Another AC anomaly?
in [Logic]
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From: Marshall on 11 Dec 2009 22:49 On Dec 11, 6:39 pm, "K_h" <KHol...(a)SX729.com> wrote: > "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote in messagenews:KuHL7p.3zE(a)cwi.nl... > > > > > In article <5ZedndTmvoBac7zWnZ2dnUVZ_qOdn...(a)giganews.com> > > "K_h" <KHol...(a)SX729.com> writes: > > > "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote in message > > >news:KuFy3L.Cxt(a)cwi.nl... > > ... > > > > Have a look at <http://en.wikipedia.org/wiki/Lim_inf> > > > > in > > > > the section > > > > 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 depends on the context. When it comes to supertasks, > > > limsup={0,1} is basically useless. That is why those > > > definitions are not good, and invalid, for evaluating > > > supertasks -- in response to WM's supertask issues. > > > I am not discussing supertasks, nor is WM here. The > > question is simply > > whether it is possible that: > > lim | S_n | != | lim S_n | > > with some form of limit. And by the definitions I gave > > (and which you > > also will find on the wikipedia page above): > > lim(n -> oo) {n} = {} > > 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. WM attributed > this definition to you: > > - 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. > > All the definitions on the wikipedia pages, and the above > definition WM attributes to you, give lim(n ->oo){n}=N. It certainly appears to me that the definitions quoted give lim(n -> oo){n} = {}. One of us is misreading it. Marshall
From: WM on 12 Dec 2009 04:32 On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > On Dec 11, 11:33 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > They *are* just the same, because your argument that above procedure > > would prove an infinite string of 3's is wrong. There is neither a > > natural nor a rational with an infinite string of digits. To be able > > to determine every digit of a number you like does not imply that > > there is a number with a never ending sequence of digits. > > If your math can't handle 1/3, it is exceedingly weak. > A third grader can do as much. You propose various > severe restrictions on math and say that everyone > else ought to adopt them, but you supply no motivation for > doing so. Every digit of 1/3 that you can handle, I can handle too. Only the claim "all digits" is incorrect. In my mathematics, this is seen, in yours it is not. > > The answer to your "why" question is because many > good and useful results come about if I do. For example, > I can handle decimal expansions of fractions such > as 1/3. I know you don't like many of those results, > however, coming up with results that are palatable > to some guy I never met is not a goal of mine. > > To get any attention to your demands, you need to > supply some motivation for people to listen. 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): > > > {1} > > {1, 2} > > {1, 2, 3} > > _________ > > {1, 2, 3} > > ======== > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > _________ > > {1, 2, 3, ...} > > ========= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > _________ > > {1, 2, 3, ...} > > ========= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > _____________ > > {1, 2, 3, ..., a} > > ============ > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ______________ > > {1, 2, 3, ..., a, aa} > > ============= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ... > > __________________ > > {1, 2, 3, ..., a, aa, ...} > > ================= > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ..., a} > > {1, 2, 3, ..., a, aa} > > ... > > {1, 2, 3, ..., a, aa, ...} > > __________________ > > {1, 2, 3, ..., a, aa, ...} > > ================= Contrary to the usual 1, 2, 3, ..., w, w+1, w+2, ..., w+w, w+w+1, ... this leads to the following sequence which stops from time to time (at limit ordinal paths): {1} {1, 2} {1, 2, 3} .... {1, 2, 3, ...} {1, 2, 3, ...} {1, 2, 3, ..., a} {1, 2, 3, ..., a, aa} {1, 2, 3, ..., a, aa, aaa} .... {1, 2, 3, ..., a, aa, aaa, ...} {1, 2, 3, ..., a, aa, aaa, ...} {1, 2, 3, ..., a, aa, aaa, ..., b} {1, 2, 3, ..., a, aa, aaa, ..., b, bb} {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb} .... {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb, ...} {1, 2, 3, ..., a, aa, aaa, ..., b, bb, bbb, ...} .... The first double-meaning, at {1, 2, 3, ...}, is responsible for the fact that the infinite binary tree can be constructed completely from finite paths (hence contains only countably many paths) and simultaneously is said to contain uncountably many infinite paths. Regards, WM
From: Marshall on 12 Dec 2009 11:19 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. Marshall
From: WM on 12 Dec 2009 12:02 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? 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. So every set of natural numbers has at most one number that is as large as its cardinal number. Using initial sets of only even numbers, this becomes immediately clear: 2 2, 4 2, 4, 6 .... These sets contain always numbers that are larger than the cardinal number of the set. But this fact does not stop set theorists from believing in actual infinity. They do not wish to see that ordinal number = cardinal number as long as only initial segments of natural numbers are involved. Therefore I have looked for a model which supplies ordinal and cardinal numbers in one and the same elements. This model is the binary tree. If you union paths, then you get the sequence > {1} > {1, 2} > {1, 2, 3} > ... > {1, 2, 3, ...} > {1, 2, 3, ...} i.e., when you union only all finite initial segments of 0.000... you get exactly the same result as when you union all finite initial segments of 0.000... *and* 0.000... 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. Of course, if only a small collection of paths from the tree is presented, then one can find paths in the tree which are not included in that collection. But this does not help to explain an uncountable set of paths: 1) We can construct the tree from its finite paths. In order to obtain one infinite "bonus" path, you need aleph_0 finite paths. But in order to explain the existence of uncountably many infinite paths, you would need to obtain at least aleph_0 infinite paths with every constructed finite path - and even that would not be enough because aleph_0 * aleph_0 = aleph_0. 2) If you construct the tree by means of infinite paths, then the ratio between constructed paths and constructed nodes is 1/aleph_0 at every step of the construction for every set of nodes covered by the construction. Do you seriously expect that this ratio can become 2^aleph_0 - after all nodes will have been constructed (not *during* construction, because as long as one node has not been constructed, the ratio remains the same)? Regards, WM
From: Jesse F. Hughes on 12 Dec 2009 12:23
"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 = {}. -- Jesse F. Hughes "It is a clear sign that something is very, very, very wrong, as human beings are, well human. Maybe some people think that mathematicians are not, but I disagree. They are human beings." -- James S. Harris |