An uncountable countable set
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From: Daryl McCullough on 27 Jun 2006 18:20 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: >> So the answer is no? There is no function f enumerating >> all the real numbers? So the reals are uncountable. > >There is a mathematical form containing all real numbers of [0, 1]. In >binary representation it is given here. Why not call it a function in >an extended sense? The question is this: does it map the naturals onto the real numbers in [0,1] such that every real number is in its image? In particular, with your representation, which natural number maps to the real 1/3? -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Jun 2006 18:25 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). -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Jun 2006 18:30 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*. It's countable because it is counted by the function g defined by g(0) = d(f) g(n+1) = f(n) -- Daryl McCullough Ithaca, NY
From: Dik T. Winter on 27 Jun 2006 20:12 In article <$ZHwUumviayC(a)eisner.encompasserve.org> briggs(a)encompasserve.org writes: > In article <vmhjr2-27492D.13460426062006(a)news.usenetmonster.com>, Virgil <vmhjr2(a)comcast.net> writes: > > In article <1151332937.425608.129380(a)i40g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > >> Dik T. Winter schrieb: > >> > In article <1151135467.925502.14460(a)b68g2000cwa.googlegroups.com> > >> > mueckenh(a)rz.fh-augsburg.de writes: > >> > > Dik T. Winter schrieb: .... > >> > > > K =3D {} > > > > What does 3D {} mean? > > It's an artifact of the MIME quoted-printable encoding. The original > message containing the "=" was encoded in quoted-printable. And the > person who replied and quoted the passage including the "=" did not > use software that understood quoted-printable. Not entirely true, but I plead guilty. The original message had a plain equals sign. The message quoting it used quoted-printable (rather than ISO 8859-1 which would have worked just as well). I re-quoted it and my newsreader does not understand quoted-printable indeed (it understands none of the MIME encodings), so there it came out. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 27 Jun 2006 20:51
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". > > > 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, 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. > > Under two conditions: > > No. These conditions are irrelevant. Yes, they are. > > 1. Cantor was right. Under my current interpretation, Cantor was wrong. A countably infinite number of transpositions can change the ordinal number and well-ordering. > > 2. Your interpretation of what Cantor wrote was right. > > I think (2) is wrong, I do not know whether (1) is correct or wrong, > > because I do not know how Cantor would interprete an infinite number > > of transpositions. > 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. 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. > For n = 1 to oo: > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, > else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) Again. > And again: > 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) And again. > For n = 1 to oo: > (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, > else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) 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. > > 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. > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |