An uncountable countable set
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From: Dave L. Renfro on 12 Sep 2006 12:55 Dave L. Renfro wrote: > If no one beats me to it, my post will be post 2006 > in this thread (by chronological order) archived by google! Got it! 1995 David R Tribble Sep 11 1996 David R Tribble Sep 11 1997 mueck...(a)rz.fh-augsburg.de Sep 12 1998 Dik T. Winter Sep 12 1999 Tony Orlow Sep 12 2000 Tony Orlow Sep 12 2001 Tony Orlow Sep 12 2002 Tony Orlow Sep 12 2003 Tony Orlow Sep 12 2004 Tony Orlow Sep 12 2005 Tony Orlow Sep 12 2006 Dave L. Renfro Sep 12 2007 Tony Orlow Sep 12 2008 Tony Orlow Sep 12 2009 Tony Orlow Sep 12 2010 Tony Orlow Sep 12 2011 Tony Orlow Sep 12 Dave L. Renfro
From: Virgil on 12 Sep 2006 14:27 In article <1158059651.877076.46040(a)e63g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1157574332.235885.113030(a)d34g2000cwd.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > Talking about 0.111... defined as: for all natural p digit p is 1, there > > are no other digtits. And the list is the list of natural numbers. > > > > > > > Your "each" means in symbols of logic: "A = (for) all". > > > > > The number is nothing than all of its digit positions. > > > > > Therefore your statement is a self contradiction. > > > > > > > > Where? In logical terms (A meaning "for all" and E meaning "there > > > > is"): > > > > (1) A{p = digit position} E{q = list item} {such that q indexes p} > > > > > > That is your definition. But what we can safely say is only: > > > (1') A{p = digit position of list item} E{q = list item} {such that > > > q indexes p} > > > > Why can we only say that? The definition of 0.111... is such that (1) > > holds. > > Then it would be in the list. Why don't you accept the axiom of > infinity? There is an infinite set of natural numbers. All they are in > the list (in unary representatition). By the axiom this is accomlished. > And there is nothing else. The axiom of infinity does not say there is nothing else, and there is no axiom which says there is nothing else. And the members of set of naturals, which is an ordinal number, can be used to index larger ordinals than itself, in fact it can index any countable ordinal. > > > If it does not hold you should be able to give an index position > > such that it is false. > > It is not in the list, although we cannot give an index position which > is responsible for that fact. All possible indexes are in the list by > the axiom of infinity. But that does not require that everything that can be indexed must also be in the list, as the set of indices is not restricted to only indexing itself. If it were, it would be useless as a set of indices. > > > > > > (2) A{p = digit position} E{q = list item} {such that q covers p} > > > > > > If your definition could be satisfied, the construction of the list > > > would imply this, yes. What we can safely say, however, is only: > > > (2') A{p = digit position of list item} E{q = list item} {such that > > > q covers p} > > > > The same here. But apparently you think my definition of 0.111... can not > > be satisfied. Why not? > > Because it is not in the list which, by definition, contains all > numbers which can be indexed because there are all numbers which can > index. There is no such definition. There is no "largest" set that can be indexed, because one can always manage to index one more element. See "Hilbert's Hotel"! > > > Now we may ask: Is it possible that a list item indexes or covers other > > > digits than those which are indexed or covered by list items? That is the wrong question. If one asks "Can unindexed items be appended to the set indexed and still have a set which can be indexed?", the answer is "YES!" > > Because you assert that all digits of 0.111... can be indexed, which is > wrong. Indexing and covering "by all list numbers" is equivalent. What do you mean by ' covering "by all list numbers" ' ? The only thing necessary for indexing is a surjective function from the index set to the set to be indexed. N such thing as ' covering "by all list numbers" ' is required. > > > > This is false. 0.111... is distinguished from all list items in that it > > does not terminate. > > "Not to terminate" is not a property which can serve to distinguish a > number from others, because we can never observe the end (because it > does not exist) neither the non-end (because it is nothing but a > negation of an unobservable property). The set of naturals concretely establishes its "unendingness" by having an injection from itself to a proper subset. That unendingness is specifically guaranteed by the axiom of infinity, so in any sysem with an axiom of infinity, "not to terminate" in the form of such an injection into a proper subset is an observable property. > There is no n in the list (and in N) which yields 1/9. Therefore 1/9 = > 0.111... cannot completely be indexed. {1/9, 1, 2, 3, ...} indexes it by position in the list. > It has been proven by showing that 0.111... is not in the list. But one can create a list containing it. According to "Mueckenh" there is only one list of anything, and anything not already in that list cannot, according to "Mueckenh", be listed in any list.
From: Virgil on 12 Sep 2006 14:31 In article <4506d1ae(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Mike Kelly wrote: > >>> Better in what sense? What mathematics are you hoping to be able to do > >>> with your new foundation that cannot be done with ZFC? > > > > Tony Orlow wrote: > >> It accounts for changes of a single element between infinite sets and > >> therefore provides for a full spectrum of ordered infinite sets, thus > >> making the Continuum Hypothesis null and void. It relates the continuum > >> to the hypernaturals formulaically, and provides for an integration of > >> sets and measure in the infinite case. It rids us of anomalies like > >> omega-1=omega. > >> > >> If you remove an element, the proper subset should ALWAYS be smaller by > >> 1. That is the case for me. For a theory to claim a proper subset is the > >> same "size" as the proper superset is an immediate deal-breaker for me. > > > > If by "different size" you mean that you cannot pair up all the > > elements from both sets, then you're going to have a difficult > > time proving that for any infinite set. (You have never show this, > > BTW.) > > > > If by "different size" you mean something other than some way of > > denumerating (counting) the elements of the set (e.g., by assigning > > them different natural indices), then you should use a different term, > > because it's confusing. Oh, and you have to prove that it works > > (you have never shown this, either). > > > > Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} > > and remove one element to get set S = {1,2,3,...}. Now show that > > the "T-size" of N is exactly one less than the T-size of S. In other > > words, find a way to show that every counting of S versus every > > counting of N always leaves one element of N (0) left over. > > > > Use IFR. No such thing can exist in standard mathematics, and TO has, as yet, not produced any system in which it can exist.
From: Virgil on 12 Sep 2006 14:36 In article <4506d1fb$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45039397(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Dik T. Winter wrote: > >>> In article <1157836470.071804.254270(a)i3g2000cwc.googlegroups.com> > >>> mueckenh(a)rz.fh-augsburg.de writes: > > > >>> > A set of n elements is *a number*. It need no necessarily be a latin or > >>> > arabic symbol. > >>> > >>> Bizarre. What number is the set of all natural numbers? > >> Hi Dik - > >> > >> Yes, it's bizarre to give that set a specific size, when it's > >> unbounded. > > > > That depends entirely on how one defines "size" of a set. If one defines > > it in terms of bijectability with other sets, as is the reasonable > > definition for finite sets, there is no problem. > > > > > > IFR works perfectly well for finite quantitative sets. TO's idiotic IFR does not work for anything except Dedekind finite sets of numbers. TO's idiotic IFR does not work for sets of vectors, even for finite sets of vectors. TO's idiotic IFR does not work for non-abelian groups, even finite ones. In fact even TO does not claim TO's idiotic IFR works for anything but TO's own idiotic vision of a line with infinitely many infinities of infinite points.
From: Virgil on 12 Sep 2006 14:44
In article <4506d3b7(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > David R Tribble wrote: > >>> Specifically, like I said above, the set of bit positions bijects with > >>> the set of bitstrings. For every natural, there is a bit index, and > >>> vice versa. For every natural, there is a bitstring, and vice versa. > >>> So every bit index is a natural, and every bitstring is a natural. > >>> Hence they are exactly the same set. > > > > Tony Orlow wrote: > >>> That's my point. You can biject the naturals with the bit posiitons, and > >>> also with the strings, but the set of strings is the power set of the > >>> bit positions. Therefore you have a bijection between a set and its > >>> power set, and no cardinality which is appropriate for the set of bit > >>> positions. > > > > Mike Kelly wrote: > >> Bullshit. There is no bijection between the naturals and the set of all > >> binary strings. > > > > Finite binary strings (only). Tony thinks that the set of all possible > > finite bitstrings is the powerset of the naturals (the finite bit > > positions or indices), so there are more (finite) bitstrings than > > bit positions. > > > > Yet he admits that there is a bijection between the two. He > > concludes that this demonstrates a flaw or inconsistency in set > > theory. > > > > In a similar vein, he thinks he has a bijection between the naturals > > (as binary strings) and all the subsets of N. Likewise, he thinks that > > this proves that set theory is flawed. > > > > Obviously he's confused between powerset cardinalities (|P(S)| = 2^|S|) > > and binary bitstrings (bit n = 2^n), probably because of the similarity > > in notation. > > > > You don't have it quite right. While there are countably infinite bit > strings in the power set of the naturals It is TO who does not have things quite right. The "power set of the naturals" does not have any bit strings a members. The members of that power set are sets of naturals, not strings. So whatever nonsense TO was trying to claim, he has mucked up from the very beginning. |