Cantor Confusion
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From: Dik T. Winter on 13 Dec 2006 23:00 In article <1166036419.914921.209020(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <457ece72(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > > Dik T. Winter wrote: > > > > In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > ... > > > > > Why does that matter? > > > > > > > > It does matter because if you do not properly define your problem, > > > > mathematics is not able to give an answer. > > > > > > It's sufficiently defined if one assumes that there is a uniform > > > probability distribution. > > > > You can assume as much as you want, that does not make it a definition. > > What *is* a uniform distribution of all natural numbers? > > > > > > > This is the same thing as your stupid ball and > > > > > vase trick. Why do you need to label anything, or know what you're > > > > > choosing from the infinite set? > > > > > > > > Because that is part of the problem setting. Giving that setten will > > > > allow mathematics to model the question and give an answer. > > > > > > That problem has a clear answer with or without the labels: the sum > > > diverges as f(n)=9n. The labels are confounding, not clarifying. > > > > Yes, but that does *not* indicate anything about the limit, as I did show > > below: > > > > > > And it is bad to think that because for a sequence of sets holds that > > > > lim{n -> oo} |S_n| = k > > > > with some particular value of k, that also > > > > | lim{n -> oo} S_n | = k > > > > because the latter statement contains something that has not been > > > > defined in mathematics. > > > > > > I'm not sure what that statement is supposed to say. Can uoi give an > > > example? > > > > You have: |S_n| = 9n and you think that > > | lim{n -> oo} S_n | = lim{n -> oo} | S_n | = lim{n -> oo} 9n. > > but first "lim{n -> oo} S_n" is not defined, and second, when you define > > it you have to prove that the first part is equal to the second part. > > > > > > But even when we define it, it is not certain > > > > that it holds. Given the following (I think reasonable) definition: > > > > lim{n -> oo} S_n = S > > > So, what, S_n is supposed to be an initial segment of the sequence? > > > > I do not understand. S_n are sets indexed by n and so form a sequence. > > > > > > if: > > > > (1) for every element a in S there is an n0 such that a is in each of > > > > the sets S_n with n > n0 > > > > (2) for every element a not in S there is an n0 such that a is not in > > > > each of the sets S_n with n > n0. > > > In (2), it sounds like a would not exist in ANY S_n if it's not in S. > > > > No. Any S_n with index larger than some n0. > > > > > > So from some particular point an element either remains in the sets in > > > > the sequence or remains out of the sets. > > > > > > You mean, at some point you can tell whether a given element a is in S, > > > because if it were, it would be there by then? > > > > and if it were not it would not be there by then. (Note that the point > > where that is the case can be different for each element.) > > > > > > With this definition (when we look at the rationals) we have that > > > > lim{n -> oo} [0, 1/n] = [0] > > > Okay that interval degenerates to 0.... > > > > > > > and so: > > > > lim{n -> oo} | [0, 1/n] | = aleph0 != 1 = | lim{n -> oo} [0, 1/n] | > > > > (I am talking standard mathematics here). > > > > > > Are the |'s supposed to denote set size? If so, how can you claim that > > > [0,0] contains aleph_0 elements? > > > > Where do I claim that? > > > > > > So taking cardinality and limits can not be interchanged except in some > > > > particular cases. But that is not unprecedented in mathematics. > > > > limits and integrals can also not be interchanged except in particular > > > > cases. And so can the interchange is not in general passoble if one > > > > of the things you interchange is a limit. Even interchanging limits > > > > is not in general possible. Consider: > > > > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy > > > > > > True, but is it relevant? > > > > Yes, because the same holds for cardinality. > > Could you please explain what same holds for cardinality? If you read, it is right above this. -- 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 13 Dec 2006 23:14 In article <1166045232.670658.72540(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > cbrown(a)cbrownsystems.com schrieb: .... > > What? The *range* of a surjection from edges onto paths is /not/ every > > edge; it is every path. The /domain/ of the surjection is every edge. > > (You actually teach Analysis, and you don't know this?!?!) > > Did you never copy and paste a wrong piece? > I corrected it already: The range is "every path". Are you stupid or do you not understand things? cbrown's correction was posted about two hours before your correction. > > That is poorly stated. The mapping that maps the first edge to the left > > to every path provides us with a mapping such that "there is no path > > without edge mapped to it"; but this mapping does not prove that the > > cardinality of the set of paths <= the cardinality ({a single edge}) = > > 1. > > You misunderstood. What I explained to Dik is another mapping than that > discussed by us. It is a mapping of one single edge on the path > representing 1/3, the next edge is mapped on the path representing pi, > and so on. I claim that for every real number I can name an edge to be > mapped on this real number. Yes, you state that, and you can. But that does *not* provide a surjection from the edges to the real numbers. And you stated you had constructed such a mapping. > And noboy can disprove it by constructing a > diagonal number, because the tree contains them all. RIght. The disprove is in the fact that whenever I state a sequence of numbers that sequence inherently does not contain all real numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Han de Bruijn on 14 Dec 2006 03:28 Virgil wrote: > In article <1166011032.914983.204230(a)79g2000cws.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >>Han de Bruijn schrieb: >> >>>William Hughes wrote: >>> >>>>Tony Orlow wrote: >>>> >>>>>Well, the proof is simple. Any finite number of subdivisions of any >>>>>finite interval will only identify a finite number of real midpoints in >>>>>that interval, between any two of which will remain more real midpoints. >>>>>Therefore, there are more than any finite number of real points in the >>>>>interval. >>>> >>>>This just shows that the number of real points is unbounded. >>>>It does not show it is infinite (unless of course you use the >>>>fact that any unbounded set of natural numbers is infinite). >>> >>>Isn't unbounded the same as infinite, i.e. = not finite = unlimited = >>>without a limit? >> >>Unbounded is potentially infinite but it is not necessarily actually >>infinite. > > AS in honest mathematics, the one implies the other, in honest math one > has neither or one has both. In ZFC and NBG, both. Let's repeat the question. Does there exist more than _one_ concept of infinity? Isn't unbounded the same as infinite = not finite = unlimited = without a limit? Please clarify to us what your "honest" thoughts are. Han de Bruijn
From: Han de Bruijn on 14 Dec 2006 04:10 William Hughes wrote: > Therefore, as there are at least as many digit positions in 0.111... > as there are primes, there are an infinite number > of digit positions in 0.111... Binary or decimal? Binary: 0.111... = lim(n->oo) 1/2 + 1/4 + .. + 1/2^(n-1) = = lim(n->oo) (1 - 1/2^n)/(1 - 1/2) - 1 = 2/1 - 1 = 1 Decimal: 0.111... = lim(n->oo) 1/10 + 1/100 + .. + 1/10^(n-1) = = lim(n->oo) (1 - 1/10^n)/(1 - 1/10) - 1 = 10/9 - 1 = 1/9 Please make your choice. But .. what is your problem, then? Han de Bruijn
From: mueckenh on 14 Dec 2006 04:34
David R Tribble schrieb: > William Hughes schrieb: > >> Do you now claim > >> the natural numbers do not exist? > > > > Mueckenh wrote: > > More than enoug do exist. (More than we will ever need could be brought > > to existence.) > > How is a number "brought into existence"? > > I'm thinking of a natural number, call it q. It has the distinct > property that it is not equal to any number you (MH) can think of. > In fact, it is larger than any natural number that has been "brought > into existence yet". So what is q? Not yet a number, unless you can specify it such that one (at least you) can decide whether q < n or q = n or q > n for any natural number n given to you. Only saying that it is not equal to any natural number given is not enough. Regards, WM |