Cantor Confusion
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From: mueckenh on 30 Jan 2007 06:42 On 28 Jan., 20:44, Virgil <vir...(a)comcast.net> wrote:> > In fact? If there is a path of lengths n then there is a path of > > length n+1. And there is a path of length 1. What is the length of > > the union of all these paths (which contains only finite paths)? > If that union is to be regarded as a path at all Yes, please regard it as such! > then it cannot be a > finite path as it would have to be be longer than every finite path > (for every path of length n, the union wold have to be of length at > least n+1 ). So its length is not finite. On the other hand, it must not be longer than every natural number because it is simply the union of all natural numers. > Those of us who do not have > coniption fits over the use of the word call that /infinite/. I have been knowing for long that the complete union of all finite numbers would contain an infinite number, if this union existed. > > > > > > A set is finite if there does not exist any injection from it into > > > any of its proper subsets. > > > Now apply this to paths. > The union of all finite paths allows the mapping of each finite path of > length n to a path of length n+1 while giving the same union, so that > that union is infinite (in the sense of being not-finite). This sense means potential infinity. > > > The length of p is the union of the lengths of all finite paths. > Which is the first limit ordinal. No, lengths measured in terms of natural numbers are natural numbers. > > > > > Correct. Think about it. > Thinking about it leads to the inescapable conclusion that infinite > binary trees must have infinitely long (endless) paths. But some have more while others have less? > > The identity function on N, idn: N --> N : x |--> x, is such a > surjection, Q.E.D. That is not a set "listed in its entirety". Here is a set S of cardinality 2^2^aleph_15 listed and by this listing well-ordered in its entirety N --> S : x |--> s, Regards, WM
From: Dave Seaman on 30 Jan 2007 07:42 On Tue, 30 Jan 2007 01:33:03 -0500, David Marcus wrote: > MoeBlee wrote: >> On Jan 27, 2:26 pm, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: >> >> > The maximum number of bits required to describe any index no. i is >> > ceiling(log(i+1,base 2)). For all i>0, i+1> ceiling(log(i+1,base 2)). >> > And for i = 0, the real number is 0.000 . So, all terms on the diagonal >> > are 0 (and, more of this later, the distance in bits between the 0 on >> > the diagonal and the first non zero bit goes as (n - log(n,base2)) for >> > bit n. >> > >> > Following Cantor, we construct an antidiagonal number, different from >> > the first row, the second row, etc. Because all the diagonal terms are >> > 0, we construct the number .1111.... >> > >> > This string is certainly not in the hypothetically complete list, but it >> > is a real number and = 1.000... , which was excluded anyway from our >> > initial list. So we can make no inference that the list is >> > incomplete/invalid - we have just generated a number. >> >> What you've shown is that applying the diagonal method to your list >> only shows a real number not in [0 1). So what? It doesn't contradict >> that for any list of real numbers, there is a real number not in the >> list. We don't have to prove: For any list of real numbers in a >> certain interval, there is a real number in that interval not listed. >> Rather, we only have to prove: For any list of real numbers, there is >> a real number not in the list. > The usual diagonal proof proves that [0,1) or [0,1] (depending on how we > do it) is uncountable. So, the fact that the anti-diagonal is in the > specified interval is important in the usual proof. The main problem > with Andy's argument is that he is using a specific list. Actually, the main problem with Andy's argument is that he is using a defective variation of the diagonal argument. When correctly done, the diagonal argument does not produce a number that has a dual representation. >> Your lack of understanding of basic logic in mathematics has lead you >> to this irrelevent red herring/strawman about [0 1). It's not irrelevant. A correct argument applied to that list shows that there is a number in [0,1) that is not in the list. One way is to represent the numbers in decimal and to carry out the diagonal argument in such a way as to produce a number, all of whose digits are either 4 or 5. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 30 Jan 2007 08:13 On 29 Jan., 03:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > If I consider the path length as the union of path length: = > > {0,1,2,...,n-1}, I am right too. > > Yes, for this path. I never did have problems when you were uniting paths > that you get infinite paths and all that that entails. But path-lengths are natural numbers! > But you are uniting > *sets* of paths. In that case you do not unite paths at all. I unit finite trees. This implies the union of sets of paths. This implies the union of paths, because the union of two or more paths need not but can be a path (notice: the indexes are of the form {0,1,2,...,n-1}). The union of trees implies the union of outmost left-hand side paths for example. This union is a path. > > This is extremely unclear. A path is a set of nodes (by your definition), > its length is the cardinality of that set. Let's say that path p_n is > the set of nodes {n_1, n_2, n_3, ..., n_n}, with path length (cardinality) > n. The union of those paths is {n_1, n_2, n_3, ...} with cardinality > aleph_0. But none of the paths in the union is infinite. If you do not > agree with this, come up with a definition of path length when a path > is a set of nodes. You are correct. The union of all natural numbers is not a natural number. Therefore you think it (the cardinality) can be infinite. The union of all paths is a path in many cases. In such a case it is simultaneousy a union of natural numbers which is a natural number. Therefore your and Cantors distinction between numbers and sets is wrong. There is no infinite union of finite numbers unless there is an infinite number. (Translate number by path, then you will see it.) > Yes, each path in T(oo) is the union of a set of finite paths. And each > path in T(oo) is not in the union of the sets of finite paths. So I did > apply the knowledge you agreed to. What now? Unions of paths (for instance the outmost left-hand sided) are paths! > > > > In case of path lengths measured as I defined, the union of all > > lengths is a length. > > We were talking about paths here, methink, not about lengths? So I wonder > why you are stating this here. Because I need the identity of path-lengths and numbers for my argument. > > If T(oo) is constructed as the union of all finite trees T(k), then > > every path in P is a path which is in the union P(1) U P(2) U ... of > > the elements of P_C. > > Prove it. The union of P(i) contains finite paths only, *by the definition*. The infinite tree contains finite trees only. Is it finite on that behalf? > On the other hand P contains infinite paths *by the definition*. So P > can not be contained in the union. Or prove that the union of sets of > finite paths can contain an infinite path. > > > Therefore P is a subset of this union P(1) U P(2) U ... of elements of > > P_C. > > Is this correct? > > No. What is the origin of P when we unite finite trees? > > > 0.010101... is not in the union of all T(n), n in N, as > > you say. > > I do not say that. I explicitly state that it is in it. But it is *not* > in the union of all P(n). Taking the union of all T(n) is analog to taking the union of all P(n). The union of all P(n) does contain elements > like "0.0", "0.01", "0.010", "0.0101", but *not* "0.010101...". The > union contains all finite initial segments of "0.010101...", but not > that one itself. Then the path 0.010101... cannot be in the tree which is consructed by taking all finite trees. By taking all finite trees you simultaneously take all finite sets of finite paths. > In a similar way, N contains all the finite initial > segements of itself, but not N itself. N is not a natural number. But the union of all path-lengths is a path- length. Regards, WM
From: mueckenh on 30 Jan 2007 08:17 On 29 Jan., 04:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > And in what way is > > > III c IV c V > > > ? > > > > In no basic way. To see the basics, you must continue IIII, IIIII, > > etc. > > Ok. Pray define what sets those things do represent. > They represent the numbers 4 and 5. And: They are the numbers 4 and 5. > > > > In that way which need no further declaratiojn. Put three nuts in your > > > > hat. Then you can see the set. > > > > > > Yes, I have a set of three nuts. The set is a set of nuts. That does not > > > mean that three is a set. Three is a set of what exactly? > > > > 3 is the set of all existinig sets with 3 elements. (Yes, this set > > exists.) > > Bizarre. So 3 is the set of all existing (what does that mean) sets with > 3 elements, The number 3 is the set of all sets of 3 elements (unless they were existing, they would not have 3 elements). But 3 is also here III. It is every set of 3 elements. > > I do not ask what possible answers there are, I only ask how you would > like to propose it. And, pray extend to 111111111. This is a set of 9 distinguishable elements, it is a set representing the number 9. And: It is the nunmber 9. Every set of 9 elements and avery set of sets of 9 elements is the number 9. Regards, WM
From: mueckenh on 30 Jan 2007 08:24
Virgil schrieb: > In article <1170085646.542472.199010(a)k78g2000cwa.googlegroups.com>, > > Of course we can show in unary representation (IET) that the sum of > > all natural numbers is countable > > You can't even show that it exists, WM. Of course I cannot show that this sum exists, because, in fact, N does not exist. > > > > Do you disagree? > > Show us that sum, and then we can discuss "countable". Show us that set N, and then we can discuss "countable". Regards, WM |