An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Dik T. Winter on 7 Jul 2006 22:10 In article <J21Gtn.BCI(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: Let me expand on this. > In article <1152282364.618937.306970(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: You start with the list of natural numbers in unary notation: A0 = 0. A1 = 0.1 A2 = 0.11 A3 = 0.111 ... calculate the diagonal number K: K = 0.111... and claim that K equals An for some value of n. Note the following: > > It is not by pure accident that I chose 0.111... . This number has also > > a meaning as decimal representation of 1/9. So we can do the > > calculations. Indeed. Let's do the calculations, and let each An also have the meaning as decimal representation, so An = (1 - 10^(-n))/9 and now you claim that for some n (1 - 10^(-n))/9 eqals K = 1/9. If that were true, there must be some n such that 10^(-n) equals 0. > > may not, simply cannot occur. We get from (2) > > A p E n : p = n > > and by construction of n = 0.111...1 we obtain that all 1's in K > > coincide with the 1's of n. > > Right. But this does *not* prove that there is an n that is equal to K. > To reformulate clearly > For all p there is an n such that An[p] = K[p] This was not entirely correct: For all p there is an n0 such that for all n >= n0 An[p] = K[p]. > this does *not* imply that > There is an n such that for all p An[p] = K[p], > which is what you are arguing. > > Pray explain how you come (in logical steps) from the first statement > to the second. Note that when we expand the An's with an infinite (countable, mind) sequence of digits 0 we can do something different. After the decimal point we change each 0 to 1 and each 1 to 0. We get the following sequence: A0 = 0.11111111... A1 = 0.01111111... A2 = 0.00111111... and so K = 0.000000... Now we see: For all p there is an n0 such that for all n >= n0 An[p] = K[p]. and your claim is that from that it follows that There is an n such that for all p An[p] = K[p], But as An = 1/(9 . 10^n), this would imply that K (the limit of the sequence) must be in the list and for some n, An must be 0. So, are you now arguing that the limit of a sequence is an element of that sequence? Or what else? And if so, for what natural number (eh, finite...) is 1/(9 . 10^n) equal to 0? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 8 Jul 2006 14:09 Virgil schrieb: > In article <1152183861.949387.268390(a)m79g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > All sequences, even all single transpositions which I apply can be > > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > > OK? And as long as I can enumerate, the number reached is finite. OK? > > > > > > But the numbers which are reachable include values larger than any given > > > natural. > > > > How should that become possible??? > > It doesn't have to become, it is. Nonsense. No natural number can be larger than any given natural. If you have identified a number, than it is given and is not larger than itself. > > > Completely wrong. All I say is that every natural number is finite. > > And WE say that there are more than any finite number of them. > Or, more briefly but equivalently, that there are infinitely many of > them. Which includes only finite numbers. And only that counts. (Infinity could not count.) Regads, WM
From: mueckenh on 8 Jul 2006 14:13 Virgil schrieb: > > > > | > > o > > / \ > > o o > > / \ / \ > > > > When every path contains more than any finite number of edges, then > there are "more" paths than edges. Then there are many edges, of course. But why should there be many more paths? > > That is even harder logic. It is nonsense to believe that the above diagram would ever become invalid. But the diagram shows that there is no branching without two new edges. Hence there *cannot* be more paths than edges. It is simply impossible as long as paths consist of edges. That is hardest logic, so hard that even HE HIMSELF could not neglect it. But Virgil himself believes to succeed? > > I , and others, have previously demonstrated a bijection between the set > of all infinite paths in a complete binary tree and the set of all > subsets of N, P(N). > That would merely show that set theory is self-contradictory. > I, and others, have also demonstrated a bijection between the set of > edges (or the set of nodes) and N. That is trivial. > > So that "mueckenh"'s claim that there are "more" edges than paths, is a > claim that there is a surjection from N to P(N) (or equivalently a an > injection from P(N) to N) but no injection from N to P(N). > If I said "more" this was in the usual sense. According to set theory it should read "not less edges than paths". Regards, WM
From: mueckenh on 8 Jul 2006 14:28 Virgil schrieb: > In article <1152191557.066779.80730(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1152133409.564175.47720(a)p79g2000cwp.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > 0.1 > > > > > > > > 0.11 > > > > > > > > 0.111 > > > > > > > > ... > > > > > > > > 0.111...1 > > > > > > > > ... > > > > > > > > > > > > > > > > But in this list the number 0.111... is not contained. Hence not > > > > > > > > all > > > > > > > > of > > > > > > > > its digits can be identified. > > > > > > > > > > > > > > You have just proved that your example supports Cantor's theorem > > > > > > > by > > > > > > > producing a number, 0.111..., not in your own list. > > > > > > > > > > > > All digits of a number must be indexed by natural numbers. Otherwise > > > > > > they cannot be identified. All digits which can be identified are > > > > > > pesent in the list which contains all unary representations of > > > > > > naturlal > > > > > > numbers. > > > > > > Those are not natural numbers at all in the list, but decimal, or other > > > base, fractions. > > > > These are unary representations of natural numbers: 1 = 0.1, 2 = 0.11, > > 3 = 0.111, ... > > > What is your point in having only a sequence of natural numbers? > > No one has claimed that the set of naturals is uncountable and no one > has claimed that a sequence of distinct natural numbers converges to a > natural number limit, so that a Cantor "anti-diagonal", or a limit value > for a sequence of distinct naturals is irrelevant. If Cantor's proof fails once, it fails once and for all, because it is an impossibility proof. Due to the claim it is impossible to have the diagonal in the list. Regards, WM
From: Virgil on 8 Jul 2006 14:28
In article <1152382187.180478.193950(a)s13g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152183861.949387.268390(a)m79g2000cwm.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > All sequences, even all single transpositions which I apply can be > > > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > > > OK? And as long as I can enumerate, the number reached is finite. OK? > > > > > > > > But the numbers which are reachable include values larger than any given > > > > natural. > > > > > > How should that become possible??? > > > > It doesn't have to become, it is. > > Nonsense. No natural number can be larger than any given natural. If > you have identified a number, than it is given and is not larger than > itself. No one of us has said that. It is those who declare that the set of naturals is finite who are saying there is one natural which is larger than all others. What we have said, and what is quite true, is that for every natural there is a larger natural. > > > > > Completely wrong. All I say is that every natural number is finite. > > > > And WE say that there are more than any finite number of them. > > Or, more briefly but equivalently, that there are infinitely many of > > them. > > Which includes only finite numbers. And only that counts. (Infinity > could not count.) Right! There are more than any finite number of finite naturals, indeed, an endless supply of them which collectively form an infinite set. |