Another AC anomaly?
in [Logic]
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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
From: Virgil on 12 Dec 2009 16:26 In article <f0de1f04-bdb7-445f-a557-6045bd97a8d9(a)m38g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. You mean that there are digits that you must leave out? > In my mathematics, this is seen, in > yours it is not. In our mathematics, we do not have to leave such things out. We can deal with ALL the digits binary digits in 1/3, represented by the the simple function f:(1,2,3,...} -> {0,1} for which f(odd) = 0 and f(even) = 1.
From: Marshall on 12 Dec 2009 16:44 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. > > 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? I probably wouldn't be willing to do so, given your record. But the good news is that first order logic is mechanizable. So you could produce a computer-readable representation of your proof and have it verified by a proof verifier. Then I wouldn't need to read the whole thing; just the premises and the conclusion. Marshall
From: Virgil on 12 Dec 2009 17:00
In article <d8ef5daa-b9d6-4c51-b40d-0ea5336b1326(a)v25g2000yqk.googlegroups.com>, WM <mueckenh(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? 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. Non sequitur. > So every set of natural numbers has > at most one number that is as large as its cardinal number. That presumes what it is alleged to establish, so is invalid. > 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 the set {2,4,6,...} does not. > But this fact does not stop set theorists from believing in actual > infinity. And their doing so causes no problems for anyone with more sense than WM. > They do not wish to see that ordinal number = cardinal > number as long as only initial segments of natural numbers are > involved. A limitation claimed but never proven. > > 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. WM's so called binary trees are all incomplete. Complete INFINITE binary trees do not have any finite paths but do have lots of paths. > If you union paths, then you get the sequence The following are not paths in an infinite tree but merely finite initial segments of such paths. And while they may be paths in finite trees, they are not paths in the same tree > > > > {1} > > {1, 2} > > {1, 2, 3} > > ... > > {1, 2, 3, ...} > > {1, 2, 3, ...} > But according to WM you cannot union those paths because the result is not a path. > 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... In ZF, the union of a family of sets is not changed when one appends any union of a subset of those sets to the set of sets. This in no way supports WM's delusions. > > 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. There is no example of a COMPLETE infinite binary tree in which any finite set of nodes is path. > > And this is a contradiction. It is a lie. > > 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. If those alleged paths are finite, then they are not paths in a COMPLETE and INFINITE binary tree. > > But this does not help to explain an uncountable set of paths Lies do not, in general, help to exxplain anything. > 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. The point is that from that set of aleph_0 finite paths it is the numbers of subsets which determines the number of infinite paths, and in ZF, the number of subsets of a set is always greater than the number of members of a set. Does WM have any evidence that in his world this is not also the case? > 2) If you construct the tree by means of infinite paths One constructs an infinite tree as an infinite sets of nodes, and a path in such a tree as a set of nodes in which no node has more than one child. Each such MAXIMAL set is a path, but no finite set is a path. If, as in an infinite binary tree, one has countably many nodes, then one has necessarily more sets of nodes than nodes, and there are simple proofs that there are MORE paths than nodes. WM claims that these proofs do not hold, but his arguments are false outside of Wolkenmuekenheim. > 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. In that case, one could have only finitely many finite paths encompassing infinitely many nodes. Which won't work even in Wolkenmuekenheim. > 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)? As you analysis is false ab initio, I do not expect it to become true ever. |