Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 24 Jan 2007 07:30 Dik T. Winter schrieb: > In article <1169547646.733664.94830(a)m58g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > > If the union of singletons {1} U {2} U {3} U... U {n} U ... is an > > > > infinite number omega, then the union of domains of sequences {1} U {1, > > > > 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an > > > > infinite seqeunce {1, 2, 3, ...}. > > > > > > Pray explain what you understand under "domain of", and what you understand > > > under "union of domains". That is complete non-standard terminology. > > > > The domain of a sequence is a natural number. The domain of an infinite > > sequence is omega. > > That is not an explication. That is obfuscation. That is, it is still > clear as mud. It is from a book on set theory. The domain of a sequence is also called its pre-image, in German Definitionsbereich. I don't know f other Enlish words for it. > > > But > > > whatever, whenever in a collection of sets A_k, none of the sets A_k > > > contains a particular element E, it is also not in the union. > > > > Therefore there is no uncuntable set of path in the union tree. But > > there are all paths in the union trees. > > Conclusion is false. Summary 1) Every complete infinite binary tree T (containing all nodes and edges) contains all paths. 2) The union tree T(oo) of all finite trees is well defined (as I have shown elsewhere) and yields the complete infinite binary tree containing all nodes and edges: T = T(oo). 3) The union of all finite trees includes the union of all nodes and, with it, the union of all such subsets which are paths (because every path is a well defined subset of the set of nodes if the structure of the tree is well defined). 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. 5) A countable union of countable sets is a countable set (according to ZF with AC). ==> The set of all path is countable. (==> The real numbers are countable.) Going on, we can say: 6) T(oo) = T contains only finite paths. 7) T(oo) = T contains all paths including all infinite paths. ==> There are no infinite paths. (There are no irrational numbers.) > > > Right. It is the union of all nodes (and edges and levels and paths). > > Wrong. If you state that T1 is the countable union of all finite sets of > finite paths, it is a set of paths, and so does not have nodes, edges or > levels as element. > > > > > And T1 = T2. > > > > > > No. T1 is a set of paths, it is not a tree. T1 does not contains nodes > > > or edges as elements. Only paths. > > > > The nodes can be enumerated in various ways. > > That does not matter. T1 as you defined above is a set of paths. No. I *define* T(oo) is an ordered set of nodes, the order being expressed by the edges. > *Not* > a tree. And a set of paths does not have nodes or edges as elements, it > has paths as elements. The union of paths is a set of nodes as a path is set of nodes. > And if you consider a path as a set of node, then > T1 *still* does not have nodes as elements but sets of nodes. T(oo) is the union of the T(n) which are sets of nodes. Therefo T(oo) is a set of nodes. > > The union of trees is a union of their nodes. The paths are merely > > subsets which are defined by the nodes and the special kind of tree. > > But you have to be careful when uniting sets of paths, that is uniting > sets of sets of nodes. That union contains all paths from both tree, > you can only omit duplicates from that union, so only paths that are > *identical*, i.e. consist of the same set of nodes. The union of trees is defined by nodes! Paths are maximal subsets following some prescription. Therefore in long trees there are no short paths, but there are long paths only. > Suppose {a, b, c} > is a path in tree T1, and {a, b, c, d, e} is a path in tree T2, than > the union of the sets of paths of T1 and T2 contains *both* paths. But not the union of the elements (nodes a, b, c, d). > > > > > The domain of the path 1,0,1 is 1,2,3. > > Still not clear. You can not define things by example. Here is a quote from Hrabacek and Jech, p. 55: We begin with some new terminology. A sequence is a function whose domain is either a natural number or N. A sequence whose domain is some natural number ne N is called a finite sequence of length n and is denoted. (By the way, remembering an old discussion of ours, they speak of "adic" numbers, not "ary".) > > > A path is a sequence. A sequence > > is defined as a mapping f from a natural number or omega into the > > reals. The set of reals (here only 0 and 1) is called the range. > > Abbreviations are usual: dom f and ran f. > > Finally now you define a the domain of a sequence. So it is now also > clear that a path is an ordered set of nodes. And the domain is the > initial subset of the natural numbers that terminates at the cardinality > of the path (as set of nodes). Still if T1 contains the path {a, b, c} and > T2 contains the path {a, b, c, d, e}, the union of the sets of paths contains > both. But not the unin of the trees T1 and T2. Tghe union contains only the path {a, b, c, d, e}. > > > > > The tree is an (ordered) set of nodes. The nodes can be used to form > > > > paths, i.e., subsets of the tree. > > > > > > As a tree is a set of nodes, any subset of it is a set of nodes. > > > > Of course, therefore the subsets which are paths must exist in the > > tree if the subsets exist there. > > Ok, if a path is defined as a set of nodes. But still see above about > the union of sets of paths. > > > >Not > > > every subset of it is a path. But whatever, the number of subsets of > > > the final tree is not countable. The union of the sets of paths of > > > the finite trees is not the set of paths of the complete tree. > > > > If all nodes and edges are in the union, then all subsets are in the > > union too. > > You misread again. The set of subsets of the complete tree is *not* > equal to the union of the set of subsets of the finite trees. No. The set of subsets (which are maximal, i.e., which are paths) of the complete tree is only a subset of the union of the set of subsets of the finite trees > The > reason is clear: that union does not contain an infinite subset, while > the complete tree *does* contain infinite subsets. The union contains all we have. There is nothing remaining. Regards, WM
From: mueckenh on 24 Jan 2007 07:37 On 23 Jan., 18:17, "MoeBlee" <jazzm...(a)hotmail.com> wrote: No, we do NOT have to suppose the list is of all real numbers. But you have to suppose that all enumerated lines and columns are there. So you have to suppose the finished presence of an actually infinishable set N. That is nonsense. Regards, WM
From: stephen on 24 Jan 2007 07:39 Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > In message <9b2er2p5taadlea28n6fb3g4nuvmqeijhs(a)4ax.com>, G. Frege > <nomail(a)invalid.?.invalid> writes >>On Tue, 23 Jan 2007 23:34:03 GMT, Andy Smith >><Andy(a)phoenixsystems.co.uk> wrote: >> >>> >>> I meant that the infinite non-repeating irrational binary expansion of >>> sqrt(2) requires an [...] infinite set of numbers to define its >>> location on the line... >>> >>Right. >> >>> >>> Since the integers are finite, ... >>> >>Non sequitur. Yes, the integers are finite, but there are infinitely >>many of them. >> >>To be precise: countable infinite many of them. And that's actually >>(sic!) enough to represent each and any real number. >> > Any systematic scheme for mapping the reals to integers will be the > same, subject only to permutations of the bit positions. So a systematic > scheme is, start with bit 1, then bit 2, then bit 3 etc. This > corresponds to a reflection of the possible set of numbers in n bits > about the binary point. Any real number has an infinite binary > expansion, and its corresponding mapping integer is infinite (=NaN) So what? "Corresponding mapping integer" is just something you have made up. It has no real meaning, and is totally irrelevant to the question of how many bits are needed to represent a real number. Consider the binary number .11101 This equals .90625. Its "corresponding mapping integer" is 10111, which equals 30. What does 30 have to do with the number of bits needed to represent .90625 in binary? I do not need 30 bits to represent .90625. What does 30 have to do with anything? Stephen
From: Dave Seaman on 24 Jan 2007 07:40 On Wed, 24 Jan 2007 09:19:22 +0100, G Frege wrote: > On Tue, 23 Jan 2007 19:45:39 -0500, David Marcus ><DavidMarcus(a)alumdotmit.edu> wrote: >>> >>> People cannot conceive of an infinite past, [...] >>> >> Why not? I believe that was the usual assumption before the Big Bang >> was discovered. >> > Not really... Remember? > "In the beginning God created the heavens and the earth. Now the earth > was formless and empty, darkness was over the surface of the deep, and > the Spirit of God was hovering over the waters." > This happened about 6000 years before Christ's birth, or so. It was supposed to be 4004 BC, according to Bishop Ussher. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: mueckenh on 24 Jan 2007 07:49
On 23 Jan., 19:27, Dave Seaman <dsea...(a)no.such.host> wrote: > On Tue, 23 Jan 2007 17:15:52 GMT, Andy Smith wrote: > By definition, a set is "finite" if it has the size of some natural > number. If a set isn't finite, then it's called "infinite". If a colour is not red, then it is called green by set theorists. > It's > obvious that the set of all natural numbers can't be finite, since that > would imply the existence of a largest natural number. It is obvious tat the set N cannot be actually infinite, because that would imply infinity to be finished . > > If you see this as straightforward it is because your mindset has been > > conditioned by your education to see this as normal. I can safely say > > that if your concepts of infinite sets was placed in front of the > > population at large 99 % would think that this is barking mad > > doublethink ... And they were right! > You can't have an actually infinite integer for exactly the same reason > that you can't have an actually 200-cm. meter or an actually 4-sided > triangle or an actually finished infinity. Regards, WM |