim{y -> oo} lim{x -> oo} > (2x + 3y)/xy > interchanging limits is not possible. Ah, I made an error. I meant: lim{x -> oo} lim{y -> oo} (2x + 3y)/(x + y). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 14 Dec 2006 07:47 In article <1166090945.526859.60550(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > You misunderstood. What I explained to Dik is another mapping than that > > > discussed by us. It is a mapping of one single edge on the path > > > representing 1/3, the next edge is mapped on the path representing pi, > > > and so on. I claim that for every real number I can name an edge to be > > > mapped on this real number. > > > > Yes, you state that, and you can. But that does *not* provide a surjection > > from the edges to the real numbers. And you stated you had constructed > > such a mapping. > > There is a mapping. This is proven by the fact that there are more > edges than paths. But the latter is simply false. Moreover, you have stated that you had constructed a mapping. That statement was wrong? > Perhaps we cannot display it in the special way you > wish. But that is the same with the well ordering of the eals. There > you believe in indirect evidence. What indirect evidence do I believe in that case? > > > And noboy can disprove it by constructing a > > > diagonal number, because the tree contains them all. > > > > RIght. The disprove is in the fact that whenever I state a sequence of > > numbers that sequence inherently does not contain all real numbers. > > But the tree does. And the tree does not contain more paths than edges. > It is pitty that you cannot combine two thoughts. But you have not proven that the tree does not contain more paths than edges. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 14 Dec 2006 08:05
In article <1166092128.058778.127950(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > A surjection onto the paths covers all elements of the *range* = every > > > edge. It is not necessary that every edge is mapped on a path, to show > > > that there are not less edges than paths. It is only necessary that > > > there is no path without edge mapped on it. > > > > Yes. But you stated that you had constructed a surjection. And if you can > > not state to what number a particular edge maps you have *not* constructed > > a sufjection. So you now state that you did *not* construct a surjection? > > I proved the possibility of a surjection. I have not yet seen such a proof. > > But again your statement "it is only necessary that there is no path > > without edge mapped to it" is blatantly wrong. Consider the infinite > > set of decimal numbers. Map a digit (from 0 to 9) to a number when it > > has somewhere in its expansion that digit. If there are more numbers > > a digit maps to, use shares. By this reasoning (which is your reasoning) > > there is a surjection of the set of digits 0 to 9 to the decimal numbers. > > You have not understood, deplorably. The shares of every edge I use add > to 1 edge. And every share is mapped on one path only. And every path > gets as many shares to restate two full edges on its own. I have no idea of the meaning of the last sentence. > According to your proposal, every decimal number gets only some shares > of the digits 1 to 9, but not as many shares as one digit is divided > into. So not one full digit is mapped on a real number. Why not here? And why is that the case in your mapping? > > > If someone makes a binary tree of all real numbers and hands it out to > > > Cantor, then Cantor cannot construct a diagonal number which is not in > > > the tree, so no contradiction of completeness is possible. > > > > You assume that some way to disprove one assertion also will work to > > disprove another assertion. > > No I simply state that the diagonal proof fails in case of the tree. Yes, because there is no diagonal. > > > The problem with Cantor's theorem aleph0 < 2^aleph0 is that the > > > function f(x) = 2^x must be discontinuous: > > > For natural x = n we have 2^n < aleph0, > > > for the least infinite x = aleph0 we have 2^aleph > alep0. > > > So there is a gap. The values between aleph0 and 2^aleph0 cannot be > > > taken by the function 2^x. > > > > Yes? What is the problem with that? There are enough naturals that > > cannot be taken by that function either. > > We can simply count the edges. They are countable. So we an count the > biginnings of separated parts of paths (because every edge is a > beginnng of the separated part of a path, notwithstanding which it may > be). Therefore the beginnings of separated parts of the paths are > countable. In the whole tree we have an uncountable set of separated > parts of paths. The tree is continuous. Therefore your arguing fails. This is close to poetry. I do not understand this at all. > > Again, this makes no sense to me. > > Look at the edges. Do you deny that every edge is the beginning of that > part of a path where it is separated from other paths? Right. > Do you believe that paths beginn to run separated wihout any edge being > inolved? No. And so what? But only looking at the paths that way you never get the full infinite paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |