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From: K_h on 14 Dec 2009 20:05 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KunF0v.9y9(a)cwi.nl... > In article <k9mdnWn11pFIAbjWnZ2dnUVZ_uWdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > > > > let L=lim(n-->oo)X_n be the specified wikipedia limit > > for > > X_n. If L exists then: > > > > lim(n-->oo)A_n = lim(n-->oo){X_n} = {L} > > > > otherwise lim(n-->oo)A_n = lim(n-->oo){X_n} does not > > exist. > > Under this definition lim(n-->oo){n}={N} > > By what definition is it {N}? If a sequence of sets, A_n, cannot be expressed as {X_n}, for some sequence of sets X_n, then lim(n-->oo)A_n is defined by one of the two wikipedia limits. Otherwise let L=Wikilim(n-->oo)X_n be the specified wikipedia limit for X_n. If L exists then define lim(n-->oo)A_n=lim(n-->oo){X_n} as follows: lim(n-->oo)A_n = lim(n-->oo){X_n} = {L} otherwise lim(n-->oo)A_n = lim(n-->oo){X_n} does not exist. I referenced the wikipedia limit here just to save time. Which wikipedia definition is selected is up to the user. If you feel that defining a limit in terms of another limit definition is bad taste then the above definition could be easily reworded to include the relevant material without reference to wikipedia. > By what definition is: > lim(n -> oo) n = N? Any of the wikipedia definitions. Here is the proof again. Use this definition: - Given a sequence of sets S_n then: - lim sup{n -> oo} S_n contains those elements that occur in infinitely many S_n. - lim inf{n -> oo} S_n contains those elements that occur in all S_n from a certain S_n (which can be different for each element). - lim{n -> oo} S_n exists whenever lim sup and lim inf are equal. Theorem: lim(n ->oo) n = N. Consider the naturals: S_0 = 0 = {} S_1 = 1 = {0} S_2 = 2 = {0,1} S_3 = 3 = {0,1,2} S_4 = 4 = {0,1,2,3} S_5 = 5 = {0,1,2,3,4} .... S_n = n = {0,1,2,3,4,5,...,n-1} .... S_N = N = {0,1,2,3,4,5,...,n-1,n,n+1...} - lim sup{n -> oo} S_n contains those elements that occur in infinitely many S_n. * Every natural, n, occurs infinitely many times after S_n so limsup=N. - lim inf{n -> oo} S_n contains those elements that occur in all S_n from a certain S_n (which can be different for each element). * Every natural, n, occurs infinitely many times after S_n so liminf=N. - lim{n -> oo} S_n exists whenever lim sup and lim inf are equal. * limsup=liminf=N and so the lim(n ->oo)n exists and is N. k
From: Jesse F. Hughes on 14 Dec 2009 20:28 Ilmari Karonen <usenet2(a)vyznev.invalid> writes: > ["Followup-To:" header set to sci.math.] > On 2009-12-14, Jesse F. Hughes <jesse(a)phiwumbda.org> wrote: >> "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: >>> >>> But the standard topology on N is the discrete topology, too! Thus, >>> the standard definition of sequence convergence on N is inherited via >>> the subspace topology from Set. That is, a sequence >>> {a_n | n in N} c N converges (in N) to m iff >>> >>> (E k)(A j > k) a_j = m. >>> >>> This is (unless I'm just butt-wrong) the same as the definition of >>> sequence convergence on Set restricted to the subspace N. >> >> Yeah, well, I am just butt-wrong, ain't I? > > Well, not really. That's not the same as the definition of general > set convergence, but I do believe the two definitions are equivalent > for sequences of natural numbers, at least under any of the usual > set-theoretic constructions of the naturals. > > In particular, under the standard construction of the naturals, where > 0 = {} and n+1 = n union {n}, I believe the two definitions of lim sup > and lim inf also match: this is due to the fact that, for the natural > numbers m and n under this construction, m is a subset of n if and > only if m <= n. Oh. Okay, but if I'm right, it was only coincidence. So, perhaps I was butt-right. -- Jesse F. Hughes "We need to counter the shockwave of the evildoer by having individual rate cuts accelerated and by thinking about tax rebates." -- George W. Bush, Oct. 4, 2001
From: cbrown on 15 Dec 2009 01:35 On Dec 14, 5:28 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Ilmari Karonen <usen...(a)vyznev.invalid> writes: > > ["Followup-To:" header set to sci.math.] > > On 2009-12-14, Jesse F. Hughes <je...(a)phiwumbda.org> wrote: > >> "Jesse F. Hughes" <je...(a)phiwumbda.org> writes: > > >>> But the standard topology on N is the discrete topology, too! Thus, > >>> the standard definition of sequence convergence on N is inherited via > >>> the subspace topology from Set. That is, a sequence > >>> {a_n | n in N} c N converges (in N) to m iff > > >>> (E k)(A j > k) a_j = m. > > >>> This is (unless I'm just butt-wrong) the same as the definition of > >>> sequence convergence on Set restricted to the subspace N. > > >> Yeah, well, I am just butt-wrong, ain't I? > > > Well, not really. That's not the same as the definition of general > > set convergence, but I do believe the two definitions are equivalent > > for sequences of natural numbers, at least under any of the usual > > set-theoretic constructions of the naturals. > > > In particular, under the standard construction of the naturals, where > > 0 = {} and n+1 = n union {n}, I believe the two definitions of lim sup > > and lim inf also match: this is due to the fact that, for the natural > > numbers m and n under this construction, m is a subset of n if and > > only if m <= n. > > Oh. Okay, but if I'm right, it was only coincidence. So, perhaps I > was butt-right. > I'd say you were right, but... Cheers - Chas
From: Dik T. Winter on 15 Dec 2009 07:32 In article <03e1afc6-ec37-4212-b958-063a237d2bb4(a)f16g2000yqm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 11 Dez., 03:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > Have a look at <http://en.wikipedia.org/wiki/Lim_inf> in the secti= > on > > > > titled "Special case: dicrete metric". An example is given with > > > > the sequence {0}, {1}, {0}, {1}, ... > > > > where lim sup is {0, 1} and lim inf is {}. > > > > > > > > Moreover, in what way can a definition be invalid? > > > > > > It can be nonsense like the definition: Let N be the set of all > > > natural numbers. > > > > In what way is it nonsense? Either that set does exist or it does not > > exist. > > If it does exist there is indeed such a set, if it does not exist there is > > no set satisfying the definition. In both cases the definition is not > > nonsense in itself. > > It is nonsense to define a pink unicorn. That statement is nonsense. Let pu be a pink unicorn is a proper definition. However there is nothing that satisfies that definition. > The set N does not exist as > the union of its finite initial segments. This is shown by the (not > existing) path 0.000... in the binary tree. As you have defined your tree as only containing finite paths it is trivial to conclude that the infinite path 0.000... does not exist in your tree. However, this does *not* show that N does not exist. > Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}. > What then is > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} ? Ambiguous as stated. What is "..." standing for? Not for "go on in the same way until", because when you go on in the same way you will never get at {1, 2, 3, ...}. > If it is the same, then wie have a stop in transfinite counting. However, if we try to attach a meaning we find that it is {1, 2, 3, ...}. The reason being that uniting commutes, so it would be the same as: {1, 2, 3, ...} U {1} U {1, 2} U {1, 2, 3} ... Formally: let S_n = {1, 2, ..., n} for n a natural number. Let S_w = {1, 2, 3, ...}. Let I be the index set N U {w}. Then the union should be: union(i in I) S_i = {1, 2, 3, ...} > > But apparently you are of the opinion that you are only allowed to define > > things that do exist. > > Most essential things in mathematics exist without definitions and, > above all, without axioms. Oh. Not for a mathematician. If there is no definition or axiom there is no proof. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 15 Dec 2009 07:52
In article <iNWdnfmPh7NpQ7vWnZ2dnUVZ_hydnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:KunEFz.913(a)cwi.nl... .... > > > >> The basic idea of what a limit is suggests that an > > > >> appropriate definition for lim(n-->oo){n} should > > > >> yield > > > >> lim(n-->oo){n}={N}: > > > >> > > > >> {0}, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ...--> > > > >> {{0,1,2,3,4,...}} > > > > Why? > > Why not? See below. > > > The sensibility of a definition is the real issue. > > > Applying > > > the so-called standard definitions to {n} leads to a > > > cockamamie limit which is at odds with the general > > > notion of > > > a limit. > > > > It is not. > > Why not? See below. > > > Otherwise > > > let L=lim(n-->oo)X_n be the specified wikipedia limit > > > for > > > X_n. If L exists then: > > > > So you wish to use different definitions of limits > > depending on what > > the sequence of sets actually is? > > No, the defintion I provided is one defintion that includes > stuff from the wikipedia definition. The definition you provided for a sequence of sets A_n depends on whether each A_n is or is not a set containing a single set as an element. Your definition leads to some strange consequences. I can state the following theorem: Let A_n and B_n be two sequences of sets. Let A_s = lim sup A_n and A_i = lim inf A_n, similar for B_s and B_i. Let C_n be the sequence defined as: C_2n = A_n C_(2n+1) = B_n Theorem: lim sup C_n = union (A_s, B_s) lim inf C_n = intersect (A_i, B_i) Proof: easy. However with your definition for a sequence of sets depending on whether the terms of the sequence are or are not a set containing a single set as element, this theorem does not hold. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |