Cantor Confusion
in [Math]
|
From: William Hughes on 21 May 2007 18:06 On May 20, 10:51 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 18 Mai, 01:13, William Hughes <wpihug...(a)hotmail.com> wrote: > > > Which of the following statments is wrong? <snip attempt to change the subject> The question now is Is every node of p in some element of R? For any natural number n, let z(n) be a zero node in position n. Let Z be the set of nodes, {z(n) | n a natural number} Which of the following statments is wrong? The set of nodes in p is Z Every element of Z is contained in a path in R. - William Hughes
From: Virgil on 21 May 2007 21:53 In article <1179777652.039258.88490(a)y2g2000prf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 20 Mai, 21:09, Virgil <vir...(a)comcast.net> wrote: > > In article <1179668114.743217.104...(a)p47g2000hsd.googlegroups.com>, > > > There cannot be in any set theory of repute any infinite set which is > > not actually infinite. > > > > Whatever WM insists on as being potentially but not actually infinite, > > it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, > > but not a set. > > So the union U(T(n)) of all finite trees T(n), formed in an obvious > way, is an actually infinite binary tree AIBTwith actually countable > sets of paths and nodes and edges. I will presume that "AIBT" should be "CIBT". and the set of paths is not countable, but otherwise, you are correct. > > Changing to the CIBT does not add any node and any edge, but extends > the set of infinite paths already present in the U(T(n)) to the > uncountable set of infinite paths. Paths consist of nodes and edges. > So nothing is actually changed. That is a theory of repute? Rather it > is a theory to refute! Except that WM is quite unable to refute such a theory without imposing assumptions quite false in ZF and NBG, in which the theory holds. > > ================================ > >> (2) ==> p is nothing but a union of finite paths. > > > Not neccessarily. That is the case only if one requires that a > > path be no more than a set of nodes. > > But every path has a soul in addition? It has a set of edges as well as a set of nodes. > > ================================= > > > On 20 Mai, 22:11, Virgil <vir...(a)comcast.net> wrote: > > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > Set theory is simply biased. Consider the list > > > > > > > 0.666... > > > > > 0.3666... > > > > > 0.33666... > > > > > 0.333666... > > > > > ... > > > > > > > If the diagonal number is defined by "replace 6 by 3", then we have > > > > > two answers none of which can be preferred by logic, but the second > > > > > of > > > > > which is suppressed by convention. > > > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > The following wm-proof certainly even in your opinion belongs to the > > > diagonal proofs considered by Cantor: > > > > > 0) mmm... > > > 1) wmmm... > > > 2) wwmmm... > > > 3) wwwmmm... > > > 4) wwwwmmm... > > > ... .......... > > > > > And if the list can be considered as a completed entity, then there > > > must be all natural numbers in the first column. > > > > As each "column" can contain only w's or m's, there are no natural > > numbers in any column. > > The first column consists of natural numbers only. They are not a part of the strings, but only arguments to the function which defines the list. > > > > The diagonal, of all w's for the given list, differs from the first > > string in place 1 and in the second in place 2, and so on, differing > > from the nth in place n, and thus differing enough to be different from > > EVERY member of the list. > > > > Thus it is not in the list. > > This is correct for a finite list. An infinite list is completely > different. An infinite list is different from a finite list in many ways but not in this one. EVERY member of the list contains an 'm', but the diagonal does not, at least in ZF and NBG and sane set theories. What goes on, or doesn't go on, in WM's MathUnRealism is irrelevant. > Compare the infinite tree. While every path covers all its > subpaths, an infinite path p cannot be covered by a single path p' =/= > p, but it can be covered by an infinite set of paths p' =/= p. Depending on what is meant by "subpath", that seems to be quite correct. > > Infinity contains more miracles of that kind. There can even be an > entry equal to the diagonal while the diagonal differes from every > entry. That may happen in WM's MathUnReralism, but not in mathematics itself. > > Regards, WM
From: Dik T. Winter on 21 May 2007 22:17 In article <1179746990.290370.268800(a)z24g2000prd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > for what is now called "simply normal". > > Please do not conclude from your lacking knowledge on that of others. > Weakly normal is a common definition. See fort instance. > http://eom.springer.de/N/n067560.htm > Or do you think that "weakly normal" is not common because it is > missing in Wikipedia? I would estimate Springer somewhat higher than > Wikipedia. I do not use Wikipedia in general for mathematical knowledge. My main online source is Mathworld. And of course the large library of my institute. > > > > > There are different notions (for instance weakly normal numbers and > > > > > absolutely normal numbers). Of course normal numbers can be > > > > > constructed, one of the simplest cases is the rational number > > > > > 0.12012012... with respect to base 3, > > > > > > > > That number is not normal to base 3. > > > > > > That number is weakly normal, namely normal to base 3. > > > > It is *not* normal to base 3. It is "simply normal to base 3", or in > > Borel's terminology "weakly normal to base 3". You cannot omit the > > base. > > I did not. I said with respect to base 3. Yes, and you omitted the "weakly" (that was my error, it was the wrong thing I said you omitted). Weakly normal does *not* mean normal to a particular base. It means normal with respect to single digits to a particular base. > > > If you don't > > > know about the definition of normal numbers you should first inform > > > you. Online for instance > > >http://eom.springer.de/N/n067560.htm > > > > Read what is written there, A number is normal to a particular base if > > *all* n-digit sequences are equi-probable. Did you read this? > > > > Do you know about the Chapernowne numbers? But be also aware that the > > > > Copeland-Erdos number is normal to base 10. A quote: > > > > "While Borel proved the normality of almost all numbers with respect > > > > to Lebesgue measure, with the exception of a number of special classes > > > > of constants, the only numbers known to be normal (in certain bases) > > > > are artificially constructed ones such as the Champernowne constant > > > > and the Copeland-Erdos constant." > > > > > > Another quote: "The weakly-normal number (to base 10) > > > 0.01234567890123456789... is of course rational." > > >http://eom.springer.de/N/n067560.htm > > > > Yes, so what? The Champerowne constants and the Copeland-Erdos constants > > are *not* rational. Read just below your quote, where they give a > > Champerowne constant. > > So what? I never said that every however normal number must be > rational. But 0.01234567890123456789... (weakly normal to base 10) > and 0,012012012... (weakly normal to base 3) *are* rational. So your statement was irrelevant as a response to my statement. But from your book: "also eine sogenannte normale irrationale Zahl, die keine erkennbares Muster der Ziffernfolge aufweist" In the first place, a normal number can have a well defined sequence of digits (as the Champerowne numbers show). And, in the second place, there are numbers without a well defined sequence of digits that are *not* normal. (Normal here meaning normal to a particular base.) The two are not equivalent. So even if no rule can be given for the digits of pi, that does *not* mean that it is normal. -- 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 21 May 2007 22:28 In article <1179747854.168331.52890(a)x35g2000prf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Mai, 04:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179654356.461792.242...(a)n59g2000hsh.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > According to current mathematics, pi is well defined. Even according > > > > > to MatheRealism pi is well defined (as an idea). > > > > > > > > Your distinction between "number" and "idea" is just terminology, and not > > > > more than that. > > > > > > Its is more. You cannot answer the question whether the numbers P = > > > [pi*10^10^100] and P' = P with the last digit replaced by 3 nsatisfy P > > > < P'. > > > > Yes, so what? Your distinction is just terminology, and not more than > > that. > > It is by far more than a difference in terminology if one will never > or always be able to answer a question. Makes absolutely no sense. You have some implicit notion of number in mind that I do not have. So it is nothing more than a difference in terminology. > > > I use it for the paths and nodes of the tree. But you keep on asking > > > for a bijection. The injection has already been shown. > > > > No. You do *not* give an injection from paths to nodes. What node does > > the path 0.0101010101... inject to? How do you define the injection? > > I define it in the same way as you define the injection for definable > numbers, namely by a catalogue. Map, for instance, node no. 17 on the > path 0.010101... That is an injection from nodes to paths. Darn, can't you read? I ask for an injection of paths to nodes. And this means that given a particular path I can find the node it maps to. > Obviously, the catalogue can contain all nodes. So let's wait whether > it also can contain all paths. You are obviously confused. For an injection from paths to nodes you need a catalogue of all paths. > > > Map every node onto the path which leaves it to the left-hand side. > > > > I would think that that is *not* an injection from paths to nodes. > > Moreover, at each node there are many paths that leave it on the left-hand > > side, so it is not even an injection. > > Here is another mapping: Map the node on one of the many paths, call > it A. The others will get their nodes when they separate from A. Again, you are mapping nodes to paths. Why can't you read? > > > Map every node onto the path which leaves it to the left-hand side. > > > > Which of the paths that leaves it on the left-hand side must I chose? > > Choose the path bunch. If the paths which you assume to be present in > the bunch will become separated then choose the due nodes by which > this happens. Again mapping nodes to paths. Now I ask again, given a path {n1 n2 n3 n4 ...} where the n_i are nodes. Which node does that path map to? > > > > Makes no sense. > > > > > > How would you count inseparated paths? > > > > You are trying to do the counting. When I count I find at every node > > uncountably many paths. > > You do not find any path. You find at most path bunches. As each path bunch starts at the root, I find that at each node the number of path bunches does not increase but diminish. And as at each node there are many paths to be found that go from that node, I do not understand how you come at the idea that I do not find any path. > > > The number of paths cannot be larger than the number of edges. > > > > Why not? > > Because every path contains more edges than it has in common with any > other path. That is not a proof. -- 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 21 May 2007 22:44
In article <1179750646.698997.275410(a)b40g2000prd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > The writers sum the *sequence* 1, 2, 3, ... with a definite and unique > > > result (no definition could lead to another result. Your claim that > > > this was not done or was impossible is wrong. > > > > No. They do not state it is a sequence. > > They write 1 + 2 + 3 + .... That is the sum of a sequence, i.e., a > series. Yes, you like to use their informal statements as formal statements. > > But definitions *can* lead to > > another result. > > For instance? Definition: sum{i in N} i = 0. > > It depends on the definitions. The sum of the set of > > all natural numbers is defined *by them*. Other definitions can give > > other results, because the result is not defined without their definitions. > > Which result do you think could be possible? Note, 2*aleph_0 = > aleph_0. > Of course you are wrong. The only result in set theory is aleph_0. The > only result in mathematics is infinite. The only result in analysis and algebra is not infinite, it is undefined. > > It is not defined in standard arithmetic. It is defined in that book using > > some set theoretical definitions, that are not common. > > You must not think that only such things are common which you happen > to know. Oh. And everything they write is common. > > > I said that the sequence of natural numbers can be summed up. This is > > > true. > > > > Depends. Did you miss the point that the axiom of choice was required? > > It is required for many tasks. Yes, so you can *not* state that the set of natural numbers can be summed up. > > But > > *their* sum is not a sum of natural numbers, but of cardinal numbers, and > > it is not a sequence. > > Their notation shows the sum of the sequence of natural numbers > > 1 + 2 + 3 + ... + n + ... (n in N) > > They write, it is easy to see that this sum is equal to aleph_0. And they also write that they are doing cardinal arithmetic. Again, you are reading too much in an informal statement than is warranted. > > They do only show it when you use *their* definitions (and when you > > consider cardinal arithmetic). > > It is a common and natural definition. That is proven by the fact that > there is no other result possible. But there are other results possible. Consider my definition: sum{i in N} = 0 now try to prove that that is not possible. > > Oh. I can define division on cardinals such that > > (aleph_0 + 1) * aleph_0 / 2 = aleph_0, > > which would perfectly fit. Or do you have some other ideas why such a > > definition would not be possible? The only problem is that such > > definitions are not common. > > The second problem is that according to he proof f the formula by > complete induction, the number appearing on the right-hand side of the > formula is in N. Oh, I did not know that. You can not prove the formula by induction. You can prove it by induction for finite 'n', and as 'aleph_0' is not a finite 'n', you cannot prove it. > Your definition would prove aleph_0 in N. Not at all. > And n fact, > if there were all the natural numbers actually existing, then an > infinite one must be among them. A common assertion by you, unproven. Pray give a proof, assuming the axiom of infinity, and with using the negation of that axiom. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |