Another AC anomaly?
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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
From: Marshall on 12 Dec 2009 12:29 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. If it is true, then you can formalize it. If you do that then you will get attention. Marshall
From: WM on 12 Dec 2009 12:45
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? > 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? Regards, WM |