Cantor Confusion
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From: Franziska Neugebauer on 15 Jan 2007 13:26 Dik! Dik T. Winter wrote: > mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: [...] > > 0.111... and its initial segments are *representations* (according > > to you, they cannot be numbers). They can represent natural numbers > > as well as certain real numbers. Obviously both sets of numbers are > > isomorphic. > > What is the isomorphism? And 0.111... does *not* represent a natural > number in unary notation. Do you claim there is an isomorphism > between the set of natural numbers and the set of real numbers: > {1/2, 3/4, 7/8, ..., 1/9} > That is false. There is a bijection, but that bijection does *not* > work through the representation. There _is_ a bijection? Between which sets? F. N. -- xyz
From: Andy Smith on 15 Jan 2007 03:27 Dave Seaman wrote: ... > > Let me clarify a bit. First of all, what I was > saying is that an > infinite ordered set may have a last member. A > *series* is something > rather different. > > It probably makes more sense in the present context > to speak of a > *sequence* rather than a *series*. Usually the word > "sequence" means a > mapping whose domain is the natural numbers, and in > that sense a sequence > never has a last member. > > However, the word "sequence" also has a more general > meaning. If alpha > is any ordinal, then an alpha-sequence is a mapping > defined on alpha. An > ordinary sequence in the sense of the previous > paragraph is thus an > w-sequence. A sequence in this more general sense > therefore has a last > member whenever the domain is a successor ordinal. > In particular, an > (w+1)-sequence is a mapping f: w+1 -> X for some set > X, and such a > sequence does indeed have a last element, namely > f(w). > OK, thanks. > > Each of the crossings is at some finite x. No > infinitesimals are > involved. > But for any finite x such that |x| > 0, there are an infinite number of zero crossings of sin(pi/x) between 0 and x ? There are an infinite number of crossings at x = 0.0.. ? There cannot be a first crossing at any finite x ? sin(pi/x) = 0 at x = o from antisymmetry of the function ? regards --- Andy
From: Dave Seaman on 15 Jan 2007 15:48 On Mon, 15 Jan 2007 13:27:58 EST, Andy Smith wrote: > Dave Seaman wrote: >> Each of the crossings is at some finite x. No >> infinitesimals are >> involved. > But for any finite x such that |x| > 0, there are an infinite > number of zero crossings of sin(pi/x) between 0 and x ? Yes. > There are an infinite number of crossings at x = 0.0.. ? For each epsilon > 0, there are infinitely many crossings in the interval (0,epsilon). > There cannot be a first crossing at any finite x ? There cannot be a first crossing at any positive x. > sin(pi/x) = 0 at x = o from antisymmetry of the function ? No, sin(pi/x) is undefined at x = 0. You can define a function f: R -> R by f(x) = sin(pi/x), if x != 0, = 0, if x = 0. and in that case, it's true that f(x) = 0 (but because the definition of f says so, not because of antisymmetry. You can deduce antisymmetry from the definition, not the other way around. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Virgil on 15 Jan 2007 16:05 In article <1168868712.421404.294530(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > 1 1 > > > > / \ / \ > > > > / \ / \ > > > > / \ / \ > > > > 2 3 2 3 > > > > / \ / \ / \ / \ > > > > 4 5 6 7 > > > > / \ / \ / \ / \ > > > But with this interpretation the union of the sets of paths is *not* the > > set of paths in the tree on the right. The set of paths in the tree on > > the right is missing the path 1-2-L for example, which should be in the > > union. Actually the set of paths on the right misses *all* paths from > > the tree on the left. So when you state that there are only 4 paths in > > the tree on the left and 8 in the tree on the right, I retract my > > remark. In that case the union of the sets of paths has 12 elements. > > In the finite trees there are ony initial segments of the infinite > paths of the infinite tree, because every path representing a real > number is the limit of a sequence of finite paths. In the infinite > union of all finite trees there are all sequences - countably many. If > there are not more limits than sequences, then the limts are countable > too. In any "complete" infinite binary tree, the set of infinite paths is clearly equivalent to the set of all functions from the actually infinite set of naturals N to the set {L,R} in an obvious way. If WM claims the set of all such functions is countable, let him count it by finding an explicit surjection form N to the set of all functions from N to {L,R}. Cantor, and others, have produced explicit proofs that no such surjection can exist. > > No initial segment of Cantor's diagonal number is infinite. Neverthelss > there is an infinite diagonal number? Yup! > > > > Pray revise how union is defined. I think I quoted the definition quite > > proper. It *is* required. If an element a is in the union of a collection > > of sets, it *must* be in at least one set from the collection. > > The limit of a sequence is not an element of the sequence. > > > > > Are you stating that the Peano axioms are nonsense because there is no > > set that satisfies the axioms? Or what else? > > If there actually is a set satisfying these axioms, then its number > omega of members is not in the set as an element. An infinite number is > created by a set of finite numbers or segments (as the limit). Depends on what one allows as a "number". An infinite set is not necessarily a number of any particular type. > > > > What is the relevance? > > The relevance is: If the union of all nodes of a path is not an > infinite path, then the union of all digits of Cantor's diagonal is not > a real number. The union of all nodes of a path is, at most, that path. So that WM is assuming his conclusion. Such circular argument is not mathematically or logically acceptable.
From: Virgil on 15 Jan 2007 16:27
In article <1168869430.273702.199810(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > See my recent posting. You need to assume that the complete union of > > > levels of the binary tree does not contain all the paths representing > > > real numbers. Therefoer something must be added, in order to get the > > > complete set of paths. But there is nothing to be added - but ghosts. > > > > Oh. But depends on what is done. But even if you unite two finite levels, > > the set of paths in the union is not the union of the sets of paths of > > the separate levels. > > Depends on definition of "finite tree". If all finite paths are > continued with sequences of 000... or 111..., then the union of paths > of two finite trees is the set of paths of the larger tree. So that in WM's definition, all finite binary trees of fixed order must be nested, so they have a common root, and one is a subtree of the other. And then his definition of union of a set of such trees is the smallest binary tree containing all of them as subtrees. That is a very specialized set of definitions, and is not a proper definition of union. it is more like the definition of a directed set or a filter. > If you unite all finite trees (finirte paths continued by 000... and > 111...), then you unite all possible nodes, you unite all possible > edges, but you don't unite all possible paths? What is the difference > between the union of all finite trees and the infinite tree in terms of > nodes? The difference is in terms of sequences of branches. In the union as described, every sequence of branchings in any path is eventually constant (either all left branchings from some node onward or all right branchings from some node on) as these are the only paths in your set of 'extended' trees being unioned. > > > > There are infinite paths in the union of the trees. But each is "eventually constant" in the sense described above, whereas "most" infinite paths ( all but countably many) in the true infinite tree are non-constant ( have both infinitely many left branchings and infinitely many right branchings). > > There is *no* infinite > > path in the union of the sets of paths. Those two things are different. > > The infinite paths are the unions (limits). The only ones in the union are the countably many which are eventually constant (either have only finitely many left branchings or only have finitely many right branchings). > > A union cannot have more elements than are united. No matter how hard WM tries to make it happen. > A set of sequencs cannot have more limits than there are sequences. Which leaves out uncountably many of them when unoining 'finite' trees. |