Cantor Confusion
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From: mueckenh on 20 Dec 2006 16:48 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > As usual you cut the relevant parts: > > ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] > | >> >> > If an unbounded diagonal exists, then obviously an unbounded > | >> >> > line > | >> >> > must exist. > | >> >> > [(*)] > | >> 6. You shall refocus on a proof of (*). > | > > | > Every element of the diagonal is n some line. Do you agree? [(1)] > | > Every element of the diagonal is a finite natural number, by > | > definition. [(2)] > | > > | > Up to the finite natural number n: Every element =< n of the > | > diagonal > | > is contained in line n. Induction shows this is valid for any finite > | > natural number. [(3)] > `---- > > >> That does not prove (*) since the whole diagonal is _not_ bounded by > >> any such n. > > > > If you doubt my proof, > > Which proof? You simply stated: > > (1) forAll elem in diag Exists line (m is in line) > (2) forAll elem in diag (elem is finite natural number) > (3) forAll natural n (forAll elem in diag <= n > is in line n) > > So please show us _step-by-step_ how (*), which states > > (*) notExists natural n (forAll m in diag (m <= n)) > -> Exists natural n' (notExists natural n(forAll m > in line n' (m <=n)), > > follows from (1), (2) and (3). See my recent reply to William Hughes. Regards, WM
From: William Hughes on 20 Dec 2006 17:13 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > Please give a counter example for a finite number n. > > > > No such counterexample exists. The claim is that despite > > the fact that given any n we can find a single line, L(n), that works > > for this n, there is no single line, L_D, that will work for all n. > > This claim is obviously false. At least my counter claim cannot be > disproved by an example.> see the example that disproves your counter claim below. (by mistake you snipped it) > > > > > Remember, even if there is not a last line, every n eps N is finite. > > > > Which means that for any n we can find a single line L(n) that works > > for this n. It does not mean that that we can find a single > > line L_D, that works for all n. > > That is impossible, beause there is no "all n". "It does not mean that that we can find a single line L_D, that works for all n." written out in full is It does not mean that there exists an L_D with the property that: if a natural number n can be shown to exist in the diagonal n must be an element of L_D. There is no claim that "all n" exists. Restoring the example you snipped (and expanding the language slightly) There exists a linearly ordered set of finite elements that does not support quantifier reversal. Consider the (potentially infinite) set of natural numbers. This is linearly ordered. Every element is finite. For any natural number n that can be shown to exist, a natural number m(n), such that m(n) > n can be shown to exist. However it is not true that: There exists a natural number m, such that for any number n that can be shown to exist, m>n. - William Hughes
From: Franziska Neugebauer on 20 Dec 2006 17:29 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> As usual you cut the relevant parts: >> >> ,----[ <4588194e$0$97248$892e7fe2(a)authen.yellow.readfreenews.net> ] >> | >> >> > If an unbounded diagonal exists, then obviously an >> | >> >> > unbounded line >> | >> >> > must exist. >> | >> >> > [(*)] >> | >> 6. You shall refocus on a proof of (*). >> | > >> | > Every element of the diagonal is n some line. Do you agree? >> | > [(1)] >> | > Every element of the diagonal is a finite natural number, by >> | > definition. [(2)] >> | > >> | > Up to the finite natural number n: Every element =< n of the >> | > diagonal >> | > is contained in line n. Induction shows this is valid for any >> | > finite natural number. [(3)] >> `---- >> >> >> That does not prove (*) since the whole diagonal is _not_ bounded >> >> by any such n. >> > >> > If you doubt my proof, >> >> Which proof? You simply stated: >> >> (1) forAll elem in diag Exists line (m is in line) >> (2) forAll elem in diag (elem is finite natural number) >> (3) forAll natural n (forAll elem in diag <= n >> is in line n) >> >> So please show us _step-by-step_ how (*), which states >> >> (*) notExists natural n (forAll m in diag (m <= n)) >> -> Exists natural n' (notExists natural n(forAll m >> in line n' (m <=n)), >> >> follows from (1), (2) and (3). > > See my recent reply to William Hughes. Please copy and paste. I am not able to guess what you have in mind. F. N. -- xyz
From: Virgil on 20 Dec 2006 19:36 In article <1166651240.822366.231880(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > Please give a counter example for a finite number n. > > > > No such counterexample exists. The claim is that despite > > the fact that given any n we can find a single line, L(n), that works > > for this n, there is no single line, L_D, that will work for all n. > > This claim is obviously false. At least my counter claim cannot be > disproved by an example. WRONG! Consider the model given within any infinite tree by any of its (infinite) paths. The path is the diagonal and each line is the section of that path from the root node to any other node. Each edge in the path is labeled by (identified with) the natural number of edges in the line of which it is the last edge. In this model, every natural appears in some line. In this model, there is no line in which every natural appears. This model disproves WM's contentions about his EIT. > > > > > Remember, even if there is not a last line, every n eps N is finite. > > > > Which means that for any n we can find a single line L(n) that works > > for this n. It does not mean that that we can find a single > > line L_D, that works for all n. > > That is impossible, beause there is no "all n". If "all n" were > somewhere, then we had no problem to put them in one single line. See the explicit example above in which WM's claim is false. > Due > to the finiteness of every n, every position of the line was finite. > And due to the fact that a number with only finite indexes is a natural > number, the whole number was a natural number. But the diagonal in my above model is not contained in any line > > Recently we had the following claim (by Virgil) > > The number of equations > > 1+1+1+...+1 = n > > with finitely many 1 is infinite. Considering the fact that there are > exactly as many 1 as different numbers, it is impossible that the > number of numbers can be infinite while the sme number of 1 is finite. > > That is obviously our old EIT. > > 1 > 11 > 111 > ... > > The number of 1 is not *actually* infinite without one line being > actually infinite. See the above model for an EIT in which the diagonal MUST contain infinitely many objects while each line MUST be finite. > > > > Therefore your claim is void, unless you can find a linear set of > > > finite elements which forbids exchange of quantifiers. Done! For every edge in the full path there exists a finite subpath staring at the root node which contains that edge. There are, in fact, infinitely many such finite subpaths. BUT: There does not exist any line (finite subpath staring at the root node) which contains every edge of the diagonal (infinite path). > > There exists a simple proof that all elements of the diagonal are in > the lines. And by the linearity and finity of the lines we find that > all elements which exist in some lines exist in one line. Not in the counterexample modeled above where the diagonal is an infinite path in an infinite binary tree. And since WM has committed himself to the existence of such trees, he is stuck with them. This is a > proof by induction covering all finite natural numbers, i.e., all > elements of the diagonal. > > Regards, WM
From: Virgil on 20 Dec 2006 19:39
In article <1166651337.524455.310670(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > If you doubt my proof, > > > > Which proof? You simply stated: > > > > (1) forAll elem in diag Exists line (m is in line) > > (2) forAll elem in diag (elem is finite natural number) > > (3) forAll natural n (forAll elem in diag <= n > > is in line n) > > > > So please show us _step-by-step_ how (*), which states > > > > (*) notExists natural n (forAll m in diag (m <= n)) > > -> Exists natural n' (notExists natural n(forAll m > > in line n' (m <=n)), > > > > follows from (1), (2) and (3). > > See my recent reply to William Hughes. Why bother, as it is almost certain to be as clearly flawed as all WM's other circular arguments. |