An uncountable countable set
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From: Dik T. Winter on 4 Aug 2006 18:27 In article <1154722206.554768.237310(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > The last conclusion requires proof (and it is false). What is the > > > > case is that for every Ai the indexing of digits will terminate at > > > > some point, and the indexing of digits of 0.111... will never > > > > terminate when we start at the first digit and go on. Because > > > > there will always be a further digit. (As there will always be a > > > > further An.) > > > > > > But each one will be finite. > > > > You mean the index of the digit and the An. Yes. > > All you agree with for finite numbers applies to all index numbers, > because each one is finite. Yes, each index number is finite, I *never* did argue otherwise. But I do *not* agree that the set of index numbers is finite, because it isn't. > > > (every digit of n can be indexed) ==> (n is in the list) > > > > Where is the proof? There is an easy proof of: > > (n is in the list) ==> (every digit of n can be indexed) > > but not of the converse. > > Indexing is a symmetric relation. A position n of number N can index a > position m of number M, if and only if n = m. What do you *mean* with "a position n of number N"? What is N here? What do you *mean* with "a position ... can index"? How do you index with positions? As far as I know you index with natural numbers in this case. The only thing I see is: a natural number n indexes digit position m of M when n = m. I see no symmetry at all. > As 0.111... has more > digit positions than the binary representation of any natural number, > your claim implies that indexing is not a symmetric relation. Let's assume that we use indexing to mean a bijective mapping from one set to another set. And further in this case, a bijective mapping from the set of naturals to the other set. What is the other set? Well, that is the set of index positions. And indeed, the set of index positions *also* is the set of natural numbers. That 0.111... has more index positions than each and every natural number in unary (not binary) notation is irrelevant. Just as N has more elements than each and every indivual natural number is irrelevant. -- 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 4 Aug 2006 18:50 In article <1154722450.815196.312700(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > In general, the ordinal number of any initial segment of ordinal numbers > > is the smallest ordinal number that is larger than all ordinal numbers in > > that segment. (This can be done because the ordinal numbers form a > > well-ordered set.) And note that 0 is also an ordinal number (of the empty > > set). So the ordinal number of {} is 0, of {0, 1, 2} is 3, etc. > > Finally, the ordinal number of {0, 1, 2, 3, 4, ..., w} is w+1. And so > > is the ordinal number of {1, 2, 3, 4, 5, ..., w}, because there is an > > order preserving bijection between the two. > > > > You apparently did not appreciate that 0 is an ordinal number. (And that > > is independent of Bourbaki.) > > I find that in the *true list* every number n gives the cardinality of > the sequence of *positive* ordinal numbers from 1 to n inclusive and > the cardinality of the 1's of the corresponding unary representation. Yes, I have no problem with that. But now you are talking about cardinality. > I find that this is no longer true for positive ordinal w of the > extended list > > 0 = 0.0 > 1 = 0.1 > 2 = 0.11 > 3 = 0.111 > ... > w = 0.111... Now wait a moment. The cardinal number of this extended list is aleph-0, the ordinal number is w+1. Removing the last element does not change the cardinality but reduces the ordinality to w. > This shows that w is not the cardinal number of the natural numbers > alone. If you really want to equate aleph-0 with w, it is. You are again confusing cardinal numbers with ordinal numbers. The cardinal numbers of all of the following sets are equal: (w, 0, 1, 2, 3, ...) (0, 1, 2, 3, ..., w) (0, 1, 2, 3, ..., w, w+1) but their ordinal numbers are all different (respectively w, w+1 and w+2). > It is independent of the question whether or not 0 is an ordinal. Well, if you are confusing cardinal and ordinal numbers... -- 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 4 Aug 2006 19:21 In article <1154722892.397252.6590(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > So the triangle gets infinitely long but not infinitely broad. This > > > means the Diagonal must bend. > > > > Eh? Where do you conclude *that* from? The triangle gets infinitely long > > and infinitely broad. Your conclusion is based on something unstated. > > But this is, again, a distraction. > > 1) Infinitely long means that there are infinitely many natural > numbers. > 2) Infinitely broad means that there are infinite natural numbers. Eh? Why? Can you provide a proof? > Of course both statements are equivalent. There is no "of course" here. Such a statement needs proof. > > It remains a straight line, see above. > > That means: an actually infinitely long triangle is an actually > infinitely broad triangle. Therefore there are either natural numbers > with infinite magnitude or there are not infinitely many natural > numbers. Why? > > Back again to your old horse. What is the relevance that I am not able to > > do that within the time that is left for me and the computer power left for > > me? > > It does not depend on your time and your computer. Even with infinite > time you would not be able to have the first 10^100 digits of pi stored > somewhere, because all the universe is to small for this purpose. > Therefore pi does not exist, neither does [pi*10^10^100]. But that is not mathematical thinking. This is the thinking that leads users of (non-symbolic) calculators to complain that when they calculate sin(pi) they do not get 0. Mathematics is not concerned with finite space, time or whatever. But, again, this is a philosophical standpoint, *not* a mathematical one. By the same reasoning sqrt(2) does not exist. It (and other square roots) comes nevertheless in pretty handy when (for instance) factoring large integers. Check for the number field sieve for factoring which abounds with such kinds of roots. You are just repeating Kronecker. > > > On the other hand, if you start with a > > > > path: > > > > o > > > > | > > > > and use the basic figure: > > > > | > > > > o > > > > / \ > > > > you get other information. > > > > > > Not at all. The number of infinite paths is always less than the number > > > of edges. > > > > But in this case neither the paths, nor the edges are terminated. So while > > in all cases the number of edges is less > > larger! > > > than the number of paths (while > > keeping finite), this does *not* mean that they will have a different > > cardinal number when the thing is compleated (if you are able to do that). (I made two typo's. "less" for "larger", and "compleated", which is quite old-fashioned.) > That is correct. But in no case the number of paths can be larger than > that of edges (which is countable). No, the number of edges will also be uncountable. > > I need to do nothing in this sense at all. You apparently think that I > > have to, in fact, change each a_nn. But there is no need for that. I > > just state that if a_nn is 4, than I can use 5, if it is not, I use 4. > > It does not bother me whether it is 4 of not, what is important to me > > is that what I find is definitely not equal to an. > > With the same arguing I can state that it is not necessary to enumerate > the number of paths in the binary tree. In any case I know that no path > can split unless there are two additional edges. What is the argument here? If you state that the number of paths is countable you need to provide a method that gives the index number when a path is given. > > > > Indeed. You do not define what "..." means. If I add 1 infinitely > > > > times to itself, is the sum finite? > > > > > > No. it is infinite. But if you have infinitely many naturals this means > > > you have infinitely many 1's as differences. And the sum of these 1's > > > shall be a finite number??? > > > > Where do I state such, or how can you conclude that? > > Because all naturals have to be finite. But infinitely many naturals > have infinitely many differences of 1 which accumulate to an infinite > number. By what reasoning do you conclude that that "infinite number" is a natural number? > > > > > And from the natural numbers we know as their most important > > > > > property: Every number is finite. > > > > > > > > And that is completely irrelevant, as there are infinitely many of > > > > them. > > > > > > And that is impossible because each number has one 1 more (normal > > > addition) than its predecessor. > > > > You are again contradicting the axiom of infinity? > > Correct. I do not refuse it but I contradict it. Nope, the plain statement above "that is impossible" is a contradiction of the axiom of infinity. So as such there is no basis for that statement. So your statement in itself is already clearly in contradiction with the axiom as infinity and so can not show that the axiom of infinity is in contradiction with itself. > > > After infinitely many predecessors the > > > number of accumulated 1's is infinite. Hence the due number is not a > > > natural number. > > > > Yes, did I contradict that? My statement was: there are infinitely many of > > finite numbers. You state that is impossible. Pray show some *proof* of > > your statements. > > Infinitely many naturals have infinitely many differences of 1 which > accumulate to an infinite number. That is just a repeat of your argument. Pray give a proof that what I actually did state was impossible. To recap, I state: There are infinitely many of finite numbers. You state: That is impossible. I want to see a *proof*. I can easily give a disproof. Suppose there are not infinitely many finite numbers. In that case there is a largest, say n. But if n is finite, so is n+1, a contradiction. > > > Here is a plain picture: Imagine a staircase with infinitely many > > > stairs. Can it have finite hight but infinite width (given equal hight > > > and width of each stair)? > > > > No. What is the problem with that? > > That is the infinite set of finite numbers. Pray explain. (On the other
From: mueckenh on 6 Aug 2006 12:44 Dik T. Winter schrieb: > This is illogic. What you are indexing is the list of digits of K. That is your definition, but it is wrong (= impossible to satisfy) > You are claiming that K is in the list of indices. That makes no sense > at all. I am claiming that every index is in the list of indexes. This list is infinite. There are infinitely many indexes. And, yes, 0.111... is not among them and therefore cannot be indexed. > > > > And there is no such proof. > > > > In order to save set theory with its "infinite set of finite numbers" > > you are forced to assert that the list of all natural numbers does not > > contain the list of all natural numbers. > > Yes, but that is obvious. As the list of all natural numbers is not a > natural number, it is not in the list of all natural numbers. My list of all natural numbers is not a set in the pathological sense of set theory. My list contains all single natural numbers, and it contains all pairs of natural numbers (though a pair is not a number) - not in one row but in pairs of rows. And my list contains the list of all natural numbers - not in one row but in all rows. Regards, WM
From: mueckenh on 6 Aug 2006 12:46
Virgil schrieb: > > > > T. Jech is one of the leading set theorist. > > > > He says up to the advent of set theory there was no use for actual > > > > infinity. > > > > > > Does he say that there is no use for actual infinity within current set > > > theory? I doubt it. > > > > He says that there is only use for actual infinity in (current and > > uncurrent) set theory > > > > Regards, WM > > So that all of "mueckenh"'s railings against actual infinities are > thereby countered. Only by assertion. Not by facts. Regards, WM |