An uncountable countable set
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From: Dik T. Winter on 19 Aug 2006 21:30 In article <ec7gek$frg$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > Inductively provable: Any set of consecutive naturals starting at 1 > always has its largest element equal to size of the set. What is the size of the empty set? What is the largest element of the empty set? -- 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 19 Aug 2006 21:42 The original did not arrive her, so I am piggybacking. In article <1156009246.842952.233770(a)i42g2000cwa.googlegroups.com> imaginatorium(a)despammed.com writes: Actually nothing that I quote. It is more about: > Tony Orlow wrote: > > A circle is a regular polygon with an infinite number of infinitesimal > > sides. The formula for the angle of the vertices of a regular polygon of > > n sides is 2*pi*(1/2-1/n). The limit of this formula as n->oo is pi. Tony, I see you did agree with Wolfgang Mueckenheim in this thread. But I should warn you. According to WM neither pi nor that limit do exist. (Gives me a deja vu about JSH agreeing with EEE. While the latter proved that FLT was false and the former proved that it was right.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Tony Orlow on 19 Aug 2006 22:16 Dik T. Winter wrote: > In article <1155999375.602088.28870(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > Again a conclusion without proof. Use my definition of K: K[p] = 1 for > > > all p in N, and there are no other digits. > > > > Why should I? > > I would think because we are talking about that number? You keep stating > that that number is not indexable, without proof. When you want to show > that you have to start with that definition. > > > Use my definition of the indexable digit positions and > > there are other digits than those of numbers of the list. > > This sentence does not parse. What is the meaning of the part after "and"? > Anyhow, the number, with my definition, does not have digit positions other > than those on the list. > > > > Show that that number can not be indexed or show that that number can > > > be covered by a natural number, give a *proof* of either. > > Again, I ask yuo that. > > > > > But you pretend it would consist of index positions which are all in > > > > the list. > > > > > > Not pretend, define. > > > > I define the list such that your definition is wrong. if all natural > > numbers do exist, my list is complete. > > What is the wrongness of the definition? Let me refrase: > Let L be the list of all natural numbers > Let K be the sequence of digits such that for each p in L, K[p] = 1. > What is *wrong* with that definition? Hi Dik - Sorry to interject, but could you tell me how long a sequence K is? If it's finite, then you only have a finite list L, and if it's countably infinite, then you have an uncountable list of naturals? Is there something in between? Could K have a length that is not possible given standard transfinite set theory? :) TO > > > > > Now find out how that can be: All index positions of 0.111... are in > > > > the list, but not in the form of 0.111... . In which form should they > > > > exist there? > > > > > > What do you mean with "in the form of 0.111..."? > > > > The required digit positions are not contained in the list in form of > > this infinite number but only disperged over many finite numbers (which > > is impossible, of course). > > Why is that impossible? There is no "of course" here, because that "of > course" is just stating "because the set of all naturals does not exist".
From: Tony Orlow on 19 Aug 2006 22:22 Dik T. Winter wrote: > In article <1155999560.728582.29520(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > But I now understand that the series: > > > sum{n = 1 .. oo} (-1)^n/n > > > does not converge according to your logic. > > > > For every natural number n we have n/n = 1. But the notation lim n --> > > oo is not clear. > > You do not know the convention that taking powers goes before division? > What *do* you know about mathematics? But I will refrase: > But I now understand that the series: > sum{n = 1 .. oo} ((-1)^n)/n > does not converge according to your logic. > And why is the notation lim n --> oo not clear. You have defined it just > a few days ago: > > > > The same is meant as in sequences and series like: > > > > lim [n --> oo] (1/n) = 0 > > > > n becoming arbitrarily large, running through all natural numbers, but > > > > always remaining a finite number < aleph_0. > > > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings > > > > where the remainings are |{n+1,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 > > > > > > That makes no sense. As I read it, we have: > > > |{1, 2, 3,..., 2n}| / |{2, 4, 6, ..., 2n}| = 1 + remainings = 1 + 1 = 2. > > > > ? > > 1 + remainings = 1 + 1 = 2 > > Which error did you see? > > Read the fifth line above the question mark. > > > > But I will allow an error here. But I was thinking about: > > > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > > > which might be written as: > > > lim{n -> oo} |{1, 2, 3, ..., n}|/|{2, 4, 6, ..., 2n}| > > > > = 1 by mathematics and by set theory. > > > > > or as: > > > lim{n -> oo} |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| > > > > = 2 by mathematics but 1 by set theory. > > As set theory does not know about limits, I fail to see that, but > even if set theory had knowledge about limits, I would state: > No, also 2 by set theory. What you fail to see is that > |{1, 2, 3, ...}| / |{2, 4, 6, ...}| > is defined by neither mathematics, nor by set theory. Both do give > *no* definition of aleph-0/aleph-0. The cardinal numbers do not > form a field, only a ring (and not even a division ring). So > arbitrary division is not defined, definition is not even possible. Yes, that's a shame, isn't it? It would seem to me that any good theory of infinite sets could be applied to infinite sets of points, such as the reals in (0,1] or those in (0,2], and be able to draw conclusions such as that there is twice as many points and twice as much space in the second as in the first, rather than coming to to useless conclusion that the infinite sets are equal in size. Transfinite cardinalities don't really measure anything, though, do they? Indeed, they don't form a field. Tsk tsk. :) TO
From: Tony Orlow on 19 Aug 2006 22:33
Dik T. Winter wrote: > In article <1156000079.633380.83620(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > Do you agree: If there are more than one can check then one cannot > > > > check all? > > > > Do you agree: If there are more than one can check then one cannot > > > > check every line? > > > > May one say so in set theory? > > > > > > But there is no need to check each and every individual line. When I > > > state: > > > sum{i = 1 .. n} i = n * (n + 1) / 2 > > > do I need to check for each and every n? > > > The *definition* of the diagonal makes clear that it is different for > > > each and every n. Like in the formula above I need not check for > > > each individual n, I need not check for a difference for each and > > > every n. > > > > Exactly. Above formula holds for every finite number, as can be shown > > by induction. But it doesn't hold for an infinite number of finite > > numbers. > > Eh? It holds for every finite number but does not hold for an infinite > number of finite numbers? What do you mean with that? If there are > infinitely many finite numbers, I would say that as it holds for > every finite number, it also holds for infinitely many finite numbers. If the formula applies to an infinite number of finites, then does the sum of the aleph_0 finite naturals equal (aleph_0^2+aleph_0)/2? In standard theory, would this not equal aleph_0, and if so, does it make sense that sum(n=1->aleph_0: 1) = sum(n=1->aleph_0: n), when n>1 for all n>1? The scond would appear to be clearly a larger sum. > > > > 100 %? But with sets I would state that a set exists if all of its > > > elements do exist and all of its subsets do exist. > > > > Is it possible? If ZFC is consistent, then it has a countable model. If > > all of its subsets would exist, then it was uncountable. > > Is ZFC a set? > > > > But you asked > > > for the *first* 10%, which is something different. > > > > Then drop the "first". I will ask only for 10 % of the elements. Or > > would you prefer the last 10 % ? > > With ordering you lose, so not the last. {1, 11, 21, ...} are an example > of a set that contains 10% of the natural numbers (whatever that may mean). I think that what WM is pointing out here is that aleph_0 does not behave like a normal number when it cannot be divided to determine a midpoint of the set. This is a problem when viewed quantitatively. > > > > > Don't misunderstand me: All subsets of |N do exist. But the set of all > > > > non-generators does not exist. > > > > > > Oh. I must have missed something, because I have not seen a proof. > > > Given an injection f: N -> P(N), why does the set > > > M(f) = {n in N | n !in f(n)} > > > not exist? > > > > Excuse me, the non-existing set is the triple {f, n, M_f(n)} > > Excuse me, Hessenberg was not talking about such triples. And I do not > even understand your notation at all. Hessenberg was talking about sets > like M(f) as I defined above. And that the triple {f, N, M(f)} does not > exist shows that there is no f that is surjective from N to P(N). And > that only because surjectivity requires such a triple. |