Cantor Confusion
in [Math]
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From: Dik T. Winter on 5 Jun 2007 22:03 In article <JJ5665.45K(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1180966798.913696.131020(a)n4g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: .... > > Therefore they are ghost paths. Hrbacek and Jech have seen that this > > line of reasoning is nonsense. They say: "A branch in T is a > > transfinite sequence b whose all proper initial segments are in T but > > itself is not." (p. 220) And they are right, of course. But the > > immediate consequence is that an infinite sequence of digits is also > > missing in Cantor's list and, therefore, not available for his > > diagonal proof. > > I have to look it up, but I trust that you thoroughly misinterprete > their position. I suspect they define T as a tree of finite paths. In > that case they are right. But in that case 1/3 is missing in the tree. You are even more insidious than I thought you were. The above quote is in the chapter about inaccessible cardinals, and T is defined to be a tree with height H, where (from memory) such H is such a cardinal. -- 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 5 Jun 2007 22:09 In article <JJ57wB.92A(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1180979008.019427.261130(a)n4g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: .... > > But probably you will not acknowledge these authors, because they also > > use, as I did, the term "adic", which you opposed to as not being > > common use. Then you were as wrong as you are here. > > Oh, that was a long time ago. I will look how they use the term "adic", > if it is in the index, otherwise provide a reference. I have found it, and indeed they use it in your sense. Which proves it is a translation from German. The term -adic rather than -ary is in English books extremely uncommon. And indeed the term "p-adic expansion" is normally used only when p is prime. But give me books by Anglo-Saxon authors that use "-adic" in your sense, and we can talk further about this subject. -- 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 5 Jun 2007 22:49 In article <1181022904.358133.224630(a)q66g2000hsg.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 5 Jun., 05:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Sum {A < L} K_A for L = aleph_0 is the sum of all cardinals less than > > > aleph_0. This sum can also be written by 1 + 2 + 3 + .... > > > > You may write it such, but there is *no* sequencing in *their* statement. > > Take L = aleph_1. Where is the sequencing? (Pray be a bit more careful in the future about not placing your new text as if it is quoted text.) > And I talked about aleph_0 being the sum of all natual numbers. Where is the sequencing? > They take aleph_0. Where is the sequencing? Their definition is independent of whether it is possible to give a sequence of all entities involved or not. The only requirement is the axiom of choice. > > > > 1 + 2 + 3 + ... > > > > is an informal notation for: > > > > sum{i in N} > > > > or > > > > sum{i < aleph_0} > > > > no sequence at all. Just summation over a set. > > > > > > Yes, summation over the ordered set of natural numbers. In general > > > this is written as a series. > > > > Their statement is *not* about natural numbers. Try to read. It is > > about cardinal numbers. > > It is about 1 and 2 and 3 and so on. *As cardinal numbers*. It is cardinal arithmetic they are doing. So your conclusion that the result is a natural number is not justified. > > > Hracek and Jech even state the identity on p. 188: > > > "It is not very difficult to evaluate infinite sums. For example, > > > consider > > > 1 + 2 + 3 + ... + n + ... (n in N). > > > It is easy to see that this sum, SUM{n in N}n is equal to aleph_0." > > > Can you read: 1 + 2 + 3 + ... + n + ... = this sum = SUM{n in N}n? > > > > Yes, that is an informal statement. Can you distinguish formal definitions > > from informal statements? > > Not in the way you believe you could. No, indeed, you can not at all. > > > But probably you will not acknowledge these authors, because they also > > > use, as I did, the term "adic", which you opposed to as not being > > > common use. Then you were as wrong as you are here. > > > > Oh, that was a long time ago. I will look how they use the term "adic", > > if it is in the index, otherwise provide a reference. > > p.98:" We conclude this section with a few remarks on decimal (or, in > general, p-adic) expansions of rationals." Yes, a translation from German. But you apparently think that is normative for Anglo-Saxon usage. Can you come up with a native Anglo-Saxon book that states "p-adic" when "p-ary" is meant? > > > > > > Definition: sum{i in N} i = 0. > > > > > > > > > > Wrong definition. > > > > > > > > What is wrong about it? > > > > > > Simply the fact that the equality is not an equality. > > > > Can you prove that? Stating something is a fact is no proof. > > I am glad to see the arguing which is typical for set theory. Here > even you should be able to recognize the nonsense involved in your > statement. > > SUM{n>0}1/2^n is not larger than 1. Pray define "SUM{n > 0} 1/2^n". According to the standard definitions "SUM{n > 0} 1/2^n" means: "lim{k -> oo} sum{n = 1..k} 1/2^n" which is exactly 1. > Do you agree? This is correct without any axiom, just by observing > that always the remaining is divided by 2 and one half of it is > removed. No. This requires the *definition* of that infinite sum, and using limits appropriately. What you are doing is not mathematics, because without that definition that sum does not exist. > SUM{n>0}1/n is not less than 1. Indeed, according to the standard definitions it is not defined. So it is also not larger than 1, nor equal to 1, it is simply undefined. The simple reason is that the limit does not exist. > Do you agree? Yes, even for the whole infinite set of natural numbers > this holds. We can prove it. Well proving things about undefined things is fairly easy. > SUM{n>0}n is not less than 1. Neither it is larger than 1, nor equal to 1. It is undefined. > So your Definition: sum{i in N} i = 0 is obviously wrong. Why? There is something that is undefined, and I give it meaning. > The same holds for Definition |{2,4,6,...}| = aleph_0. That is not a definition, it is a theorem. > We can argue > with the same limits as above. What limits? > For every set {2,4,6,...,2n} we have larger numbers than n = | > {2,4,6,...,2n}|. Taking he limit as in SUM{n>0}1/2^n or SUM{n>0}1/n we > get the result > |{2,4,6,...}| = aleph_0 cannot be true without an infinite natural > number (which dos not exist). In all those cases the limit in natural numbers does not exist. We could postulate the *definition* (with natural k and a_k): lim{k -> oo} a_k = Z if for any natural n, there is a k_0 such that for all k > k_0, a_k > n. In that case the proof that lim{k -> oo} |{2,4,6,...,2k}| = Z is left as an exercise for the reader. This does *not* make Z a natural number. Within cardinal arithmetic, H&J define the concept of limits with respect to cardinal numbers, but there they do not imply ordered sets, but take limits with respect to all the elements of a set. But also in that case we have that the limit is aleph_0 without there being an infinite natural number. I may note that with the definition I gave: lim{k -> oo} k = lim{k -> oo} 2k. But for some reasons you have extreme difficulties with non-terminating sequences (that exist due to the axiom of infinity). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 6 Jun 2007 03:54 On 6 Jun., 04:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <JJ57wB....(a)cwi.nl> "Dik T. Winter" <Dik.Win...(a)cwi.nl> writes: > > In article <1180979008.019427.261...(a)n4g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > ... > > > But probably you will not acknowledge these authors, because they also > > > use, as I did, the term "adic", which you opposed to as not being > > > common use. Then you were as wrong as you are here. > > > > Oh, that was a long time ago. I will look how they use the term "adic", > > if it is in the index, otherwise provide a reference. > > I have found it, and indeed they use it in your sense. Which proves it > is a translation from German. It is not. The book was first published in USA. Before there existed a Czech versin, Never German. > The term -adic rather than -ary is in > English books extremely uncommon. And indeed the term "p-adic expansion" > is normally used only when p is prime. But give me books by Anglo-Saxon > authors that use "-adic" in your sense, and we can talk further about this > subject. > -- No, Dik, we cannot furher talk about this subject, because you are unable to understand or to confess to understand the most simple facts. (Also your argument about the tree which happens to appear in their chapter about inacessile cardinals but nevertheless means what I said, is completely wrong if not a lie.) So I give up to believe that your mind will ever become accessible to logical arguments outside of the world of messy axioms and dogmas. Regards, WM
From: WM on 6 Jun 2007 04:51
On 6 Jun., 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1181022904.358133.224...(a)q66g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 5 Jun., 05:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > Sum {A < L} K_A for L = aleph_0 is the sum of all cardinals less than > > > > aleph_0. This sum can also be written by 1 + 2 + 3 + .... > > > > > > You may write it such, but there is *no* sequencing in *their* statement. > > > Take L = aleph_1. Where is the sequencing? > > (Pray be a bit more careful in the future about not placing your new > text as if it is quoted text.) > > > And I talked about aleph_0 being the sum of all natual numbers. > > Where is the sequencing? > Look here: 1 + 2 + 3 + .... Can you see the increasing sequence which is summed? It is obvious. On p. 190 hey formulate even infinite products over "an increasing sequence of cardinals". This implies the product over all natural numbers: 1*2*3*...*n*... Can you see the sequence? They do not write 2*4*3*80*9*.... No they write, 1*2*3*...*n*... as I said . But you cannot recognize a seuqnece. > > They take aleph_0. > > Where is the sequencing? Their definition is independent of whether it is > possible to give a sequence of all entities involved or not. The only > requirement is the axiom of choice. Their *result* is the sum of the sequence of all natural numbers as well as the product. They *say* "sequence" and they *write* the sum in the sequential order. But you are not able to recognize it. Doesn't that make you sceptical about your further capabilities of recognition, in the tree for instance? > > > > > > 1 + 2 + 3 + ... > > > > > is an informal notation for: > > > > > sum{i in N} > > > > > or > > > > > sum{i < aleph_0} > > > > > no sequence at all. Just summation over a set. > > > > > > > > Yes, summation over the ordered set of natural numbers. In general > > > > this is written as a series. > > > > > > Their statement is *not* about natural numbers. Try to read. It is > > > about cardinal numbers. > > > > It is about 1 and 2 and 3 and so on. > > *As cardinal numbers*. It is cardinal arithmetic they are doing. So > your conclusion that the result is a natural number is not justified. All natural numbers are cardinal numbers. The sum over the sequence (which in this case is simply an ordered set) of all natural numbers is the sum over all finite cardinal numbers. There is not the slightest difference. > > > > > > > > Definition: sum{i in N} i = 0. > > > > > > > > > > > > Wrong definition. > > > > > > > > > > What is wrong about it? > > > > > > > > Simply the fact that the equality is not an equality. > > > > > > Can you prove that? Stating something is a fact is no proof. > > > > I am glad to see the arguing which is typical for set theory. Here > > even you should be able to recognize the nonsense involved in your > > statement. > > > > SUM{n>0}1/2^n is not larger than 1. > > Pray define "SUM{n > 0} 1/2^n". According to the standard definitions > "SUM{n > 0} 1/2^n" means: "lim{k -> oo} sum{n = 1..k} 1/2^n" which is > exactly 1. Of course, why should I give another definition? > > > Do you agree? This is correct without any axiom, just by observing > > that always the remaining is divided by 2 and one half of it is > > removed. > > No. This requires the *definition* of that infinite sum, and using > limits appropriately. What you are doing is not mathematics, because > without that definition that sum does not exist. That is wrong. "SUM{n>0}1/n is not less than 1" is true without any axioms and further definitions. > > Indeed, according to the standard definitions it is not defined. So it > is also not larger than 1, nor equal to 1, it is simply undefined. The > simple reason is that the limit does not exist. Wrong. This limit and sum exist for over 2000 years without your presumed "standard definitions". > > > Do you agree? Yes, even for the whole infinite set of natural numbers > > this holds. We can prove it. > > Well proving things about undefined things is fairly easy. > > > SUM{n>0}n is not less than 1. > > Neither it is larger than 1, nor equal to 1. It is undefined. It is aleph_0, "easily provable" according to Hrbacek and Jech. > > > So your Definition: sum{i in N} i = 0 is obviously wrong. > > Why? There is something that is undefined, and I give it meaning. I think this point is so important for the understanding of what set theorists are thinking that I will make up another thread: "Dik T. Winter says: SUM{n in N}n = 0". I am curious what people will say to your bold claim. It is similar a discussion as about the tree, but I think it is more suggestive. Regards, WM |