Cantor Confusion
in [Math]
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From: Virgil on 2 Mar 2007 15:38 In article <1172834875.244086.99100(a)31g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Mrz., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172740041.611220.208...(a)8g2000cwh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 1 Mrz., 01:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1172650983.733643.316...(a)m58g2000cwm.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > What is misleading about it if every mathmatician calls them > > > > "numbers"? > > > > > > It veils the physical restrictions. > > > > Physical restrictions are not something mathematicians are concerned with. > > Environmental questions were no something politicians were concerned > with. This is now changing. Mathematics is not politics. > > > > > > sqrt(2) is in trichotomy with the rational numbers. > > > > > > In fact not even all rational numbers are numbers. You know, there are > > > less than 10^100 bits ... > > > > In that case not all natural numbers are numbers. And being a number > > depends > > on the whim of the observer. But that is only your finitistic approach. > > It is he MatheRealism. Not in the view of real mathematicians. > Finitism is not well defined. There are > finitists who claim a largest number, others do not. Further there is > a negative after taste with finitism. Realism will have better chances > to succeed. Except that what WM is trying to sell, is, at best, pseudorealism, as it argues against logic. > > > > > > So in your mathematical courses you call what we call the > > > > "complex numbers", "complex ideas" or something like that? > > > > > > I use the current words. I do not talk about these things there. > > > > So you are lying (in your opinion) to your pupils? > > No, I use another language. One in which "lying" has a different name?
From: Virgil on 2 Mar 2007 16:06 In article <1172835359.498019.275350(a)64g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Mrz., 04:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > No, you have not. Review the definition of 2^omega. It is only valid > > when that cross section of the unit tree can be well-ordered. Pray show > > a well-ordering of the cross section of the unit tree. (Ordinal numbers > > require well-ordering.) > > The cross section of the tree is the same as the numkber of unit > fractions in the last parenthesis used by Oresme. The "cross section" of an infinite tree cannot, in the manner of finite trees, be equated with a set of nodes, but only with maximal sequences of nodes in which each is a child of its predecessor, and there are uncountably many such sequences. > Well-ordering of all unit fractions is as easy as well-ordering of all > nodes enumerated by natural numbers. If Wm is claiming a well ordering of the set of all paths of a complete infinite tree, then WM is claiming a well ordering of the reals, which previously he claimed could not exist. > > But in order to save you from showing that 2^omega is countable, > simply consider that the cross section of U(T(n)) cannot be larger > than the set of all nodes of U(T(n)). And that set is countable. But 2^|omega|, which is the cardinality of the set of all paths of a complete infinite binary tree, is not countable. Why WM keeps harping on his demonstrably incomplete tree when the complete tree is demonstrably different is a puzzlement. > > > I said: The union of some finite paths is an infinite path. There are > > > only countable many unions of finite paths which > > > yield infinite paths > > > > You stated something different. > But I meant what I said above. Since what you said above is wrong, and provably wrong in , say, ZFC or NBG, it is quite probable that you meant it. But there is a bijection between the set of paths in a complete binary tree of depth n and the set of finite binary sequences of length n. And if we allow the "union" whenever we have a set of such sequences of increasing length in which every member except the first is a one-place extension of its predecessor, then there is one such "union" for every endless binary sequence. And the set of all such unions is, like the set of all such endless sequences, uncountable. WM's ability to deny facts which contradict his beliefs defies belief.
From: Dik T. Winter on 2 Mar 2007 20:09 In article <1172834875.244086.99100(a)31g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 2 Mrz., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172740041.611220.208...(a)8g2000cwh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 1 Mrz., 01:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1172650983.733643.316...(a)m58g2000cwm.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > What is misleading about it if every mathmatician calls > > > > them "numbers"? > > > > > > It veils the physical restrictions. > > > > Physical restrictions are not something mathematicians are concerned > > with. > > Environmental questions were no something politicians were concerned > with. This is now changing. This has already been changing a long time. But I fail to see the relevance. > > > > sqrt(2) is in trichotomy with the rational numbers. > > > > > > In fact not even all rational numbers are numbers. You know, there are > > > less than 10^100 bits ... > > > > In that case not all natural numbers are numbers. And being a number > > depends on the whim of the observer. But that is only your finitistic > > approach. > > It is he MatheRealism. Finitism is not well defined. There are > finitists who claim a largest number, others do not. Further there is > a negative after taste with finitism. Realism will have better chances > to succeed. Perhaps. In that case, pray start a form of mathematics that *does* use it. But do not use it to show contradictions in current mathematics if they are only contradictions with MatheRealism. > > > > So in your mathematical courses you call what we call the > > > > "complex numbers", "complex ideas" or something like that? > > > > > > I use the current words. I do not talk about these things there. > > > > So you are lying (in your opinion) to your pupils? > > No, I use another language. And I do not go into a depths the students > would not understand (and not need). But here you use your own personal language to state what mathematicians state is wrong. So you are teaching your pupils something wrong. > > > > At least those are not in > > > > trichotomy with the "really real numbers". > > > > > > Yes. I would prefer another word, but that will be impossible to > > > introduce. You may have noticed that I talk of "irrationale Zahl" even > > > in my book. I had considered to use only "Irrationalit�t". But it > > > sounds too heavy-handed. > > > > Not only that. > > Oh, "Irrationalit�t" is used in German synonymously to > "Irrationalzahl", but using it exclusively would show some > inflexibility and stubbornness which I always try to avoid. You are trying to avoid stubborness? -- 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 2 Mar 2007 20:02 In article <1172774506.011332.315500(a)p10g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 1 Mrz., 01:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > You know that every natural number like 2n is finite and therefore > > > > > less than aleph_0 while |{2,4,5,...,2n}| in the limit is aleph_0. > > > > > > > > And 2n in the limit is aleph_0, when you define such limits. > > > > > > That is what I say, but according to set theory it is wrong. > > > > Where is that stated? Set theory does not define limits, so it is not > > stated in set theory. > > Every nondecreasing sequence of real numbers bounded from above has a > limit. There is a limit superior and a limit inferior proven by set > theory. Not defined by set theory. > In order to get to higher ordinals, Hrbacek aqnd Jech say: It is easy > to continue the process after this "limit" step is made. The meaning of 'limit' here is soemthing different, that is why it is in quotes. > > > N is the > > > infinite set of all *finite* numbers. Accordinfg to set theory there > > > is no infinite number in it although the limit aleph_0 of numbers is > > > defined by set theory. > > > > No, that is *not* defined as "limit". > > In fact there are limits in set theory. "Es ist sogar erlaubt, sich > die neugeschaffene Zahl omega als Grenze zu denken, welcher die Zahlen > nu zustreben, wenn darunter nichts anderes verstanden wird, als da� > omega die erste ganze Zahl sein soll, welche auf alle Zahlen nu folgt, > d. h. gr��er zu nennen ist als jede der Zahlen nu." No there are no limits defined in set theory. What this states it that you can look at it as such, but that it is not such, because limits are not defined. > > It is stated that from the > > axiom of infinity the set N does exist (and does contain finite > > elements only, because of the way it is defined). It further > > *defines* the cardinality of that set as aleph-0. It does not say > > anything about a limit at all. > > Only because you clever fellows recognized that limits lead to > contradictions you abolished any sensible meaning of omega and let it > "exist" in the vacuum with no relation to natural numbers. What vacuum? It is larger than any natural number. And it is the order type of set N. Where is the lack of sensible meaning? Perhaps not sensible to you, but it makes perfect sense for mathematicians. > > > > So in the > > > > same way I could state: > > > > You know that |{2,4,5,...,2n}| is finite and therefore less than > > > > aleph_0, while 2n in the limit is aleph_0. > > > > You are actually "proving" here that lim{n -> oo} n/n =3D 0. > > > > Bizarre. > > > > > > Yes, it is a mess. But it is exactly what everybody asserts when being > > > talking about the infinite set of finite numbers. > > > > You are making a mess of it because of lack of understanding. > > I am in a good company. But as you even outperform Fraenkel and Levy > et al. I cannot cope with you. Where do they define limits in set theory? > > Assume > > we define limits of sequences of natural numbers such that if the elements > > grows without bound the limit is aleph-0. But if we do that, the old > > theorem: > > lim [ a_n / b_n ] = lim a_n / lim b_n > > if both limits exist and lim b > a_n != 0 > > no longer holds, which you assume. And that is easy to show, because in > > the proof of the theorem use is made of the fact that all limits are > > finite (in the context of that theorem a limit exists only if it is > > finite). > > So in the context of the extended definition we have to adjust the > > condition on the limit to: > > if both limits exist and are finite and lim b_n != 0. > > You are making a mess of it because you use a theorem unchanged while a > > basic definition has been changed. > > I think Cantor started this game. What game? What makes you think that you can use a theorem unchanged while the definition has been changed? To use that theorem with the new definition you have first to *prove* that definition. Further you have to *define* what aleph_0/n, n/aleph_0 and aleph_0/aleph_0 actually *are* before you can even apply such a theorem. > > > In case of Cantor's argument you apply the principle: "Compare with > > > all numers simultaneously." In case of N differing from its final > > > segments you allow only comparison one after the other. That is bad > > > logic. > > > > The comparisons can be done all together, but they will not all yield the > > same result, which is not needed, they yield all yield a result, and that > > is needed. > > If the comparison can be done all together, then there must be one > result (for linear sets). Why? Can you prove that? > > No. I do not say that. I say that S can not be distinguished by any > > finite elemen n from all the finite sets. This does inhibit S to > > exist and being different from all finite sets, that is why I ask for > > a proof. > > It does inhibit S to exist. In fact, S leads to a contradiction if > postulated by axiom. I ask for a *proof*. A valid, mathematical proof. As you have not yet given one, I am still waiting. > > > Why do you use such ridiculous examples to demonstrate your "non- > > > quantifier dyslexia"? > > > > What is ridiculous about it? > > These are not linear sets Indeed. > > It clearly shows that for a set S to exist > > and being different from a collection of other sets it is *not* necessary > > that there is a single element where it differs from all those other sets. > > But this is necessary for linear sets, i.e, such sets which obey A1 c > A2 c A3 c ... A *proof*, please. -- 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 2 Mar 2007 20:18
In article <1172835359.498019.275350(a)64g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 2 Mrz., 04:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Here we have the same situation as with the number 2^omega which has > > > cardinality aleph_0. There is a one-to-one map to the cross sections > > > of the trees and hence to the cross section of the unit tree. > > > > No, you have not. Review the definition of 2^omega. It is only valid > > when that cross section of the unit tree can be well-ordered. Pray show > > a well-ordering of the cross section of the unit tree. (Ordinal numbers > > require well-ordering.) > > The cross section of the tree is the same as the numkber of unit > fractions in the last parenthesis used by Oresme. I missed something. What is the definition of the cross section of the "unit tree"? What is the definition of the "unit tree"? > Well-ordering of all unit fractions is as easy as well-ordering of all > nodes enumerated by natural numbers. > > But in order to save you from showing that 2^omega is countable, > simply consider that the cross section of U(T(n)) cannot be larger > than the set of all nodes of U(T(n)). And that set is countable. How do you define the cross section of U(T(n))? Until now you have only defined cross sections for the finite trees and you are unwilling to define the cross section C(oo). So I really can not answer this question, because I have no idea what it is about. > > > > A proof, please, that there are only countably many unions of > > > > finite paths which yield infinite paths. > > > > > > The set of unions cannot be larger than the cross section of the tree > > > U(T(n)) which is Card(2^omega). > > > > Proof please that that cross section is well ordered. Anyhow, a cross > > section that you refuse to define? > > C(oo) =< Card(U(T(n))) So you refuse to define it? As long as you do not define it, it can not be ascertained whether the statement above is true or false. > > > > But what are you now stating, that those > > > > infinite paths are *not* in the infinite tree? Where are they? > > > > > > I said: The union of some finite paths is an infinite path. There are > > > only countable many unions of finite paths which > > > yield infinite paths > > > > You stated something different. > > ? > But I meant what I said above. So I should not consider what you state but what you mean? Do you think I am a mind-reader? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |