An uncountable countable set
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From: mueckenh on 28 Jun 2006 10:36 Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: > > > >> >f: 1 --> {1} and a --> { } with K_f = { }, > >> >g: 1 --> { } and a --> {1} with K_g = {1}. > >> >Here we have certainly no problem with lacking elements. > >> > >> The claim made by Cantor is that for any set A, there is > >> no surjection from A to P(A). Of course, if you construct > >> a new set B by adding extra elements to A, then there might > >> be a surjection from B to P(A). But why is that relevant? > > > >To show you that the condition K is impossible, independent of any > >surjectivity. > > The fact that f is not a surjection from A to P(A) *follows* from > the fact that K(f) is not in the image of f. So surjectivity is > *not* independent. > > >Map |R (including |N) on P(|N) with the only condition that a natural > >number has to be mapped on that set K e P(|N) which contains all > >natural numbers which are not mapped on sets containing them. You see: > >It is impossible a condition. > > You are just repeating once again that for any function f with > any domain A, the set K(f) = { x in A | x is not in f(x) } > is not in the image of f. Yes, we all agree. That means that > if f is a function from A to P(A), then f is *not* a surjection. > The impossibility of K(f) being in the image of f *implies* > the impossibility of there being a surjection from A to P(A). No. It shows the impredicable definition of K, but not more. It does not stronger imply the impossibility of a surjection from N to P(N) than it implies the impossibility of a surjection from R to P(N). It implies only any surjection including this impredicable condition. Regards, WM
From: mueckenh on 28 Jun 2006 10:40 Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Virgil schrieb: > > > > > >> It is not "Cantor's list" but ANY list. > >> You are challenged to find a list which lists all reals, then Cantor > >> presented a rule for showing any list you provide is incomplete. > > > >I have already shown that any list of all or even some rationals and > >its diagonal number is uncountable. > > No, you didn't, since it is false. > > 1. Start with a function f that is a surjection from > N to Q (Q is the rationals). > 2. Compute the diagonal d(f). This is a real number that is > guaranteed to not be in the image f. > 3. Form a new set M(f) = Q union { d(f) }. > 4. This new set is *countable* like any subset of P(N), unless you require the *definition* of K being included (the corresponding *set* K may be included, but in the form of an unconstructed set. As we know, there are many elements of P(N) which cannot be constructed. Regards, WM Regards, WM
From: mueckenh on 28 Jun 2006 10:55 Dik T. Winter schrieb: > In article <1151441001.426436.208850(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > But I still can be wrong, because the "ordinal > > > number" *can* change with a countably infinite number of interchanges. > > > Consider N with ordinal number w. Interchange 0 with 1, 0 with 2, ... > > > at the end you will have a well-ordered set with ordinal number w+1. > > > So to me the quote remains quite unclear. > > > > That is correct too, but the quote is irrelevant, because it is clear > > that a countable number of transpositions cannot change a well-ordered > > set into a not well-ordered set: There would have to be a first one to > > destroy the well-ordering. Perhaps Cantor meant that. > > This makes no sense. The quote specifically states that an infinite number > of transpositions does not change the "Anzahl" ("ordinal number"), but that > is false, as I did show above. And the quote was only talking about changing > "Anzahl". Your example makes use of k + n in the limit n --> oo. That may be prohibited. > > > > > Hence "Anzahl" is connected with the special kind of well-ordering. In > > > > particular different wel-orderings lead to different numbers > > > > (Anzahlen). > > > > > > Yes, this leads to the conclusion that it is the ordinal number of a set. > > > Every infinite set that can be well-ordered can be well-ordered in > > > different ways, leading to different ordinal numbers. But also, every > > > infinite set that can be well-ordered can also be ordered in such a way > > > that it is no longer well-ordered. > > > > But not by a countable number of transpositions. > > You state so (and I think it is true), but a quote from Cantor is not a proof, Cantor may have been wrong here. But my example is not. > more so if the quote is not even about this particular concept, it is about > "Anzahl", and there it is false (under some reasonable interpretation of > infinite number of transpositions). > > > > I do not know how to interprete it. A finite number of changes does not > > > change the ordinal number and well-ordering. A countably infinite number > > > of changes, with order type w, does not change well-ordering (I think) but > > > changes the ordinal number (but it depends on the meaning of a countably > > > infinite number of interchanges). But it is unclear to me how to > > > interprete an infinite number of interchanges. > > > > That is unimportant, because for my example there is only a countable > > number of transpositions required. > > How do I interprete an infinite number of transpositions? You say it is > unimportant, but it is very important. Unless I know that I can only come > up with reasonable interpretations, but not with defined interpretations, > whether the number is countably infinite or uncountably infinite. It is nothing else but the mapping between two infinite sets. And if the elements of inifinite sets do exist, then the transpositions, interpreted as elements of a countably infinite set, do exist. > > > Interprete it as a mapping. > > > > For n = 1 to oo: > > (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, > > else: (q_2n-1, q_2n) --> (q_2n, q_2n-1) > > This makes no obvious sense. If I interprete it right, this is not a > mapping, but a change of ordering. Yes? (With maps, ordering is > irrelevant.) Also the "<" used is apparently not the "<" used in the > ordering, but is related with the mathematical order. "<" means order by size, yes. > Moreover, there > is no "for n = 1 to oo", I have no idea what you mean with that. Please > be a bit more precise in the future. It is a mapping. Example: The set {1,6,5,4,8} is mapped on {1,6,4,5,8} by the first procedure which reorders the pairs with indices 2n-1 and 2n like indices 1 and 2 and 3 and 4. The next mapping maps {1,6,4,5,8} on {1,4,6,5,8}. It reorders the pairs with indices 2n and 2n+1 like 2 and 3 and 4 and 5. Then again use the first mapping and so on. > And again. > > > repeat aleph_0 times. > > Yeah. When are we done? This is a sequence of transpositions of order type > w * w. I have shown how that would destroy the well-orderedness of the > natural numbers. No, there is always one set indexed by the natural numbers. When are we done? When are we done with the mapping of an infinite set? When are we done with Cantor's diagonal? > > > > But your picking the rationals is really too much. If we accept the > > > intuitive "definition" of the limit and infinite "permutation" given > > > above, we could define the following: > > > P(k) = lim{n->oo} (k,k+1)(k,k+2)...(k,k+n) > > > as a permutation, and with the same method we could define: > > > lim{k->oo} P(0)P(1)...P(k) > > > which would reverse the ordering of the naturals and so is not a > > > well-order. > > > On the other hand, the number of transpositions is still countably finite > > > (but the order-type is not w, but w * w). > > > > I am not sure whether you can define k+n for n --> oo. > > Why do you think I define that? I just give an intuitive meaning to an > infinite number of transpositions. If you have a way to define such, pray > give it. Above you used the expression k+n in the limit n -> oo. > > > However, my mapping, given above, has always a well defined first > > element, and each subset has such a well-defined first element. > > Same with my sequence of transpositions of N given above. Then it is not an example for the the set w*w. Regards, WM
From: mueckenh on 28 Jun 2006 10:59 Virgil schrieb: > > > > And each time I implement the diagonal number K(f) into the list f > > > > (insert it before the first ordinary list number, for instance), you > > > > will construct another diagonal number K(g) and say, look here, this > > > > diagonal is not contained in your g. > > > > > > As that K(g) was not in M(f) , that is irrelevant. > > > > That the diagonal of Cantor's list is not in the list is irrelevant. > > That is the one thing about Cantor's anti-diagonal number that IS > relevant. That has turned out wrong. It is you special taste to think that it was a point, but it it is not. Regards, WM
From: Dik T. Winter on 28 Jun 2006 10:51
In article <1151505404.420349.107060(a)y41g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Daryl McCullough schrieb: .... > > >To show you that the condition K is impossible, independent of any > > >surjectivity. .... > No. It shows the impredicable definition of K, but not more. Id does not show the impredictable definition. Given any f, K(f) is well-defined. > It does not stronger imply the impossibility of a surjection from N to > P(N) than it implies the impossibility of a surjection from R to P(N). It implies the impossibility of a surjection from N to P(N) because it is not in the image (by its very definition). It does *not* imply the impossibility of a surjection from R to P(N) because it *can* be in the image. (Given a mapping f: R -> P(N), and the set K(f) = {x, x in N and x !in f(x)} there is nothing that makes it impossible that say f(1/3) = K(f).) > It implies only any surjection including this impredicable condition. Eh? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |