Cantor Confusion
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From: Newberry on 28 Dec 2006 09:30 Virgil wrote: > In article <1167281308.269041.104540(a)48g2000cwx.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > David Marcus wrote: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > I did not calculate |N|/|R| but edges per path. This calculation yields > > > > a real number for any finite path - and the infinite path is nothing > > > > else but the union of all finite paths. > > If all properties of the finite trees carry over to infinite trees that > infinite trees must be finite, as that is one of those properties. > > So that an argument that something holds for all finite cases can never > be sufficient to prove it in any infinite case. > > Any such proof would need far better arguments than any yet provided. > > > > > > And, as you've said many times, anything true for finite is also true > > > for infinite. Oh, wait. There is no theorem that says that. So, how does > > > the fact that the infinite path is the union of its finite segments tell > > > you anything about the edges per infinite path? > > > > Because there is a one-to-one correspondence between integeres and > > nodes of the infinite paths. And it can be proved that > > An(1 <= (2*2^n - 2)/2^n <= 2) > > That is for ALL levels in the infinite tree: #edges >= #paths. > > For NO "levels of an infinite tree" is it true, but it is true for > finite trees. It is true for ALL nodes, blue, green, red, finite, infinite, all nodes of the infinite path, simply all. > > > > > > > > -- > > > David Marcus
From: Virgil on 28 Dec 2006 13:41 In article <1167316205.832225.238540(a)48g2000cwx.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1167281308.269041.104540(a)48g2000cwx.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > David Marcus wrote: > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > I did not calculate |N|/|R| but edges per path. This calculation > > > > > yields > > > > > a real number for any finite path - and the infinite path is nothing > > > > > else but the union of all finite paths. > > > > If all properties of the finite trees carry over to infinite trees that > > infinite trees must be finite, as that is one of those properties. > > > > So that an argument that something holds for all finite cases can never > > be sufficient to prove it in any infinite case. > > > > Any such proof would need far better arguments than any yet provided. > > > > > > > > And, as you've said many times, anything true for finite is also true > > > > for infinite. Oh, wait. There is no theorem that says that. So, how > > > > does > > > > the fact that the infinite path is the union of its finite segments > > > > tell > > > > you anything about the edges per infinite path? > > > > > > Because there is a one-to-one correspondence between integeres and > > > nodes of the infinite paths. And it can be proved that > > > An(1 <= (2*2^n - 2)/2^n <= 2) > > > That is for ALL levels in the infinite tree: #edges >= #paths. > > > > For NO "levels of an infinite tree" is it true, but it is true for > > finite trees. > > It is true for ALL nodes, blue, green, red, finite, infinite, all nodes > of the infinite path, simply all. Then prove it. I can prove that in an infinite tree there are a countably infinity of nodes or edges but an uncountable infinite of paths. Since every edge terminates in a unique node, there is at least one node for each edge. One can biject the nodes with the naturals by: root node <--> 1 2nd level nodes <--> 2,3 3rd level nodes <--> 4...7 4th level nodes <--> 8...15 ... nth level nodes <--> 2^n ...(2^(n+1)-1) ... One can biject the the set of infinite binary sequences with the set of infinite paths by matching each 0 in a sequence with a left branching edge of a path and 1 with a right branching edge. And while there are injections from the set of naturals to the set of binary sequences, a simple diagonal argument shows that there are no surjections from the set of naturals to the set of binary sequences. Thus there are "more" paths than nodes (or edges).
From: Newberry on 28 Dec 2006 15:38 Virgil wrote: > In article <1167316205.832225.238540(a)48g2000cwx.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > Virgil wrote: > > > In article <1167281308.269041.104540(a)48g2000cwx.googlegroups.com>, > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > David Marcus wrote: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > > > > I did not calculate |N|/|R| but edges per path. This calculation > > > > > > yields > > > > > > a real number for any finite path - and the infinite path is nothing > > > > > > else but the union of all finite paths. > > > > > > If all properties of the finite trees carry over to infinite trees that > > > infinite trees must be finite, as that is one of those properties. > > > > > > So that an argument that something holds for all finite cases can never > > > be sufficient to prove it in any infinite case. > > > > > > Any such proof would need far better arguments than any yet provided. > > > > > > > > > > And, as you've said many times, anything true for finite is also true > > > > > for infinite. Oh, wait. There is no theorem that says that. So, how > > > > > does > > > > > the fact that the infinite path is the union of its finite segments > > > > > tell > > > > > you anything about the edges per infinite path? > > > > > > > > Because there is a one-to-one correspondence between integeres and > > > > nodes of the infinite paths. And it can be proved that > > > > An(1 <= (2*2^n - 2)/2^n <= 2) > > > > That is for ALL levels in the infinite tree: #edges >= #paths. > > > > > > For NO "levels of an infinite tree" is it true, but it is true for > > > finite trees. > > > > It is true for ALL nodes, blue, green, red, finite, infinite, all nodes > > of the infinite path, simply all. > > Then prove it. Proof: 1) lim{n-->oo} (2*2^n - 2)/2^n = 2 2) for n = 1 we get (2*2^1 -2)/2^1 = 1 3) (2*2^n - 2)/2^n is a monotonic function of n Hence An(1 <= (2*2^n - 2)/2^n <= 2), that is for all n 1 <= #edges/#paths <= 2 All n means ALL levels and ALL edges in any infinite paths. Again, all means all regardless if it is left or right, terminal or nonterminal or what not. > > I can prove that in an infinite tree there are a countably infinity of > nodes or edges but an uncountable infinite of paths. > > Since every edge terminates in a unique node, there is at least one node > for each edge. > > One can biject the nodes with the naturals by: > root node <--> 1 > 2nd level nodes <--> 2,3 > 3rd level nodes <--> 4...7 > 4th level nodes <--> 8...15 > ... > nth level nodes <--> 2^n ...(2^(n+1)-1) > ... > > One can biject the the set of infinite binary sequences with the set of > infinite paths by matching each 0 in a sequence with a left branching > edge of a path and 1 with a right branching edge. > > And while there are injections from the set of naturals to the set of > binary sequences, a simple diagonal argument shows that there are no > surjections from the set of naturals to the set of binary sequences. > > Thus there are "more" paths than nodes (or edges).
