An uncountable countable set
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From: mueckenh on 17 Aug 2006 08:44 Dik T. Winter schrieb: > In article <J42BHx.IzE(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > In article <1155664930.866986.157410(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > 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. > > Let's have an easier example. The blocks are 1/2^n high and 1-1/2^n wide. I propose the following stair 1 - 1/2^n high and 1 - 1/2^n wide up to the n-th step, in order to have symmetry. > When you make a stair of them, when you complete, you have a stair with > height 1 and width 1. But there is neither a block that is 1 wide, nor > a block that is at height 1. On the other hand, for every positive k, > there are blocks that are beyond the height of 1-k and beyond the width > of 1-k. Both in height and in width the blocks just do not reach the > boundary line. And note that in both "infinity is reached". What you argue is a potential infinity. Infinity, aleph_0, here reduced to 1, is present, actually existing, according to Cantor, in width. All steps do exist. But, according to him, infinity, here 1, is not present in height. Regards, WM
From: Dik T. Winter on 17 Aug 2006 09:33 In article <1155817776.184870.134560(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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 Yes, I know this al already. Just the same argument. But: > 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. Right. > 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. Wrong. Indeed, there is no n such that 0.111... can be indexed by the numbers 1 to n. This does *not* mean that 0.111... can not be indexed by the complete 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. I do not only *claim* that 0.111... can be indexed by the natural numbers; I can *prove* it, as I have done again and again. The only things you are doing is handwaving and circular reasoning. Indeed, the discussion is useless if you are not able to formulate definitions and proofs. > > 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. So, in your opinion the axiom of infinity is purest nonsense. You have a right to opiniate that. But do *not* claim that you have found an inconsistency with the axiom of infinity, because you have not. > > > 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. What is that 'n' in the last equation? It can not be the 'n' in 'all n' because that is an arbitrary 'n'. > > > 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? Yes. > Do you agree that every natural number which can index a digit position > is in the list? Yes. > If a number is not in the list, what does this fact say about its digit > positions? Nothing. If all digit positions are natural numbers, they can all be indexed by numbers from the list. But in that case the number of digit positions is infinite, as there are infinitely many natural numbers. So the number is not in the list because the list contains only numbers with finitely many 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. Why the 'therefore'? Can you provide a proof? > > > 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. That is quite non-standard terminology. In standard terminology something that is not finite is infinite and the reverse. However, I have no idea what a 'potentially infinite set' is. Can you provide a definition? > > 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. Elementary mathematics with equivalence classes is false? What is false about it? > > > 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. *My* error? Where did I ever state that? Already before 8 august I have written an article where I wrote: > Eh? Where do you conclude *that* from? The triangle gets infinitely long > and infinitely broad. Your conclusion is based on something unstated. Please keep the record correct. > > > > > 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 a
From: Dik T. Winter on 17 Aug 2006 10:51 In article <1155817978.033024.136310(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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. That is nonsense. Each finite initial segment can be covered by a natural number. The complete number can not be covered by a natural number. > > > 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. Yes. But why should it be in the list? It is not an index position as it is not a natural number. -- 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 17 Aug 2006 10:57 In article <1155818167.396269.42480(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Fine, I now understand. lim{n -> oo} > 0. An interesting assertion. > > > > > > > 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) .... > > 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. Yes, omega = omega. This does *not* imply that omega / omega = 1. That depends on how you define that operation. In mathematics that operation is not defined. > > > > > If a > b, then a has a sequence or subset equal to b. Now you can > > > define a/b = 1 + remainings. Your definition. > > 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. Your definition was about sets, not about cardinalities. What are the remainings? According to the definition above I have no idea what the remainings are. Can you define remainings in terms of cardinalities? > > > 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. You are repeating the simple argument I used just above. Do you not *read* what I write? -- 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 17 Aug 2006 11:02
In article <1155818502.182962.180850(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1155674426.389672.172370(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > > Dik T. Winter schrieb: > > > > > > > Nonsense. Cantor invented the list and constructed the > > > > > > > first one. .... > > > 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. Yup. > > > 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! The sequence of naturals 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. Edge 1 leads to 0.0 Edge 4 leads to 0.01 Edge 9 leads to 0.010 Edge 20 leads to 0.0101 What edge leads to 1/3? That was not what I asked. You assigned a natural number to a subset of the edges, not to all edges. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |