Cantor Confusion
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From: Franziska Neugebauer on 17 Nov 2006 08:02 mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > >> > > > I see. But recently you used the word "completed infinity". >> > > >> > > I don't think I ever said that. Do you have a quote? >> > >> > Here it is: >> >> In the below post, I was just trying to paraphrase what you are >> saying. > > Please don't try to paraphrase what I said, because I don't believe > that you understand it sufficiently. This is due to your conceptional weakness. [...] >> I didn't say I would say that or that I understood what you were >> trying to say. > > The question was whether you "ever said that" it. I hope this > question as been settled now. > >> In fact, I don't know what you you mean by the phrase. Did you >> really misunderstand what I wrote? > > There is no misunderstanding possible. You refuted Lester's > interpretation, by proposing to have a better one: > > "That doesn't seem to be what WM is saying. He seems to be saying that > the notion of a completed infinity leads to either absurdities or > contradictions. Perhaps he thinks the way to avoid these absurdities > is to only consider things that can be physically produced." > > I [k]now that cranks never admit having made an error. Introspection? > But do you think that obvious lies like this are a way to reach your > aim?> I cannot spot any lie. >> I don't know what you mean by "completed anywhere". > > The completed initial segment contains every natural number. http://mathworld.wolfram.com/InitialSegment.html A "completed initial segment" is not an *initial* segment. > Another segment contains not every natural number. No *initial* segment contains every element of its underlying set. F. N. -- xyz
From: Franziska Neugebauer on 17 Nov 2006 08:29 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: [...] >> ISOTFC := { {1}, {1, 2}, {1, 2, 3}, ... } >> L := { 1, 2, 3, ... } >> >> B := { <{1}, 1>, <{1, 2}, 2>, <{1, 2, 3}, 3>, ... } >> >> The bijection between the initial segments of the first column >> (ISOTFC) and the lines (L) is explicitly contructed. I have named it >> "B". I cannot see any reason to debate about the existence of B. > > This bijection contains only finite sets. This is in perfect agreement with set theory and with the usual natural numbers. Omega has only finite elements. Omega has only finite initial segments. > If you add one element to a finite set, then you get a finite set. Surely. > If you add an element [call it x] to an (ordered) infinite set [call > it A] you get a set of ordinal number omega + 1. If A = { a_1, a_2, ... } has order type omega then { a_1, a_2, ..., x } order type omega + 1. But there are other ways to "add" an element to an ordered set. > Therefore a set of ordinal number omega must exist (it is the first > column) omega "is" not "the first column". What you may write is, that in your sketch the first column _represents_ omega. > but it does not appear in your bijection. There is no reason, why it should appear. Omega is not an initial segment of the ordered set { omega, e }. > The ISOTFC is no part of your B. Of course not. Two sets X and Y are "bijected" by "pairing" the *elements* of the sets not by "pairing" the sets itself. >> > If this is possible, and if the ordinal of the first column is >> > omega, >> >> What does "possible" mean? Sets are or are not. > > The opposite of impossible. What does the category Possibility precisely mean? >> There is no set theoretical notion of "possibility of sets". > > But an explanaion: The set of all sets is an impossible set. This mathematically reads: There is no set of all sets (in some axiomatic systems). Shall I conclude from your "definition by example", that you define possible := exists impossible := does not exist >> Sets (including bijections) are neither possible nor impossible. They >> exist or do not exist. > > If they exist, then they are possible. If not, then they are > impossible. Shall I interpret this as "possible := exists, impossible := does not exist"? So why don't you simply use the established terms "exist" and "not exist"? Maliciousness? Now the part which you have not answered: ,----[ <455c4019$0$97239$892e7fe2(a)authen.yellow.readfreenews.net> ] | Adding one element (named "x") to every initial segment gives us: | | ISOFTC' := { {1, x}, {1, 2, x}, {1, 2, 3, x}, ... } | | I have no clue what precisely you mean by adding one element to every | line. Do you mean the set | | L' := { 1, 2, 3, ..., x } ? | | > The fact that it is not maintained proves that your asserted | > bijection does ot exist. | | There exists a bijection between ISOTFC' and L' : | | B' := { <{1, x}, x>, <{1, 2, x}, 1>, <{1, 2, 3, x}, 2>, ... } | | proving |ISOTFC'| = |L '|. `---- F. N. -- xyz
From: William Hughes on 17 Nov 2006 09:55 mueckenh(a)rz.fh-augsburg.de wrote: > The natural numbers count themselves. Bijection of initial segments of > column and lines > > 1 > 2 > 3 > ... > n <--> 1,2,3,...n > > If there is no infinite number then there are not infinitely many > numbers. This is clearly the point of contention. Consider, N, the set of all natural numbers. By definition N only contains natural numbers. Cases i: There is a largest natural number,n_L, then N={1,2,3,...,n_L} In this case there are n_L natural numbers. ii: There is no largest natural number. We will write this as N={1,2,3,...} (the ... represent only natural numbers). Set N has infinitely many elements. iii: There are a finite number of natural numbers, but there is no largest natural number. Case i is highly counterintuitive because of "What about (n_L+1)?". I claim case 2 (existence of N and no largest number by the axiom of infinity, infinitely many numbers the fact that all the natural numbers have been used to provide sizes of other sets). You claim case iii (finite number based on an argument that only the integers that will actually be named can be said to exist, no largest number based on the fact that the largest number is unknowable before the end of the universe). If case ii is correct then there can be infinitely many lines without any line that does not correspond to a natural number. In this case the bijection exists. You want to show two things. a: case iii makes sense, i.e. there can be a finite number of natural numbers but no largest natural number. b: case ii does not make sense. To do a you need to define finite and show that this definition does not lead the existence of a largest natural. (note that an argument that there are only a finite number of natural numbers because there are only a finite number of available bits does not help. You still need to define finite). To do b you need to assume case ii and show this leads to a contradiction. However, when you do this you cannot use case iii. In particular, we know that asumming case ii means that we can find a bijection between a matrix with an infinite number of lines, but for which every line is finite. (There is a line for every natural number, but no other line. Therefore by ii there are an infinite number of lines. Line n will be {1,2,3,...,n} and have length n, which is finite) - William Hughes
From: Tony Orlow on 17 Nov 2006 11:15 William Hughes wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > >> The natural numbers count themselves. Bijection of initial segments of >> column and lines >> >> 1 >> 2 >> 3 >> ... >> n <--> 1,2,3,...n >> >> If there is no infinite number then there are not infinitely many >> numbers. > > This is clearly the point of contention. > > Consider, N, the set of all natural numbers. > By definition N only contains natural numbers. > > Cases > i: There is a largest natural number,n_L, then N={1,2,3,...,n_L} > In this case there are n_L natural numbers. > > ii: There is no largest natural number. We will > write this as N={1,2,3,...} (the ... represent only natural > numbers). Set N has infinitely many > elements. > > iii: There are a finite number of natural numbers, but > there is no largest natural number. > > > Case i is highly counterintuitive because of "What > about (n_L+1)?". > > I claim case 2 (existence of N and no largest number by the > axiom of infinity, infinitely > many numbers the fact that all the natural numbers have been used > to provide sizes of other sets). > > You claim case iii (finite number based on an argument that only the > integers that > will actually be named can be said to exist, no largest number based on > the > fact that the largest number is unknowable before the end of the > universe). > > If case ii is correct then there can be infinitely many lines without > any line that does not correspond to a natural number. In this > case the bijection exists. > > You want to show two things. > > a: case iii makes sense, i.e. there can be a finite number > of natural numbers but no largest natural number. > > b: case ii does not make sense. > > > To do a you need to define finite and show that this > definition does not lead the existence of a largest natural. > (note that an argument that there are only a finite number > of natural numbers because there are only a finite number > of available bits does not help. You still need to define finite). > > To do b you need to assume case ii and show this leads > to a contradiction. However, when you do this you cannot > use case iii. In particular, we know that asumming case ii > means that we can find a bijection between a matrix with > an infinite number of lines, but for which every line is finite. > (There is a line for every natural number, but no other > line. Therefore by ii there are an infinite number of lines. > Line n will be {1,2,3,...,n} and have length n, which is > finite) > > - William Hughes > Hi William - nice summary. The difference boils down to potential (countable) infinity vs. actual (uncountable) infinity. The set of naturals is countable, and therefore only potentially infinite. If we define an infinite set as one where expressing the indexes of all elements in a finite base digital system requires unending strings for most elements, then only uncountable infinity fits that bill. Where each element is the sum of a sequence of increments which is finite, no element is in an infinite position within the set. In internal set theory, the naturals are not an infinite set, or even a set, because it's an external set. Internally, there is no infinite distance between any two elements.
From: Tony Orlow on 17 Nov 2006 11:25
Franziska Neugebauer wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > >> David Marcus schrieb: >> >>>>>> I see. But recently you used the word "completed infinity". >>>>> I don't think I ever said that. Do you have a quote? >>>> Here it is: >>> In the below post, I was just trying to paraphrase what you are >>> saying. >> Please don't try to paraphrase what I said, because I don't believe >> that you understand it sufficiently. > > This is due to your conceptional weakness. > > [...] > >>> I didn't say I would say that or that I understood what you were >>> trying to say. >> The question was whether you "ever said that" it. I hope this >> question as been settled now. >> >>> In fact, I don't know what you you mean by the phrase. Did you >>> really misunderstand what I wrote? >> There is no misunderstanding possible. You refuted Lester's >> interpretation, by proposing to have a better one: >> >> "That doesn't seem to be what WM is saying. He seems to be saying that >> the notion of a completed infinity leads to either absurdities or >> contradictions. Perhaps he thinks the way to avoid these absurdities >> is to only consider things that can be physically produced." >> >> I [k]now that cranks never admit having made an error. > > Introspection? > >> But do you think that obvious lies like this are a way to reach your >> aim?> > > I cannot spot any lie. > >>> I don't know what you mean by "completed anywhere". >> The completed initial segment contains every natural number. > > http://mathworld.wolfram.com/InitialSegment.html > > A "completed initial segment" is not an *initial* segment. > >> Another segment contains not every natural number. > > No *initial* segment contains every element of its underlying set. > > F. N. How exactly d you define "initial segment"? It would seem to me to be a subset of an ordered set which, if it contains element x, also contains every element with an index in the ordered set which is less than x's. If this is the definition, then the complete set IS an initial segment of itself. If you disagree with thi definition, can you provide a better one? Tony |