An uncountable countable set
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From: David R Tribble on 11 Sep 2006 19:43 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.
From: mueckenh on 12 Sep 2006 07:14 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. > 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. > > > > (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. > > > 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? The > > answer is: no. > > You are repeating yourself again, and I have already agreed to that. Why > then repeat again and again? Because you assert that all digits of 0.111... can be indexed, which is wrong. Indexing and covering "by all list numbers" is equivalent. Both is true (can be done) or both is false (cannot be done). > > > And we may further ask: How can 0.111... be distinguished from all list > > items? The answer is the only one possible: By digit positions occupied > > by 1's which are neither indexed nor covered by list items. Why? > > 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). What can serve to distinguish two numbers in unary representation is a 1 at a digit position. All numbers which can be distinguished by such an 1 are in the infinite list. Numbers are not processes. > Consider them as decimal fractions. The list consists > of the numbers (1-10^(-n))/9, 0.111... is 1/9. There is no n in the list (and in N) which yields 1/9. Therefore 1/9 = 0.111... cannot completely be indexed. > > > because 0.111... does not have any other component. > > It does not have any other component, but also it does not terminate, > in contrast to all numbers on the list. This cannot occur other than by more digits 1 than available by list numbers. > > > So we may finally ask: Is it possible that all digit positions of > > 0.111... are indexed or covered by list items? > > The answer is obvious: no. > > Every digit position can be indexed and covered. Otherwise state which > digit position can not be indexed, use my definition: > Talking about 0.111... defined as: for every natural p digit p is 1, > there are no other digtits. > but there is no natural that covers all digits. > > And if you think that that definition is wrong, please *prove* that. It has been proven by showing that 0.111... is not in the list. Regards, WM
From: Dik T. Winter on 12 Sep 2006 09:49 In article <1158059651.877076.46040(a)e63g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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 do you state that without proof? > 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. Indeed, and 0.111... is *not* a natural number, so why should it be in the list? Why do you think 0.111... is a natural number? > > 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. It is not in the list because it is not a natural number. > > > > > > (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. This is nonsense. By the definition of your list the list contains all finite natural numbers. That has nothing to di with indexing. > > > 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? The > > > answer is: no. > > > > You are repeating yourself again, and I have already agreed to that. Why > > then repeat again and again? > > Because you assert that all digits of 0.111... can be indexed, which is > wrong. If not all digits can be indexed, there should be a digit that can not be indexed. But *by the definition* I gave above, all digits can be indexed. Why is that wrong? > Indexing and covering "by all list numbers" is equivalent. Both > is true (can be done) or both is false (cannot be done). You again do not clearly state what you mean. But the statement as it stands is *false*. > > > And we may further ask: How can 0.111... be distinguished from all list > > > items? The answer is the only one possible: By digit positions occupied > > > by 1's which are neither indexed nor covered by list items. Why? > > > > 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). What can serve to distinguish > two numbers in unary representation is a 1 at a digit position. You are not talking nonsense. As 0.111... is different from al sequences 0.111...1 because it is infinite in length, there is for each 0.111...1 an digit position where 0.111... is different from that specific 0.111...1. > > Consider them as decimal fractions. The list consists > > of the numbers (1-10^(-n))/9, 0.111... is 1/9. > > There is no n in the list (and in N) which yields 1/9. Therefore 1/9 = > 0.111... cannot completely be indexed. You are talking nonsense. > > > because 0.111... does not have any other component. > > > > It does not have any other component, but also it does not terminate, > > in contrast to all numbers on the list. > > This cannot occur other than by more digits 1 than available by list > numbers. This is nonsense. > > > > > So we may finally ask: Is it possible that all digit positions of > > > 0.111... are indexed or covered by list items? > > > The answer is obvious: no. > > > > Every digit position can be indexed and covered. Otherwise state which > > digit position can not be indexed, use my definition: > > Talking about 0.111... defined as: for every natural p digit p is 1, > > there are no other digtits. > > but there is no natural that covers all digits. > > > > And if you think that that definition is wrong, please *prove* that. > > It has been proven by showing that 0.111... is not in the list. Proof that it should be in the list. But remember: 0.111... 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: Tony Orlow on 12 Sep 2006 11:26 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. N maps to S using f(n)=n+1. The inverse of that function is g(x)=x-1. So, over the range of 0 to N, |S|=|N|-1. Map N to the Evens E using f(n)=2n. The inverse function is g(x)=x/2, so over the range of N, the evens have |N|/2 elements. Isn't that intuitively satisfying? And gee, it works for finite sets accurately too!! :) Tony
From: Tony Orlow on 12 Sep 2006 11:28
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. Tony |