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From: WM on 8 Dec 2009 16:31 On 8 Dez., 19:36, A <anonymous.rubbert...(a)yahoo.com> wrote: > > A function f is said to be continuous at a point x in its domain if > the limit of f(a), as a approaches x, is equal to f(x); in others > words, the limit of the values of f is equal to the value of f at the > limit, speaking loosely. Of course, not every function is continuous > at every point in its domain, and some functions are not even > continuous at any point in their domains at all. > > The situation for sets and cardinality is no more mysterious than > that. The cardinality of a limit of subsets of the integers is not > guaranteed to be the limit of the cardinalities of those subsets. You > don't expect an arbitrary function to always be continuous, so perhaps > it's unreasonable to expect the cardinality "function," defined on > subsets of the integers, to be continuous.- That depends on the circumstances. If infinite sets exist, then they have a cardinality. Then the limit cardinality is the cardinality of the limit set. If they do not exist but are merely an arbitrary, perhaps inconsistent delusion, then everything is possible. I argue, based on Cantor's claim, that infinite sets exist (in order to show that they do not). Regards, WM
From: K_h on 8 Dec 2009 17:03 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuAGqH.FrI(a)cwi.nl... > In article <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > ... > > > > > When you mean with your statement about N: > > > > > N = union{n is natural} {n} > > > > > then that is not a limit. Check the definitions > > > > > about > > > > > it. > > > > > > > > It is a limit. That is independent from any > > > > definition. > > > > > > It is not a limit. Nowhere in the definition of that > > > union a limit is used > > > or mentioned. > > > > Question. Isn't this simply a question of language? > > Not at all. When you define N as an infinite union there > is no limit > involved, there is even no sequence involved. N follows > immediately > from the axioms. I disagree. Please note that I am not endorsing many of WM's claims. There are many equivalent ways of defining N. I have seen the definition that Rucker uses, in his infinity and mind book, in a number of books on mathematics and set theory: On page 240 of his book he defines: a_(n+1) = a_n Union {a_n} and then: a = limit a_n. He writes "...that is, lim a_n is obtained by taking the union of all the sets a_n". The text book I have on set theory defines N as the intersection of all inductive subsets of any inductive set. So clearly there are many equivalent approaches to defining N. In Rucker's approach, we could define N as a limit: a_0 = {} //Zeroth member is the empty set. a_(n+1) = a_n Union {a_n} and then: N = limit a_n. My text book on set theory also explicitly states that we can have a limit of a set of ordinals, for example: "...the phase successor ordinal for an ordinal which is a successor and limit ordinal for an ordinal which is a limit". In fact, one of the problem sets is to prove the bi-conditional: If X is a limit ordinal then UX=X (U is union) and if UX=X then X is a limit ordinal. > > My > > book on set theory defines omega, w, as follows: > > > > Define w to be the set N of natural numbers with its > > usual order > > < (given by membership in ZF). > > > > Now w is a limit ordinal so the ordered set N is, in the > > ordinal sense, a limit. Of course w is not a member of > > N > > becasuse then N would be a member of itself (not allowed > > by > > foundation). > > Note here that N (the set of natural numbers) is *not* > defined using a > limit at all. That w is called a limit ordinal is a > definition of the > term "limit ordinal". It does not mean that the > definition you use to > define it actually uses a limit. (And if I remember > right, a limit > ordinal is an ordinal that has no predecessor, see, again > no limit > involved.) N can be defined as a limit or not as a limit. These are really equivalent approaches. k
From: Virgil on 8 Dec 2009 20:23 In article <3501073d-9187-4242-b925-37824eb682e7(a)m26g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 8 Dez., 19:36, A <anonymous.rubbert...(a)yahoo.com> wrote: > > > > > A function f is said to be continuous at a point x in its domain if > > the limit of f(a), as a approaches x, is equal to f(x); in others > > words, the limit of the values of f is equal to the value of f at the > > limit, speaking loosely. Of course, not every function is continuous > > at every point in its domain, and some functions are not even > > continuous at any point in their domains at all. > > > > The situation for sets and cardinality is no more mysterious than > > that. The cardinality of a limit of subsets of the integers is not > > guaranteed to be the limit of the cardinalities of those subsets. You > > don't expect an arbitrary function to always be continuous, so perhaps > > it's unreasonable to expect the cardinality "function," defined on > > subsets of the integers, to be continuous.- > > That depends on the circumstances. If infinite sets exist, then they > have a cardinality. No one says otherwise, but that in no way implies that the cardinality of the limit of a sequence of sets as defined by DIk need equal the limit, if it even exists, of the cardinalities of those sets. Such a claim requires proof, which WM has been unable to provide, and such a claim cannot stand in view of the counterexample which Dik provided, so that WM is doubly wrong, in (1) not having a proof of his claim and (2) having found no flaw in Dik's counterproof. > Then the limit cardinality is the cardinality of the limit set. Often claimed by WM but not proved by him nor has WM, or anyone else, refuted Dik's counterexample. > > If they do not exist but are merely an arbitrary, perhaps inconsistent > delusion, then everything is possible. In Wolkenmuekenheim, WM may declare to be possible whatever he likes and declare to be impossible whatever he likes, but outside of his private world, he is not master over what is and is not possible, however often he may claim to be. I argue, based on Cantor's > claim, that infinite sets exist (in order to show that they do not). At which argument, as with so many others, WM fails miserably.
From: Virgil on 8 Dec 2009 20:29 In article <Hv2dnXQ7LtSxUIPWnZ2dnUVZ_hSdnZ2d(a)giganews.com>, "K_h" <KHolmes(a)SX729.com> wrote: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:KuAGqH.FrI(a)cwi.nl... > > In article <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dnZ2d(a)giganews.com> > > "K_h" <KHolmes(a)SX729.com> writes: > > ... > > > > > > When you mean with your statement about N: > > > > > > N = union{n is natural} {n} > > > > > > then that is not a limit. Check the definitions > > > > > > about > > > > > > it. > > > > > > > > > > It is a limit. That is independent from any > > > > > definition. > > > > > > > > It is not a limit. Nowhere in the definition of that > > > > union a limit is used > > > > or mentioned. > > > > > > Question. Isn't this simply a question of language? > > > > Not at all. When you define N as an infinite union there > > is no limit > > involved, there is even no sequence involved. N follows > > immediately > > from the axioms. > > I disagree. Please note that I am not endorsing many of > WM's claims. There are many equivalent ways of defining N. > I have seen the definition that Rucker uses, in his infinity > and mind book, in a number of books on mathematics and set > theory: On page 240 of his book he defines: > > a_(n+1) = a_n Union {a_n} > > and then: > > a = limit a_n. > > He writes "...that is, lim a_n is obtained by taking the > union of all the sets a_n". The text book I have on set > theory defines N as the intersection of all inductive > subsets of any inductive set. So clearly there are many > equivalent approaches to defining N. In Rucker's approach, > we could define N as a limit: > > a_0 = {} //Zeroth member is the empty set. > > a_(n+1) = a_n Union {a_n} > > and then: > > N = limit a_n. > > My text book on set theory also explicitly states that we > can have a limit of a set of ordinals, for example: "...the > phase successor ordinal for an ordinal which is a successor > and limit ordinal for an ordinal which is a limit". In > fact, one of the problem sets is to prove the > bi-conditional: If X is a limit ordinal then UX=X (U is > union) and if UX=X then X is a limit ordinal. > > > > My > > > book on set theory defines omega, w, as follows: > > > > > > Define w to be the set N of natural numbers with its > > > usual order > > > < (given by membership in ZF). > > > > > > Now w is a limit ordinal so the ordered set N is, in the > > > ordinal sense, a limit. Of course w is not a member of > > > N > > > becasuse then N would be a member of itself (not allowed > > > by > > > foundation). > > > > Note here that N (the set of natural numbers) is *not* > > defined using a > > limit at all. That w is called a limit ordinal is a > > definition of the > > term "limit ordinal". It does not mean that the > > definition you use to > > define it actually uses a limit. (And if I remember > > right, a limit > > ordinal is an ordinal that has no predecessor, see, again > > no limit > > involved.) > > N can be defined as a limit or not as a limit. These are > really equivalent approaches. In ZF, unions are defined only for sets of sets and for such a set of sets S, the union is defined a the set of all elements of elements of S. Unless some other definition replaces this one, it is impossible to form the union of a family of sets unless that family are already the members of a set. In which case, N cannot be defined as the limit you suggest, as it would have to exist before it exists.
From: K_h on 9 Dec 2009 01:25
"Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-2C0D8F.18292508122009(a)newsfarm.iad.highwinds-media.com... > In article > <Hv2dnXQ7LtSxUIPWnZ2dnUVZ_hSdnZ2d(a)giganews.com>, > "K_h" <KHolmes(a)SX729.com> wrote: > >> "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message >> news:KuAGqH.FrI(a)cwi.nl... >> > In article >> > <yvCdnW28VrXBqIXWnZ2dnUVZ_s-dnZ2d(a)giganews.com> >> > "K_h" <KHolmes(a)SX729.com> writes: >> > ... >> > > > > > When you mean with your statement about N: >> > > > > > N = union{n is natural} {n} >> > > > > > then that is not a limit. Check the >> > > > > > definitions >> > > > > > about >> > > > > > it. >> > > > > >> > > > > It is a limit. That is independent from any >> > > > > definition. >> > > > >> > > > It is not a limit. Nowhere in the definition of >> > > > that >> > > > union a limit is used >> > > > or mentioned. >> > > >> > > Question. Isn't this simply a question of language? >> > >> > Not at all. When you define N as an infinite union >> > there >> > is no limit >> > involved, there is even no sequence involved. N >> > follows >> > immediately >> > from the axioms. >> >> I disagree. Please note that I am not endorsing many of >> WM's claims. There are many equivalent ways of defining >> N. >> I have seen the definition that Rucker uses, in his >> infinity >> and mind book, in a number of books on mathematics and >> set >> theory: On page 240 of his book he defines: >> >> a_(n+1) = a_n Union {a_n} >> >> and then: >> >> a = limit a_n. >> >> He writes "...that is, lim a_n is obtained by taking the >> union of all the sets a_n". The text book I have on set >> theory defines N as the intersection of all inductive >> subsets of any inductive set. So clearly there are many >> equivalent approaches to defining N. In Rucker's >> approach, >> we could define N as a limit: >> >> a_0 = {} //Zeroth member is the empty set. >> >> a_(n+1) = a_n Union {a_n} >> >> and then: >> >> N = limit a_n. >> >> My text book on set theory also explicitly states that we >> can have a limit of a set of ordinals, for example: >> "...the >> phase successor ordinal for an ordinal which is a >> successor >> and limit ordinal for an ordinal which is a limit". In >> fact, one of the problem sets is to prove the >> bi-conditional: If X is a limit ordinal then UX=X (U is >> union) and if UX=X then X is a limit ordinal. >> >> > > >> > > My >> > > book on set theory defines omega, w, as follows: >> > > >> > > Define w to be the set N of natural numbers with >> > > its >> > > usual order >> > > < (given by membership in ZF). >> > > >> > > Now w is a limit ordinal so the ordered set N is, in >> > > the >> > > ordinal sense, a limit. Of course w is not a member >> > > of >> > > N >> > > becasuse then N would be a member of itself (not >> > > allowed >> > > by >> > > foundation). >> > >> > Note here that N (the set of natural numbers) is *not* >> > defined using a >> > limit at all. That w is called a limit ordinal is a >> > definition of the >> > term "limit ordinal". It does not mean that the >> > definition you use to >> > define it actually uses a limit. (And if I remember >> > right, a limit >> > ordinal is an ordinal that has no predecessor, see, >> > again >> > no limit >> > involved.) >> >> N can be defined as a limit or not as a limit. These are >> really equivalent approaches. > > In ZF, unions are defined only for sets of sets and for > such a set of > sets S, the union is defined a the set of all elements of > elements of S. Check out: http://planetmath.org/encyclopedia/SequenceOfSetsConvergence.html > Unless some other definition replaces this one, it is > impossible to form > the union of a family of sets unless that family are > already the > members of a set. > > In which case, N cannot be defined as the limit you > suggest, as it would > have to exist before it exists. Here is one way to define N as a limit. - Given a set x, the successor of x is the set x'=xU{x}. - A set y is inductive if x' is in y whenever x is in y. - Given an initial set x, the inductive set comprised of the successors of x is called a limit set of the sequence of sets x'=xU{x}, x''=x'U{x'}, ... . - Let N be the limit set formed from the initial set {}. In this case N is a convergent set: http://planetmath.org/encyclopedia/SequenceOfSetsConvergence.html It should be pointed out that N is a limit set even if N is initially given by a definition that doesn't involve the notion of a limit. Here is another way to see that N is a limit even if you consider it bad taste to define it in those terms: Let \/ = union Let /\ = Intersection Define infimum and supremum as follows: liminf X_n=\/(n=0, oo)[/\(m=n, oo) X_m] (n-->oo) limsup X_n=/\(n=0, oo)[\/(m=n, oo) X_m] (n-->oo) If these two are the same then the limit exists and is both of them. So, let's consider the naturals: X_0 = 0 = {} X_1 = 1 = {0} X_2 = 2 = {0,1} X_3 = 3 = {0,1,2} X_4 = 4 = {0,1,2,3} X_5 = 5 = {0,1,2,3,4} .... Evaluate the liminf case: /\(m=0, oo) X_m = X_0 /\ X_1 /\ X_2 ... = {} /\ {0} /\ {0,1} /\ {0,1,2} /\ ... = {} /\(m=1, oo) X_m = X_1 /\ X_2 /\ X_3 ... = {0} /\ {0,1} /\ {0,1,2} /\ ... = {0} /\(m=2, oo) X_m = X_2 /\ X_3 /\ X_4 ... = {0,1} /\ {0,1,2} /\ {0,1,2,3} /\ ... = {0,1} .... etc. Now, we \/(n=0, oo) to union all these together and we get N: \/(n=0, oo) = {} \/ {0} \/ {0,1} \/ ... \/(n=0, oo) = {0,1,2,3,...} \/(n=0, oo) = N Next, evaluate the limsup case: \/(m=0, oo) X_m = X_0 \/ X_1 \/ X_2 ... = {} \/ {0} \/ {0,1} \/ {0,1,2} \/ ... = {0,1,2,3,...} = N \/(m=1, oo) X_m = X_1 \/ X_2 \/ X_3 ... = {0} \/ {0,1} \/ {0,1,2} \/ ... = {0,1,2,3,...} = N \/(m=2, oo) X_m = X_2 \/ X_3 \/ X_4 ... = {0,1} \/ {0,1,2} \/ {0,1,2,3} ... = {0,1,2,3,...} = N .... etc. Now, we /\(n=0, oo) intersect all of these to get: /\(n=0, oo) = N /\ N /\ N /\ ... /\(n=0, oo) = N Limsup and liminf both give N and so the limit is N. k |