From: Dik T. Winter on 28 Dec 2006 20:47 In article <1167254961.814852.121540(a)73g2000cwn.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > > Indeed, it can be done *when AC is true*. But perhaps we have a > > > > different view on the meaning of the sentence "can be done"? > > > > > > It has been proved that there is *no definable well-ordering* of the > > > reals. So it cannot be *done*. The only thing that can be done is to > > > prove that a well-ordering exists. Zermelo did it. > > > > Oh well, I was slightly incorrect. You need also V=L. > > You were incorrect - but not slightly. I were, slightly. > > "...; it is also possible to show that the ZFC axioms are not sufficient > > to prove the existence of a definiable (by a formula) well-order of the > > reals. However it is consistent with ZFC that a definable well- > > ordering of the reals exists -- > > I have not read the proof (because I don't like to spend my time with > useless topics), but a real expert of set theory told me that it has > been proved that there is no definable well ordering in ZFC Yes. Where is the contradiction with what I wrote? ZFC is not sufficient to give a definable well-ordering, but a definable well-ordering is not inconsistent with ZFC. > > it follows from ZFC+V=L that a particular formula > > well-orders the reals, or indeed any set." > > Instead of casting assertions: Why don't you simply give the formula > for the definable well ordering? The formla could be evaluated, and, > step by step, we would get a list of all reals. You are too lazy to look it up? If I understand it well enough, V=L immediately gives a well-ordering of the reals. But whatever, you stated that a definable well-ordering of the reals would lead to a contradiction in ZFC. But you have nowhere shown any proof of that, and it is false. It has been proven that a definable well-ordering of the reals is consistent with ZFC. -- 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 28 Dec 2006 21:09
In article <1167256020.615407.109810(a)42g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > The diagonal digits a_nn of Cantor's argument have to be > > > multiplierd by 10^-n in order to yield the diagonal number SUM a_nn * > > > 10^-n. But there is no problem with yielding zero for n --> oo? > > > --- Remarkable. > > > > Apparently you do not understand the working of limits. > > Apparently your limits work only for real numbers in lists but no for > real numbers in trees. Oh, they do. If the trees are properly defined. > > It has been > > proven that *every* Cauchy sequence yields a real number. And it is > > easy to see that *every* decimal expansion is in fact a Cauchy sequence. > > 10^-n is *never* zero. You need limits. > > Yes. And they say: As long as 10^-k is not zero, the sum {n=1 to k} a_n > + 10^-n is not an irrational number. Right. And as 10^-k is never zero, that sum is never an irrational number. You need limits to get irrational numbers. > > > There is no edge in the *finite* domain which is "passed in full". In > > > the infinite series there is an edge passed in full. > > > > Which edge is passed in full to the path that goes at each step to > > the left? > > I don't claim the existence of this representation of 0. Do you claim that is an answer to my question? Again, re-read my question. I am not talking about representations, I am talking about paths. You were talking about paths. Infinite paths. Hence my question. > > > Otherwise the path > > > does not exist at all. > > > > That is what you are trying to prove. But you can not just state it. You > > hav to *prove* it. > > I need not prove that two entities which have no different property, > cannot be distinguished. Your statement was that if there is no edge in the infinite series that is passed in full (meaning that the edge does belong only to that path), the path does not exist. That is what you have to prove. You again fail to do so. > > > The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no > > > smallest term 2^-n. > > > > Indeed. > > > > > It can be applied to calculate the edges per path. You should know that > > > absolutely converging series can be reordered. > > > > You should know that you can *not* reverse absolutely converging series. > > Proof? The infinite series has no last term, so the reversal does not have a first term, and so is not an infinite series. > > And that is what you are doing. Because that would result in a series > > without start. > > No reordering of an absolutely converging series changes its sum. But reversal makes it something quite different. What is the sum of the first 10 terms of the reversal of the above series? What is the sum of the first 20 terms? > > > It lasted rather a while. > > > > Consider it a complaint about your clarity and lack of definitions. > > Here are some readers who understood it quite easily. Perhaps. But that did not include me. > > > > But in the complete tree > > > >the parts of edges that are assigned to an infinite path are not 1, > > > >1/2, 1/4 etc. There is *no* edge that is completely assigned to a > > > > path, > > > > > > Then the tree is incomplete. You fail to understand infiniteness > > > already on this low level? > > > > Circular reasoning abounds here. Care to give a *proof* that the tree > > is not complete in that case? This statement is completely similar to > > the statement that the completed set of natural numbers should have a > > last element. > > There is no largest element, but the infinite path is nothing than the > union of all finite paths. Therefore we can calculate the union of all > finite sequences 1 + 1/2 + 1/4 + ...+1/2^n. Interesing. How do you calculate the union of sequences? But again my question: care to give a "proof" that the tree is not complete if there is no edge that is completely assigned to a path? > > > > Note that in each finite tree, the complete edge assigned to a path is > > > > the last edge on the path. As there is no last edge in an infinite > > > > path, this does not hold in the infinite tree. > > > > > > So there are different paths without different edges? > > > > No. I never did state that. > > So at least one path has its own edge? No, I never did state that. And it is false. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |