An uncountable countable set
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From: Virgil on 13 Jul 2006 15:24 In article <1152790523.616184.139450(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152735810.530270.91150(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The axiom says "there is an infinite set". It does not say that > > > 0.111... does belong to that set, in particular because all numbers > > > which in fact do belong to the set are different from 0.111... . > > > > > > But 0.111... in effect IS that set to which it does not belong. > > Yes. 111... represents aleph_0 or omega. But we see that it is > impossible to have aleph_0 or omega natural numbers. I see no such thing. > Every smaller > number is finite. If aleph_0 is the first transfinite cardinal, then > the set of natural numbers alone is not actually infinite. If the cardinal of the set of naturals is aleph_0, as he admits, and aleph_0 is transfinite, as he admits, then how does "mueckenh" manage to conclude that the set of natural numbers is "not actually finite"? "Mueckenh" needs a severe course in logic.
From: Virgil on 13 Jul 2006 15:31 In article <1152790782.066417.225300(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152736018.461492.108050(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > The binary tree: All paths, not yet disperged, start through the root > > > edge a. > > There is no need for a root edge at all. the tree can quite comforably > > be rooted in a root node from which two edges branch out to two cild > > nodes, and so on. > > > > > > > > (Let us denote them as a single path as long as they are > > > together.) > > > > As paths are sets of edges, this does not work. > > Call it "bunch of paths cross sections", if you like. I abbreviate that > by path. That is what nodes are for. One has the set of paths through any particular node as the paths which are "together" from the root to the given node. > > > > Map this root edge a on this single path. In the next level > > > the path splits in two paths. Map half of edge a on each of them. The > > > right one passing through edge b, gets b mapped on it and it inherits > > > half of a. After splitting again, each of the paths gets the next edge, > > > say c, half of b and quarter of a. > > > > > > | a > > > o > > > / \ b > > > o o > > > / \ / \c > > > > This tree contains 4 paths indicated by sequences of two branchings, > > left or right right, so that {LL,LR, RL, RR} represents the set of all > > paths for the tree as shown. > > > > For larger binary trees one gets more and longer strings of left to > > right branchings. > > > > For infinite binary trees each such list of branchings is an infinite > > sequence. And there are uncountably many possible such infinite > > sequences. > > But we know that there are nearly twice as much edges. We only know that for finite trees. "Twice as much" doesn't work for infinite sets. > There are > infinitely many of them too. And the the set of edges is countable. Then show me an ennumeration of them. I have already shown, indirectly, a bijection between the set of all paths and the set of all subsets (power set) of the set of edges.
From: Virgil on 13 Jul 2006 15:32 In article <1152790900.301649.192680(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152638895.480011.23630(a)h48g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > But not all unary sequences are representations of natural numbers. > > > > 0,111... is a sequence which does not represent a natural. > > > > > > And it cannot be completely indexed by natural numbers. > > > > That is precisely wrong, it is only the set of all natural numbers which > > CAN index it. > > It cannot be completely indexed by all natural numbers. That is no > question. Therefore the set of all does not exists. Perhaps not in your philosophy, Horatio.
From: Virgil on 13 Jul 2006 15:34 In article <1152791156.508057.8940(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > But it does not guarantee what is existing there. In order to find out > > > you must count. > > > > I do not know what "axiom of infinity", 'mueckenh" is referring to , but > > the ones in ZF, or ZFC or NBG guarantee existence of all of the > > "naturals" ( as finite ordinals) without having to count anything. > > Perhaps those naturals guaranteed by the axiom even cannot be used for > counting? They are, after all, no numbers. Why does "mueckenh" think any natural numbers are not numbers?
From: Virgil on 13 Jul 2006 15:35
In article <1152791019.872491.320950(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1152639432.511655.220040(a)m73g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > > > Except that after each (1,2), there will only be a finite initial > > > > subsequence in numerical order and an infinite terminal sequence not yet > > > > ordered. > > > > > > > > Using 1 origin indexing, the nth occurrence of (1,2) occurs at the > > > > (n^2 + n)/2 th position in the list of "transpostions". > > > > > > > > At which (n^2 + n)/2 th operation is the entire list ordered? > > > > > > At which number can I find the last element of Cantor's diagonal? > > > > The successor to the one which finishes your ordering by magnitude. > > > > Which, since neither exists, works correctly > > Both do no exist. That is the reason why Cantor's arguing is correct > and mine is not. Right! |