Another AC anomaly?
in [Logic]
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From: Virgil on 12 Dec 2009 17:07 In article <e263b0aa-9581-4a72-8128-8da5131e0d4b(a)m26g2000yqb.googlegroups.com>, WM <mueckenh(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. You last posting does not make any false claims true however vociferously it claims to do so. > But those who try but > cannot not understand these few sentences will not understand the > proof in either form. Have you tried? One cannot understand what does not exist, and the claimed proof does not exist as it is not a proof at all. > > > If it is true, then you can formalize it. If you do that > > then you will get attention. > > Would you be willing to go through those roughly 20 pages? And if so, > would you be able to understand it then? If those pages were WM's, there is little point in trying, as WM cannot produce proofs of things already proved false. In fact, there is considerable evidence that WM cannot usually produce proofs of things known to be true either, unless he copies them from others. > > Regards, WM
From: George Greene on 12 Dec 2009 18:16 On Dec 12, 12:45 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > Would you be willing to go through those roughly 20 pages? And if so, > would you be able to understand it then? We all know how to understand first-order logic. If you are not being understood then it is because you are TOO DAMN STUPID TO DO first-order logic.
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. |