Another AC anomaly?
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From: K_h on 12 Dec 2009 21:24 "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message news:87vdgcksy2.fsf(a)phiwumbda.org... > "K_h" <KHolmes(a)SX729.com> writes: > >> The only way lim(n ->oo){n}={} is if limsup and liminf >> both >> equal {}. If limsup and liminf are different then the >> limit >> does not exist and cannot equal an existing set like {}. >> Since the empty set exists, and since you are claiming >> that >> lim(n ->oo){n}={}, you need to show that limsup and >> liminf >> are both {} by the definition you are using. > > Yes, he needs to show that, but it is utterly trivial and > obvious from > the definition of limsup and liminf given here. I'm not > sure why you > think it's not obvious, but here's the proof. > > Now, let X_n = {n}. Thus, n is in X_k <-> n = k. > > n in lim sup X_k iff n is in infinitely many X_k, but we > see from the > above that n is in only one X_k. Thus, lim sup X_k = {}. > > n in lim inf X_k iff there are only finitely many X_k such > that n not > in X_k, but again, we see that this is false for every n. > Hence > lim inf X_k = {}. So, you're claiming that he is not using { and } just to bracket the argument (i.e. the X_n to be limited) but {X_n} refers to a set containing the one set X_n. Then it seems like the meaning has changed because in previous posts he writes that lim|S_n|=/=|limS_n| follows from a wikipedia definition applied to sequences of natural numbers n -- not to the non-naturals {n}. For example: > > I have explicitly defined the limit of a sequence of sets. With that > > definition (and the common definition of limits of sequences of natural > > numbers) I found that the cardinality of the limit is not necessarily > > equal to the limit of the cardinalities. Okay, if {X_n} refers to a set containing the single set X_n then lim(n-->oo){n} is not a limit of the natural numbers since the naturals are not the sets {n} but the sets n. In this case my proof shows that lim(n-->oo)n=N. Applying the wikipedia definitions to n is sensible but applying them to {n} makes a mockery of the notion of a limit. The basic idea behind a limit is that things in one state tend to some final state and a good definition and application of a limit should embody that. In looking at the sequence {n}, with 0={} and 1={0}, saying that it tends to 0={} is a betrayal of the core idea behind a limit: 1, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> 0 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}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> {{0,1,2,3,4,...}} In other words, applying the wikipedia definitions to {n} is an abuse of those definitions. The definition that is used for a limit should make sense for the kind of object it is applied to. k
From: K_h on 12 Dec 2009 21:24 "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message news:87vdgcksy2.fsf(a)phiwumbda.org... > "K_h" <KHolmes(a)SX729.com> writes: > >> The only way lim(n ->oo){n}={} is if limsup and liminf >> both >> equal {}. If limsup and liminf are different then the >> limit >> does not exist and cannot equal an existing set like {}. >> Since the empty set exists, and since you are claiming >> that >> lim(n ->oo){n}={}, you need to show that limsup and >> liminf >> are both {} by the definition you are using. > > Yes, he needs to show that, but it is utterly trivial and > obvious from > the definition of limsup and liminf given here. I'm not > sure why you > think it's not obvious, but here's the proof. > > Now, let X_n = {n}. Thus, n is in X_k <-> n = k. > > n in lim sup X_k iff n is in infinitely many X_k, but we > see from the > above that n is in only one X_k. Thus, lim sup X_k = {}. > > n in lim inf X_k iff there are only finitely many X_k such > that n not > in X_k, but again, we see that this is false for every n. > Hence > lim inf X_k = {}. So, you're claiming that he is not using { and } just to bracket the argument (i.e. the X_n to be limited) but {X_n} refers to a set containing the one set X_n. Then it seems like the meaning has changed because in previous posts he writes that lim|S_n|=/=|limS_n| follows from a wikipedia definition applied to sequences of natural numbers n -- not to the non-naturals {n}. For example: > > I have explicitly defined the limit of a sequence of sets. With that > > definition (and the common definition of limits of sequences of natural > > numbers) I found that the cardinality of the limit is not necessarily > > equal to the limit of the cardinalities. Okay, if {X_n} refers to a set containing the single set X_n then lim(n-->oo){n} is not a limit of the natural numbers since the naturals are not the sets {n} but the sets n. In this case my proof shows that lim(n-->oo)n=N. Applying the wikipedia definitions to n is sensible but applying them to {n} makes a mockery of the notion of a limit. The basic idea behind a limit is that things in one state tend to some final state and a good definition and application of a limit should embody that. In looking at the sequence {n}, with 0={} and 1={0}, saying that it tends to 0={} is a betrayal of the core idea behind a limit: 1, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> 0 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}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> {{0,1,2,3,4,...}} In other words, applying the wikipedia definitions to {n} is an abuse of those definitions. The definition that is used for a limit should make sense for the kind of object it is applied to. k
From: Jesse F. Hughes on 12 Dec 2009 23:22 "K_h" <KHolmes(a)SX729.com> writes: > "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message > news:87vdgcksy2.fsf(a)phiwumbda.org... >> "K_h" <KHolmes(a)SX729.com> writes: >> >>> The only way lim(n ->oo){n}={} is if limsup and liminf >>> both >>> equal {}. If limsup and liminf are different then the >>> limit >>> does not exist and cannot equal an existing set like {}. >>> Since the empty set exists, and since you are claiming >>> that >>> lim(n ->oo){n}={}, you need to show that limsup and >>> liminf >>> are both {} by the definition you are using. >> >> Yes, he needs to show that, but it is utterly trivial and >> obvious from >> the definition of limsup and liminf given here. I'm not >> sure why you >> think it's not obvious, but here's the proof. >> >> Now, let X_n = {n}. Thus, n is in X_k <-> n = k. >> >> n in lim sup X_k iff n is in infinitely many X_k, but we >> see from the >> above that n is in only one X_k. Thus, lim sup X_k = {}. >> >> n in lim inf X_k iff there are only finitely many X_k such >> that n not >> in X_k, but again, we see that this is false for every n. >> Hence >> lim inf X_k = {}. > > So, you're claiming that he is not using { and } just to > bracket the argument (i.e. the X_n to be limited) but {X_n} > refers to a set containing the one set X_n. Er, yes. Though, something seems to be wrong with your notation. At issue is the set X_n = {n}, not the set {X_n}. In summary: |lim X_n| = |lim {n}| = |{}| = 0. lim |X_n| = lim |{n}| = lim 1 = 1. > Then it seems like the meaning has changed because in previous posts > he writes that lim|S_n|=/=|limS_n| follows from a wikipedia > definition applied to sequences of natural numbers n -- not to the > non-naturals {n}. For example: > > > > I have explicitly defined the limit of a sequence of > sets. With that > > > definition (and the common definition of limits of > sequences of natural > > > numbers) I found that the cardinality of the limit is > not necessarily > > > equal to the limit of the cardinalities. And that's absolutely correct, as we see above. > Okay, if {X_n} refers to a set containing the single set X_n > then lim(n-->oo){n} is not a limit of the natural numbers > since the naturals are not the sets {n} but the sets n. Er, yes. Of course. > In this case my proof shows that lim(n-->oo)n=N. Applying the > wikipedia definitions to n is sensible but applying them to {n} > makes a mockery of the notion of a limit. You have some very odd notions yourself. It's a simple application of a perfectly sensible definition of limit. > The basic idea behind a limit is that things in one state tend to > some final state and a good definition and application of a limit > should embody that. In looking at the sequence {n}, with 0={} and > 1={0}, saying that it tends to 0={} is a betrayal of the core idea > behind a limit: > > 1, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> 0 > > 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}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> > {{0,1,2,3,4,...}} > > In other words, applying the wikipedia definitions to {n} is > an abuse of those definitions. The definition that is used > for a limit should make sense for the kind of object it is > applied to. You're welcome to your own cockamamie opinions about whether a particular definition is sensible or not, but they're utterly irrelevant to the issue at hand. The fact is that with this *perfectly standard* definition of limits, we see that lim |X_n| != |lim X_n|. That's all there was at issue. -- "Sorry, wakeup to the real world. You're on your own dependent on me as your guide. Luckily for you, I'm self-correcting to a large extent, so if the proof were wrong, I'd tell you. It's not wrong." --- James Harris confirms that his proof is correct.
From: WM on 13 Dec 2009 09:56 On 12 Dez., 22:44, Marshall <marshall.spi...(a)gmail.com> wrote: > On Dec 12, 9:45 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 18:29, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > On Dec 12, 9:02 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 12 Dez., 17:19, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > On Dec 12, 1:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > > > Showing > > > > > > > a contradiction would qualify, but it's been well > > > > > > > established that you don't know how to do that. > > > > > > > Consider how a union of paths is counted (I copy > > > > > > from another posting, therefore the quotation symbols): > > > > > > Ascii diagrams don't qualify as a contradiction. > > > > > Why should pictures, diagrams, acoustic signals etc. qualify less than > > > > sequences of symbols? > > > > With pictures, diagrams, etc. the possibilities for tomfoolery > > > are endless. A formal proof is more resistant to human > > > error. Anyway, if your diagrams are sound, translating > > > them into formal proofs should not be out of reach. > > > > > Every bit of information, in what form ever, can > > > > be used in proofs. But here you are: > > > > > The union of all natural numbers is, according to set theory, omega.. > > > > If *actual* infinity is meant, this is plainly impossible, because the > > > > natural numbers count themselves. > > > > No natural number counts how many natural numbers there are. > > > > > This leads to the result that the same structure, namely the tree with > > > > all its nodes, contains only a countable set of paths and > > > > simultaneously it contains an uncountable set of paths. > > > > > And this is a contradiction. > > > > That actually would be a contradiction if it were true. > > > It is true as can be seern from my last posting. But those who try but > > cannot not understand these few sentences will not understand the > > proof in either form. > > Have you tried? > > Yes. It was riddled with obvious errors. Which error was obvious? Regards, WM
From: Virgil on 13 Dec 2009 14:49
In article <4934d93b-3b3f-4e04-a831-8fc6ee4465ab(a)m26g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Dez., 22:44, Marshall <marshall.spi...(a)gmail.com> wrote: > > On Dec 12, 9:45�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 12 Dez., 18:29, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > On Dec 12, 9:02�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 17:19, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > On Dec 12, 1:32�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > > > > > Showing > > > > > > > > a contradiction would qualify, but it's been well > > > > > > > > established that you don't know how to do that. > > > > > > > > > Consider how a union of paths is counted (I copy > > > > > > > from another posting, therefore the quotation symbols): > > > > > > > > Ascii diagrams don't qualify as a contradiction. > > > > > > > Why should pictures, diagrams, acoustic signals etc. qualify less than > > > > > sequences of symbols? > > > > > > With pictures, diagrams, etc. the possibilities for tomfoolery > > > > are endless. A formal proof is more resistant to human > > > > error. Anyway, if your diagrams are sound, translating > > > > them into formal proofs should not be out of reach. > > > > > > > Every bit of information, in what form ever, can > > > > > be used in proofs. But here you are: > > > > > > > The union of all natural numbers is, according to set theory, omega. > > > > > If *actual* infinity is meant, this is plainly impossible, because the > > > > > natural numbers count themselves. > > > > > > No natural number counts how many natural numbers there are. > > > > > > > This leads to the result that the same structure, namely the tree with > > > > > all its nodes, contains only a countable set of paths and > > > > > simultaneously it contains an uncountable set of paths. > > > > > > > And this is a contradiction. > > > > > > That actually would be a contradiction if it were true. > > > > > It is true as can be seern from my last posting. But those who try but > > > cannot not understand these few sentences will not understand the > > > proof in either form. > > > Have you tried? > > > > Yes. It was riddled with obvious errors. > > Which error was obvious? Among many others, your claim that in a complete infinite binary tree there are as many nodes as sets of nodes. |