Cantor Confusion
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From: mueckenh on 22 Oct 2006 07:53 Dik T. Winter schrieb: > In article <1161435575.019298.164830(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1161378187.155995.290420(a)f16g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > It *is* continuous like a staircase. > > > > X(t) = 9 for t = 1 until t = 2 where X(t) switches to 18. That is > > > > enough to excflude X(omega) = 0. > > > > > > How do you *define* X(omega). As far as I know X is only defined for real > > > numbers, and omega is not one of them. And I see no reason to exclude > > > X(omega) = 0, = 1, = -1 at all from this reasoning. > > > > lim {t --> oo} X(t) = X(omega) > > Yes, if you define X(omega) like that your conclusion is obvious. Not I define omega as the limit ordinal number. That is a matter of set theory. > But > do you not see that you are putting the cart before the horse? You state > (1) that lim{t -> oo} X(t) != 0 because X(omega) != 0 > (2) next you define X(omega) = lim{t -> oo} X(t) > Even when we disregard the non-existence of that limit, this is just a > self-fulfilling prophesy. Not I define omega as the limit ordinal number. That is a matter of set theory. I only use the continuity of a function which is already required to find lim 1/n = 0. > > > > > > That is still not a mathematical formulation. More is required. You > > > > > need to actually state what you mean with 'number of balls in the vase > > > > > at noon' or 'natural numbers in the set at noon'. > > > > > > > > The number of transactions t is then t = omega at noon. > > > > > > Is *that* a mathematical definition? Pray provide a real mathematical > > > definition. > > > > lim {t --> omega} t = omega > > Oh. Provide a mathematical definition of that limit, please. In standard > mathematics that limit is undefined. Cantor used omega with two slightly different meanings. omega is the set N and omega is the first infinite ordinal number, i.e., the smallest number larger than any natural. These two definitions yield: lim {t --> oo} t = omega and lim {t --> oo} {1,2,3,...,t} = omega. > > > You think so. The irrational numbers are defined to be the limits of some > > > particular sequences (or rather as equivalence classes of sequences). I > > > > Equivalence classes of sequences with same limit like > > lim {t --> oo} a_t. > > Wrong. Wrong is wrong. The limit *is* the irrational number. You can use these and only these numbers in a Cantor list, not the equivalence classes of sequences. > > > (7) assume sequences of rationals. Create equivalence classes amongst > > > those sequences (a_n ~ b_n if |a_n - b_n| goes to 0; but this is > > > losely speaking and quite a few other methods are known, all > > > equivalent). > ... > > > So, at what stage in this process is the limit of a function used to > > > define the irrationals? > > > > At (7). The equivalence classes of sequences of rationals with same > > limit. > > Wrong. At that point you can not talk about sequences of rationals with > the same limit, because many of such sequences do not have a limit in the > rationals. So (7) is formulated as I wrote it (in one of the forms to > define the reals from the rationals). It is not the *limit* that is the > irrational, it is the equivalence class of sequences. The sequences belong to Q. So your irrational numbers belong to Q? That is nonsense. In Q we have sequences with Cauchy-convergence and, therefore, perhaps without a limit in Q. But the irrational numbers are definitely *not* in Q. > And (again, > losely speaking) the equivalence classes are built in such a way that > all members of the classe *ought* to have the same limit in the extended > system. Exactly. And that is the irrational number, *not* loosely speaking. You can use this and only this number in a Cantor list, not the equivalence class of sequences (because the due terms are not uniquely defined). Summarizing the original question: You need the limit omega to construct the irrational numbers. Regards, WM
From: mueckenh on 22 Oct 2006 07:57 Dik T. Winter schrieb: > In article <virgil-E8EF11.13483421102006(a)comcast.dca.giganews.com> Virgil <virgil(a)comcast.net> writes: > > In article <1161435318.373825.152830(a)e3g2000cwe.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > ... > > > Therefore lim [n-->oo] {1,2,3,...,n} = N. > > > > Depends on how one defines lim [n-->oo] {1,2,3,...,n}. It is certainly > > true if one takes the limit to be the union of all of them, as > > guaranteed by the axiom of union in ZF. > > I would not like that as a definition. That would make: > lim{n -> oo} {n, n + 1, n + 2, ...} = N Obviously wrong, because 1, 2, 3, .., n-1 all belong to N lim [n-->oo] {-1,0,1,2,3,...,n} = N is obviuously wrong too. Therefore lim [n-->oo] {1,2,3,...,n} = N. > I gave sometime ago a definition of the limit of sets that is (in my > opinion) workable, but Mueckenheim did not allow that definition. The > reason being that under some formulations of the vase problem that > definition would make the vase empty at noon. So it must be wrong and needs no further attention. Why? Because the contents of the vase increases on and on. Such a process cannot lead to emptiness in any consistent system - independent of any "intuition". Regards, WM
From: Han.deBruijn on 22 Oct 2006 12:10 Franziska Neugebauer schreef: > Han de Bruijn wrote: > > > Franziska Neugebauer wrote: > >> Han de Bruijn wrote: > >>>Franziska Neugebauer wrote: > >>>>Han de Bruijn wrote: > >> > >> [...] > >> > >>>>>Good! Now given two such members x and y. What does x = y mean? > >>>> > >>>>x = y :<-> Az(z e x <-> z e y) > >>> > >>>Of course. Because all members are sets. But I think this is an > >>>infinite recursion with the equality definition. Does it end > >>>somewhere? > >> > >> Where do you spot recursion? > > > > Two sets are equal if they have the same members. > > The "two sets" are one set if they have the same members. Let > A und B refer to the "two sets". If A = B then the "two sets" > are one set. > > > Two members are equal if they have the same members ... Right? > > Two variables x and y refer to the same member (set) if x = y. Once > again: Where do you spot recursion? Or rather call it a vicious circle definition. See the response to this by Virgil. Han de Bruijn
From: Han.deBruijn on 22 Oct 2006 12:14 Virgil schreef: > In article <b008d$453887ef$82a1e228$31075(a)news1.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > Virgil wrote: > > > > > In article <6cf73$45387e07$82a1e228$27759(a)news1.tudelft.nl>, > > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > >>Ha! Mathematicians can't even define their most frequently used symbol, > > >>which is the equality ' = '. And that is a prerequisite for their "is". > > > > > > The equal sign, "=", has many meanings, which differ depending on > > > context, so that there cannot be one Procrustian meaning that fits all > > > contexts. > > > > > > For sets A and B, A = B means that for > > > all x, x is a member of A if and only if x is a member of B. > > > > Good! Now given two such members x and y. What does x = y mean? > > In ZF it means > "for all z, z is a member of x if and only if z is a member of y". > > And so on, ad infinitum. That's precisely what I thought. You need the machinery of infinity even for a basic thing like a = b : two things being equal. Han de Bruijn
From: Han.deBruijn on 22 Oct 2006 12:16
MoeBlee schreef: > "built in" provision that equality be interepreted as the actual > identity relation. Okay. Define your "identity relation". Han de Bruijn |