An uncountable countable set
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From: mueckenh on 21 Aug 2006 15:48 Virgil schrieb: > TO fails to note that induction does not apply. > The form of induction is: > Let F(x) be a predicate whose argument, x, is a member of a minimal > inductive set S.* > If F(x) is true for the first member of S > and > if whenever F(x) is true then F(successor(x)) is also true. > then > the principle of induction says that F(x) is true for each member of S. In particular induction says that each member of the set of natural numbers is a member of a finite set, i.e. is covered by a finite sequence. This holds for all members. Therefore it is illusory to pretend any set of naturals would require more to be covered. > > It does not say anything about what is true for the set S itself, only > for its members. The set itself is "its members". The set of natural numbers is "all natural numbers" and nothing else. Or what is the difference between "all natural numbers" and the set? Regards, WM
From: mueckenh on 21 Aug 2006 15:49 Virgil schrieb: > Except that ordinal numbers cannot work that way, > as it would require each ordinal to be a member of itself, > which is not possible for ordinals. For the set of finite positive whole numbers ordinal and size are equal. The first element has value 1. Any problem about that? The set does not exist? Regards, WM
From: mueckenh on 21 Aug 2006 15:52 Virgil schrieb: > In article <ec8jpl$t2j$1(a)ruby.cit.cornell.edu>, > For every edge in an infinite binary tree, there are uncountably many > non-termnating paths passing through (or including) that edge. That is an assertion without justification and without logic. > The number of terminating paths is irrelevant. > > WE already have a satisfactory proof that there are injections but no > surjections from any set to its power set, so that the cardinality of > any set is strictly less that the cardinality of its power set. > Wrong. It would be the case if set theory was undisputedly consistent. But as we just investigate consistency, you cannot presuppose it. With your attitude it is impossible to find any inconsistency even in an inconsistent theory. Deplorably you are too simple to recognize that. Regards, WM
From: Virgil on 21 Aug 2006 15:54 In article <1156149324.184708.191200(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Nevertheless they are all countable, just like the edes of my infinite > tree. > > And by rational relation we find the set of paths being *not* larger > than the set of edges. > > Have you got it now??? > > Regards, WM The set of edges of an infinite binary tree are certainly countable, and it is not difficult to construct an explicit bijection between them and the set of binary representations of the naturals. The set of paths, however is not countable, as may is shown by constructing an explicit bijection between the set of all such paths and the set of all subsets on the set of naturals. Both the bijections referred to above have been presented here with no one able to show either to be anything other than as advertised.
From: David R Tribble on 21 Aug 2006 15:55
Dik T. Winter wrote: >> No that does not make sense. sum{n = 1 .. oo} 1 is not defined, the same >> holds for sum{n = 1 .. oo} n. If you want to use them you have to >> provide a definition for them. > David R Tribble wrote: >> But Tony thinks he has provided a definition, based on his "Big'un" >> number. The problem, of course, is that he simply assumes that >> arithmetic operations on Big'un work "es expected" without providing >> any proof of that whatsoever. >> >> It's one thing to provide a definition, it's quite another to prove >> that it is a consistent definition. > Tony Orlow wrote: > I think the responsibility lies with you to point out an inconsistency > that arises from my assumptions. You're kidding, right? > Set theory is not "proven true". It > cannot prove itself consistent. It took years of trying, with some > succcess in detecting failures, to refine set theory so that it was > somehow actually consistent. But, it cannot "prove" itself so. It's true that in any sufficiently powerful axiomatic system a theorem cannot be formed that proves the consistency of the that system. However, within the framework of the theory (the axioms and theorems that it comprises), you can prove theorems true and prove false statements to be false. In that sense, the theorem that, for example, the set of all naturals is infinite is a provably true theorem within standard set theory. > Please > show where I am being inconsistent, not with set theory, but within my > own assumptions. Remember, I don't claim to believe in transfinite set > theory, and don't intend to be consistent with it. The problem is that your own assumptions are not consistent with each other - they don't form a coherent theory. Many of these inconsistencies have been pointed out to you before, but you choose to not believe them or simply to ignore them. |