Cantor Confusion
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From: Dik T. Winter on 28 Jan 2007 21:39 In article <1169987266.609036.283220(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 28 Jan., 03:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > We can boil down the discussion about trees to the following simple > > > question, considering only one path, for instance the path p on the > > > outmost left hand side of the tree. This path p (in terms of nodes) is > > > the union of all paths of finite trees with length n, n in N. > > > > If you consider the union of paths as the union of sets of nodes, you are > > right. > > 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 you are uniting *sets* of paths. In that case you do not unite paths at all. > > > then at least > > > one of the paths in the union must be infinite. > > > > Wrong. The union of a collection of finite sets can be infinite, and in > > this case is. > > Here we are not concerned with the cardinal number of the union of > infinitely many elements but with the length of the union of all > finite path lengths, where each path length is measured from the > common origin, namely 0 at the root of the tree. 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. > > No. You fail to see a crucial difference. Going back at natural numbers: > > union(n in N) {1, 2, 3, ..., n} is N > > but > > union(n in N) {{1, 2, 3, ..., n}} does not contain N. > > As your paths are equivalent to initial segments {1, 2, 3, ..., n} we find: > > union(n in N) p_n is p_oo > > but > > union(n in N) {p_n} does not contain p_oo. > > Fine. Apply this knowledge to the paths of the infinite tree T(oo). 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? > > > That is set theory. But it is wrong. The union of all finite paths of > > > length n is infinite, but not actually. It has no upper bound. > > > > A fundamentally misunderstanding. The union of a set of things is not > > the set of the union of things. > > 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. > > And T(oo) does not contain any path? But how do you *define* "finite > > without an upper bound"? Can you come up with mathematical definitions > > of "actual infinitiy" and "potential infinity"? > > Read my chapter 8. There (nearly) all is said what can be said about > that topic. Still to do that. I begin to wonder. > > Wrong, and that is not what you claimed. You claimed that P was a subset > > of P_C. You claimed: > > > 4) The set of paths in T(oo) is a subset of the countable set > > > of finite sets of all paths in the finite trees. > > P was the set of paths in T(oo). P(k) was the set of paths in the finite > > tree T(k). P_C was the set of finite sets P(k). > > 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*. 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. > > But even this statement is indeed wrong: > > > It is a subset of the union set P(1) U P(2) U ... > > 0.010101... is a path in the complete tree, but is not a path in any of > > the P(k). > > > 0.010101... is in the complete tree T = T(oo). (Correct me, if I > remeber that wrong, but here are as much opinions as are set > theorists.). Right. > 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). 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. In a similar way, N contains all the finite initial segements of itself, but not N itself. > Therefore T(oo) = {T(n) | n in N} =/= T(oo). Isn't that a > contradiction? Would be if I ever stated such a thing. -- 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 28 Jan 2007 22:00 In article <1169988074.025610.14700(a)v45g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 28 Jan., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > For natural numbers c is the same as <. 2 c 3 is the same as 2 < 3. > > > This is proven by II + I = III. > > > > Perhaps. You are focused at notation. Mathematics is *not* concerned with > > notation, it is concerned with the basics behind it. > > An abuse which yields infinite finity and finite infinity. That is what you appear to think. > > 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. > > > 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 existnig 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, 2 is something similar with two elements, and 1 is something similar with a single element. Under that definition, what is the union of 1 and 2? Not a natural number, because that is the set that contains all existing sets with either one or two elements. Moreover, what is card(3)? I would think the number of elements it contains, but that is not 3 at all. Also defining 2 < 3 as 2 subset 3 does not work, because 2 is not a subset of 3. And finally, the definition is recursive. > > > In unary representation it has. In other representations it has to be > > > defined. > > > > Oh. So you think that subsetting comes before ordering in unary, but it > > requires definition in other representations? Remarkable. > > Subsetting is the most basic operation. You can do it with sets of > match sticks before you know any ordinal number. III is more than II > because II c III. But the meaning of c is, on this level, the same as > the meaning of <. Both are closely connected. Therefore I informed > Fanziska that here first impression about "what comes first" was > wrong. As you do not provide definitions, it is hard to decide whether you are wrong or not. At least with the definition above of 3, you are wrong. > > Set theory > > and the Peano axioms are not interested in representations at all. They > > are only concerned with the abstract concept behind it. Moreover, when > > you want to apply set theory to unary representations you will get > > problems. Consider: > > 111 > > how do you propose to define it as a set? > > There are different answers. Applying curly brackets, {111}, this can > be understood as a set witht one element {1} only. In case without > brackets, you have three elements, left, mddle, right, or first, > second, third. I do not ask what possible answers there are, I only ask how you would like to propose it. And, pray extend to 111111111. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 28 Jan 2007 22:30 In article <JCM0DD.9Bz(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1169988074.025610.14700(a)v45g2000cwv.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > On 28 Jan., 02:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > For natural numbers c is the same as <. 2 c 3 is the same as 2 < 3. > > > > This is proven by II + I = III. > > > > > > Perhaps. You are focused at notation. Mathematics is *not* concerned > > > with > > > notation, it is concerned with the basics behind it. > > > > An abuse which yields infinite finity and finite infinity. > > That is what you appear to think. > > > > 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. > > > > > 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 existnig sets with 3 elements. (Yes, this set > > exists.) Not in ZF or ZFC it doesn't. The equivalence class definition of a natural number does not work in ZF or ZFC, since nothing in ZF or ZFC guarantees that any of these exist. It will work in some versions of New Foundations set theory, but then, as Dik points out below, unions of naturals are not naturals. > > Bizarre. So 3 is the set of all existing (what does that mean) sets with > 3 elements, 2 is something similar with two elements, and 1 is something > similar with a single element. Under that definition, what is the union of > 1 and 2? Not a natural number, because that is the set that contains all > existing sets with either one or two elements. Moreover, what is card(3)? > I would think the number of elements it contains, but that is not 3 at all. > Also defining 2 < 3 as 2 subset 3 does not work, because 2 is not a subset > of 3. And finally, the definition is recursive.
From: G. Frege on 29 Jan 2007 01:06 On 28 Jan 2007 13:01:30 -0800, davidmarcus(a)alum.mit.edu wrote: >> >> I had understood that countability or otherwise was independent of a >> specific arrangement. So if a specific arrangement clearly doesn't >> allow all the reals to be countable (if they were), then that seems >> like good evidence for the initial hypothesis that the reals were >> countable to be false. >> > If by "arrangement" you mean enumeration (or listing), then this is > trivially false. Cranks use non-standard terminology without defining > it. Non-cranks use standard terminology if such exists and/or define > the words that they use. > @Andy Smith: The rationale behind using standard terminology (in a standard way) is that this helps to communicate you ideas to others. (This way they have a real chance to understand what you mean.) Moreover sometimes terminology shapes thought: using standard terminology then helps to keep track (while the usage of non-standard terminology may lead you astray). This is especially helpful if you are not an expert in a certain field. F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 29 Jan 2007 10:40
On 28 Jan., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Jan 28, 7:21 am, mueck...(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > You claim that if something is true for every set L_n, > > > then it is true for N > > I do not talk about N. This symbol has become the aim of heaviest > > abuse.You talk about the union of all natural numbers. N is > the union of all natural numbers. Something that is valid for each n in N need not be valid for N. I am interested in properties which are valid for every n. For instance, every n is finite. Every segment {1,2,3,...,n} is finite. Every path being he union of paths (segments) is finite. > > > > > > We know something about the maximum that is true > > > for every set L_n. > > > > So you do want to prove something about the maximum > > > of the set N. > > I want to see whether the union of all finite numbers can be an > > infinite number. > A union of numbers is a set of numbers. It is > also a number. The union of all finite numbers is an > infinite number. No. > > > This question was raised in the framework of the > > infinite tree. Set theorists asserted that a union of finite paths > > cannot be / contain any infinite path. > A union of finite paths is a set of finite paths. A set > of finite paths is not a path. A union of two sets of nodes is a set of nodes. A union of two paths need not be a path, but can be a path. > A set of finite paths > does not contain an infinite path. That is the problem. The union of all paths having only nodes with value 0, is a path. Is it infinite? Potentially infinite, yes. Actually infinite, no. > > Do not confuse paths and nodes, they are > not the same thing. > > Each path p has a set of nodes N(p). > The union of finite paths p1 and p2 contains the set > of nodes (union of N(p1) and N(p2)). > (note that both N(p1) and N(p2) are finite). > The union of all finite paths contains a set > of nodes that is the union of finite sets. > We know that the union of finite sets > can be a (potentially) infinite set. The union > of all finite paths contains a (potentially) > infinite set of nodes. Yes. But the same holds for path-lengths. All finite path-lengths are natural numbers. This is your last sentence: The union of all natural numbers contains an infinite natural number. As there are no actually infinite finite numbers, there is only the potential infinity. You are right. Regards, WM |