Cantor Confusion
in [Math]
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From: Virgil on 6 Jun 2007 14:50 In article <1181116461.877767.54300(a)o5g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > 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. AS WM's "facts" are often our falsehoods, It is WM who cannot talk futher. > 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. That world of axioms is what is producing almost all the mathematics that is being produced, and supports even those few areas produced without specifically relying on axiomatics. That WM does not like it does not man it is not so.
From: Virgil on 6 Jun 2007 15:07 In article <1181119911.759974.186410(a)n4g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 6 Jun., 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > 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. By that argument, the sum of two odd naturals must be an odd natural. All odd naturals are naturals so there is not the slightest difference. > > > > 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. WM is wrong to think that things which are not defined must have any intrinsic meaning. > "SUM{n>0}1/n is not less than 1" > is true without any axioms and further definitions. It is equally true that, until there is a definition of what _number_ "SUM{n>0}1/n" means is given, it is equally true to say "SUM{n>0}1/n is not greater than 1" as only numbers can be compared or size with 1 and sans definition, n o comparison can result in "true". Now it may be that in WM's MathUnRealism, comparing undefined objects for size is allowed, but mathematics marches to a better drummer. > > > > 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". Only in such backward corners of the nonworld as MathUnRealism. > > > > > 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. They will continue to say that WM is wrong.
From: Dik T. Winter on 6 Jun 2007 16:16 In article <JJ6zB2.2Ep(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > 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. Not German, Czech. -- 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 7 Jun 2007 20:05
In article <1181119911.759974.186410(a)n4g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 6 Jun., 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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? Yes, but in the definition of that sum there is *no* sequencing. It is the cardinal number of an infinite union of sets, by definition. Within infinite unions there is no sequencing. And there is a very good reason to do this, because the definition *also* holds for uncountable sets of cardinal numbers. > It is obvious. On p. 190 hey formulate even infinite > products over "an increasing sequence of cardinals". But in their *definition* of that infinite product there is, again, no sequencing, so it also works for uncountable sets. > 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. I recognize one. And they write it that way so that it is easily recognised what is meant. But their *definition* is done by: sum{n in N} and prod{n in N} No sequencing at all. > > > 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. Yes, but the sum is *not* calculated sequentially, but all at once. > 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? I seriously doubt your capabilities with respect to mathematics. > > > > > > 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. The difference is that the sum is a cardinal number that is not a natural number, because there are cardinal numbers that are *not* natural numbers. They define cardinal arithmetic, so you should not expect as result a cardinal number that is also a natural number. > > > > > > > > Definition: sum{i in N} i = 0. .... > > > 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? So you follow the standard definition here. > > 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. As that sum is undefined without any further axioms or definitions, you cannot state that it is not less then 1. Or are you now comparing undefined things with numbers? > > 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". So, what *is* sum{n > 1} 1/n? By what definitions and limits? > > > 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. Using *their* definition of that infinite sum in cardinal arithmetic. You can not prove it without their definition. Did you actually *do* that easy proof? I think not. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |