An uncountable countable set
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From: mueckenh on 17 Aug 2006 08:29 Dik T. Winter schrieb: > > Because it is very relevant. You state that an infinite string could be > > indexed by finite strings. Complete indexing includes covering. > > Nope. You claim the last, but it is false. No, it is true. The list of sequences of 1's 1 0.1 2 0.11 3 0.111 .... n 0.111...1 n+1 0.111...11 .... is constructed such that the segment from 1 to n does not contain numbers which can completely index or cover the digit positions of a number which does not belong to the segment, like the sequence n+1 and all sequences which are larger, i.e., which have more 1's. As 0.111... does not belong to the list, it does not belong to any segment of the list. Hence its digit positions cannot be completely indexed or covered by the sequences of the list. Now you may claim, that the mathematics of finity does not hold in infinity, but that there everything is different. But that is not mathematics in my opinion. May be that it is mathematics in some one other's opinion. Nevertheless it is not useful to continue the discussion on this topic between us. > Each finite segment can be > covered, and each finite index can be covered. But the total is not > finite, but contains only finite digit positions. That is, in my opinion, purest nonsense. But the believer will never give up his belief. So let's finish this discussion. > > > > > Because your assertion that K could be completely indexed and covered > > is equivalent to K being in the true list. > > My assertion is that K can be completely indexed but not completely > covered. Again you misquote my assertion. Indexing n implies covering all digit positions m =< n. Indexing all n implies covering all m =< n. > > > > > > That is just opinion. .. > > > > Likewise a number with infinitely many digit positions can not be > > indexed by a natural number. > > Why not? Each and every digit position is finite, so can be indexed, > and, by *your* definition the number can be indexed. But there are > infinitely many digit positions, as there are infinitely many > natural numbers. > Yes. And all of them are in the true list. If n can be indexed, then also n+1 can be indexed, because with n also n+1 is in the list. This holds for every natural number. Do you agree that every digit position which can be indexed by a natural number is in the list? Do you agree that every natural number which can index a digit position is in the list? If a number is not in the list, what does this fact say about its digit positions? > > It is obvious that an infinite > > number of differences of 1 leads to an infinite number. > > Yes. When you can add all those numbers. But the result is not a natural > number. Exactly. Therefore there cannot exist an infinite number of natural numbers. > > > > > Important: There is no largest number. The axiom says only that the set > > > > is infinite. > > > > > > I said, there is no largest number. The axiom says the set does exist. In > > > and of itself the axiom does *not* state that the set is infinite. It is > > > a theorem that the set is not finite, and so must be infinite. > > > > Why should a set which is not finite be actually infinite (i.e. have a > > cardinal number)? > > A set that is not finite is infinite, that is the definition of the word > infinite. No. Every (potentially) infinite set is finite. > So yuor question is actually, why are there cardinal number? > Well, they come in handy when comparing infinite sets. And (by definition) > a cardinal number is the equivalence class of sets that can be put in > bijection with each other. A set being in bijection with another set is > easily shown to be an equivalence relation, and so we can split the sets > in equivalence classes. That is all elementary mathematics. And those > equivalence classes are called cardinal numbers. Just as the equivalence > classes of Cauchy sequences are called real numbers and the equivalence > classes of ordered pairs of integers are called rational numbers (with > some particular equivalence relation in mind). All extremely elementary. And all extremely false too. In particular if you consider the infinite binary tree and call its set of edges uncountable. > > > > > > Again an assertion, without proof. > > > > > > > > The staircase is a proof. > > > > > > No. > > > > You may deny this, nevertheless it is true. If there is an actual > > number of steps, then the width is equal to the lenghts is aleph_0. > > You keep asserting that. If you say the length is infinite but the width is not infinite, then simply exchange the axes of size of numbers and number of numbers. You will easily see your error. > > > What I do not understand is how you can believe that someone (except, > > perhaps, Virgil) could share your opinon that only in one dimension the > > infinity was actually realized, but not in the other. > > Just because there is no last line. If there were a last line, it would > have width aleph-0, but there is no last line. And the line next to the last line, and the line next to the next to the last line? They all do not exist. I have the impression that almost every line does not exist, except a few, but in no case more than finitely many. Regards, WM
From: mueckenh on 17 Aug 2006 08:31 Dik T. Winter schrieb: > > > Nobody ever has said that the diagonal is ever completed. There is a > > > definition of that number, and it is easy to show, with that definition, > > > that that number (when completed) would be different from all numbers on > > > the list. > > > > But if it is impossible to complete it, then the result is void. > > There are many real numbers that can not be completed. But by the > definition of "real number", each sequence of decimal digits defines > a real number, and it is easily shown, by the above, that that real > number is not in the list because it is different from each number > in the list. Oh no. It is not different for any infinite line but only for some finite lines. None of them, however, is decisive. If the real cannot be completed, then it is unknown, whether it differs from every entry of the list. > (There are indeed also many real numbers that are not > computable..., but they are nevertheless real numbers, by the definition > of "real number".) Like the devils which are devils by the definition of devil? > > > > > > But the axiom of infinity asserts that > > > > > the set of all aleph-0 lines does exist. I think you are asserting a new > > > > > axiom: 10% of the lines does exist. > > > > > > > > I think that 10 % must exist if the whole set = 100 % does exist? No? > > > > > > No. > > > > Here you see the inconsistence of set theory. The set shall exist > > actually = completely, but the first 10 % shall not exist. > > What is the inconsitency? Can you show what axiom of set theory or what > proposition from set theory it violates? If you tell me what is to be understood by "set" and by "existence" in set theory. I do not know your opinion about that matter. Zermelo refrained from defining what a set is. Regards, WM
From: mueckenh on 17 Aug 2006 08:32 Dik T. Winter schrieb: > All digits can be covered, but there is no n that covers all digits. What is the result of your assertion? There is a list number which covers only the first digits of 0.111..., while another covers not the first digits but some more behind. > > > > > You assert that every digit of 0.111... can be covered, but that > > > > 0.111... cannot be covered. That assertion is void of meaning. It makes > > > > no sense. > > > > > > To you, apparently. I did show a proof, what was wrong with the proof? > > > > You showed it for digit positions only which can be indexed by natural > > numbers. > > But *all* digit positions can only be indexed by natural numbers. > There are no digit positions that can not be indexed by natural numbers. These positions are all in the list - by definition. 0.111... is not in the list by mathematical proof. Regards, WM
From: mueckenh on 17 Aug 2006 08:36 Dik T. Winter schrieb: > In article <1155674284.887539.128610(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > Every natural number is finite. > > > > > > Still makes no sense. > > > > But it is a fact (by definition). > > You snipped context. It made no sense in the context (and was not an answer > to my question). My question was: > > How do I apply that to the "limits" you are using above? > your answer was: > > It is always the same: n takes the values of all finite natural numbers, > > one ofter the other, without ever becoming infinite. > And my question is unanswered. When I try to extract a definition from > that, I find: > lim{n -> oo} 1/n > 0. > because 1/n never becomes 0. That is correct. There is no natural number of infinite size, therefore for every natural we have 1/n > 0. > > > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 > > > > > > (because omega is a fixed, well defined and well deteremined > > > > > > quantum, according to Cantor) > > > > > > > > > > You think so. > > > > > > > > No, I do not think so. Cantor did - and you do. > > > > > > Not in the way you think. It is the "because" I am objecting to. > > > > Is omega a fixed, well defined and well determined quantum or not? > > As I said (pray read), it is the word "because" I objected to. And I stated, > quoted just below, that it *is* a well defined quantum. So why ask for > something that I have already answered days ago? omega / omega = 1 because Cantor stated that it is a fixed quantum and, therefore, omega = omega. > > > If a > b, then a has a sequence or subset equal to b. Now you can > > define a/b = 1 + remainings. > > Let us try it with: > > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 > let's say a_n = {2, 4, 6, ..., 2n} and b_n = {1, 2, 3, ..., n}. What is > the subset equal to b_n in a_n? What are the remainings? Both sets have same cardinality. |b_n| = |a_n|. There are no remainings. Therefore the result of the division is 1. > > Do you see your formal objection as the only chance to escape? > > Nope. You use "because" in your terms, and *that* is the word I objected > to. I ask you to justify that word. > > > > > > lim [n --> oo] |{2, 4, 6, ..., 2n}| / |{1, 2, 3, ..., n}| = 1 > it is much simpler than that. For each n the quotient is 1, and so, using > standard terminology for limits, it is also 1 in the limit. This is not > > > > > > (because omega is a fixed, well defined and well deteremined > > > > > > quantum, according to Cantor) > (and neither does it prove that aleph-0/aleph-0 can be defined and is 1, > yes, I prefer aleph-0 when talking about cardinalities). We should not divide alephs. Omegas are better defined. But there is a simple argument. As long as we are in the finite domain we have well defined: 2n / |{2, 4, 6, ..., 2n}| = 2. In the limit we have n remaining a finite number but aleph_0 being larger than any finite number. >From lim [n --> oo] 2n / 2n = 1 and aleph_0 > 2n we obtain lim [n --> oo] 2n / |{2, 4, 6, ..., 2n}| < 1. Regards, WM
From: mueckenh on 17 Aug 2006 08:41
Dik T. Winter schrieb: > In article <1155674426.389672.172370(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1155640812.322879.187040(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > > > > > Dik T. Winter schrieb: > > > > > > > > > > Nonsense. Cantor invented the list and constructed the first one. > > > > > > > > > > Quote, please, especially for the latter remark. > > > > > > > > =DCber eine elementare Frage der Mannigfaltigkeitslehre. > > > > [Jahresbericht der Deutsch. Math. Vereing. Bd. I, S. 75-78 (1890-91).] > > > > > > Do you have an URL? Is it in his "Werke"? Otherwise I have no access > > > to it. > > > > Werke, page 278. In particular the list is on p. 279. > > That is not a specific list, but an arbitrary list. It is a list, the first one. > Anyhow, that assumed list can be enumerated by definition. That definition is the following: If for any line a natural number can be determined, uniquely, then the lines are countable. > > > > > Are you really believing that the number of edges was uncountable? > > > > Take any edge you like. You can attach a natural number to it. That and > > > > nothing else is the definition of countability. > > > > > > That is not the definition of countability. The definition of countability > > > is that you can attach a (different) natural number to each and every > > > edge. > > > > Yes, of course. And what is your problem with the edges (how do they > > differ from the lines of Cantor's list)? > > The lines are countable (by assumption). But pray assign natural numbers to > each edge in your tree. I will even allow you to assign finite sequences of > digits 0 (go left) and 1 (go right) to your tree (the mapping to natural > numbers is quite standard in that case). What is the finite sequence of > digits assigned to the edge that leads to 1/3? You claim that the edges > are countable, so you should have an answer to my question. The sequence of edges is countable but it is not finite! 1/3 has the sequence of nodes 0.010101... in binary notation. 0. /1 \2 0 1 /3 \4 /5\6 0 1 0 1 /7\8/9\10........ The edges of 1/3 are enumerated 1, 4, 9, 20, ... . For any edge a natural number can be determined, somewhat cumbersome, but it can be done. > > And before you mutter: I just assign 0 to it and change all other assignments > so that it fits. Be prepared that I will ask you to give such a sequence > for another number. I will not tell you which, but if your mapping does > not map that other number to a finite sequence, your mapping is shown > incomplete. And this works *regardless* the mapping you are using. > Your mapping really should contain each and every real number before I > consider further. It does - obviously. Regards, WM |