Cantor Confusion
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From: Sebastian Holzmann on 24 Oct 2006 06:03 mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > If f(0) is undefined, i.e., if 0 is not element of the domain of f, > then we can find f(0) = 0 by l'Hospital' s rule. Ever heard of? In your Do you recall the precise constraints imposed on using l'Hospital's rule? If f(0) is undefined, f(0) is undefined. Full stop. This is completely unrelated to the question whether lim_{x -> 0} f(x) exists. > example we have no limit because of lacking continuity. For every > interval delta covering the value x = 0, we cannot find an epsilon < > 1/3. And as you may know, there is no x > 0 which is closest to 0. Try learning something about quantifiers.
From: mueckenh on 24 Oct 2006 06:09 jpalecek(a)web.de schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > jpalecek(a)web.de schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de napsal: > > > > jpalecek(a)web.de schrieb: > > > > > > > > > > > > > > > > > > 0. > > > > > > > > > > /a \ > > > > > > > > > > 0 1 > > > > > > > > > > /b \c / \ > > > > > > > > > > 0 1 0 1 > > > > > > > > > > ..................... > > > > > > > > > > > > > > > > > > > > An edge is related to a set of path. If the paths, belonging to this > > > > > > set, split in two different subsets, then the edge related to the > > > > > > complete set is divided and half of that edge is related to each of the > > > > > > two subsets. If it were important, which parts of the edges were > > > > > > related, then we could denote this by "edge a splits into a_1 and a_2". > > > > > > But because it is completely irrelevant which part of an edge is > > > > > > related to which subset, we need not denote the fractions of the edges. > > > > > > > > > > Sorry, but your "proof" doesn't work. Imagine an infinite path in the > > > > > tree. Which is the edge it inherits as a whole? Whenever you give me > > > > > that edge, I can tell you're lying because if a path inherits an edge > > > > > as a whole, it means that the path terminates by that edge. > > > > > > > > How should I be able to name the last term of a sum which has no last > > > > > > Indeed, INDEED. But that means, that there is no term 1 in the series, > > > too. > > > > No. That does it not mean. Unless you agree that no real numbers do > > exist. > > Come on, real numbers do exist. Where? > > > > Moreover, it means that there is no positive term in the series. > > > (Because > > > 1 would be counted for the last edge, which does not exist, etc.). > > > > > > > term? But while we cannot name any individual edge we can prove: No > > > > path splits into two paths without the supply of two new edges, one > > > > edge for each path. This implies there cannot be less edges than paths. > > > > > > No. This holds for the finite case only. > > > > Why? If you follow a path you can see that where ever it splits off > > another path, this splitting happens by an edge. As this situation does > > never change, it remains in infinity for the whole path. For every > > partial sequence (i.e. for every finite sequence of the path with n > > edges, starting from the top), we have the "load of edges" accumulated: > > > > (1 - (1/2)^n)/(1 - 1/2). > > > > We cannot construct a finite sequence with less edges, even if we would > > like to do so. In the limit n --> oo we obtain the whole load 2. This > > is the correct calculation. All other arguing is nonsense, with or > > without marriage. > > You cannot take the limit. By taking the limit, you say "All paths with > however large, but FINITE length happen to behave somehow uniformly". Absolutely not. All finite segments lead to a sum of less than 2 edges. > There is nothing about infinite paths in the statement. Or otherwise, > if there is a positive sum, there should be at least one positive term. What we need is the number of terms. We know it is omega. Therefore we can sum up the geometric series. The aleph_0 terms 1 + 1/2 + 1/4 + ... yield exactly the sum 2. 1 + 1/2 + 1/4 + ... = ... + 1/4 + 1/2 + 1. You know that in absolutely converging series infinitely many terms can be replaced without changing the sum? It is not important where we start or end. Or what about recursion? The square root of 2 can be calculated by the recursion f(n+1) = f(n)/2 + 1/f(n) For infinite paths we have the number of edges given by the recursion f(n+1) = 1 + f(n)/2 with f(1) = 1 and n -> oo. The limit is 2, even for arbitrary f(1). A very stable recursion. > You say you can't name it - but at the same time, you say that you > can name all edges by their number. Do you think that the edges are uncountable? That is wrong. > > > > > Or the other way round: Assume there were more paths than edges, then > > > > at least two paths could no be distinguished. (A path can be > > > > distinguished from every other path by at least one edge.) > > > > > > Which doesn't mean anything. > > > > Oh yes, a different edge is obviously a necessary condition to > > distinguish two paths. > > I've written it badly, I meant "Assume there were more paths than > edges, > then at least two paths could no be distinguished" statement is false. It is false. It would not be false, if the paths could split and reunite and so on. But that is impossible. Once split means split forever. > > So for a series sum 1/2^n, we have > > s_1=1 > s_2=1+1/2 > s_n=1+1/2+...+1/2^(n-1) > > and finally > > s_oo=1+1/2+1/2^2+... > > But for the edges, we have just the same. For infinite paths we have the number of edges given by the recursion f(n+1) = 1 + f(n)/2 with f(1) = 1 and n --> oo. > > Or, put another way, "You cannot read infinite series backwards". 1 + 1/2 + 1/4 + ... = ... + 1/4 + 1/2 + 1. You know that in absolutely converging series infinitely many terms can be reordered without changing the sum? Regards, WM
From: mueckenh on 24 Oct 2006 06:19 Sebastian Holzmann schrieb: > mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > >> I wouldn't neccessarily call ZFC - AoI "finite set theory". > > > > It is a theory without the actual infinite, a theory without omega. One > > may execute any operation. The finite domain will never be left. Which > > of the remaining axioms should yield infinity? > > Which should prohibit it? > > > How would you construct an actually infinite set? Pair, power, union? > > They all stay in the finite domain if you start with existence of the > > empty (or any other finite) set. Comprehension or replacement cannot go > > further. So, how would you like to achieve it? > > Let's assume that ZFC - AoI is consistent, otherwise the statement is > void. Let \phi_n be a sentence in the language of set theory that > encodes the statement "There exists a set that has at least n elements" > (you can formulate this one as a homework problem). Perhaps you do another homework first: Why should such such a sentence be correct? It is not difficult to encode the sentence: Card(Q) > Card(R). Is that a proof? > > Since for every finite subset of the set of \phi_n, there exists a model > of ZFC - AoI where this subset is true (any model of ZFC - AoI should do > the trick), by compactness, there exists a model of ZFC - AoI where > _all_ \phi_n are simultaneously true. ("True" might not be the correct > English word for it, please do someone correct me in this.) A set that > has, for any n, at least n elements cannot be finite. Therefore it must > be infinite. Further in this case it would be nonsense to introduce the axiom of infinity. Regards, WM
From: mueckenh on 24 Oct 2006 06:21 Lester Zick schrieb: > On Mon, 23 Oct 2006 02:00:06 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > wrote: > > >In article <1161518242.756958.103660(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > [. . .] > > > Virgils definition, although not wrong, definitions > >can not be wrong, does not lead to desirable results. > > Since when can definitions not be wrong? Since they got to now that set teory is correct only by definition. Regards, WM
From: mueckenh on 24 Oct 2006 06:24
Randy Poe schrieb: > Lester Zick wrote: > > On Mon, 23 Oct 2006 02:00:06 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> > > wrote: > > > > >In article <1161518242.756958.103660(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > [. . .] > > > > > Virgils definition, although not wrong, definitions > > >can not be wrong, does not lead to desirable results. > > > > Since when can definitions not be wrong? > > A definition is a statement by a person that a certain > symbol will be used by them to stand for some concept. > > What is there in such a statement that can be wrong? > > Example: I will use the term "gleeb" to refer to an integer > which is divisible by 2. > > How can that statement be wrong? How would you > define "wrong" for such a statement? Example: I will use the term "gleeb" to refer to an integer which is divisible by pi. This statement is wrong. And it is nonsense. And don't come up with the empty set. Regards, WM |