An uncountable countable set
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From: Virgil on 20 Jun 2006 21:23 In article <1150836068.607034.205340(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Give us an example of such an f, K_f and M_f, to show that such an f, > > K_f anf M_f can actually exist. > > If the set {f, k, K} could actually exist, then M would be countable. > But it is not, isn't it? If M_f exists at all it is countable, and if it doesn't exist the question of its countability is disappears. > > > > If they can exist at all, it is easy enough to show that there is a > > bijection beween |N and M_f, and if they can't then there is no M_f to > > worry about. > > In order to define K_f you first have map all natural numbers n of |N > on the finite subsets of |N. Then consider K_f. But there is no element > k of |N remaining, which could be mapped on K. Then try a different mapping, say g:N --> M_f. I have previously shown that if K_f and M_f exist for a non-bijective f:N --> M_f, then M_f must be countable. I have also shown that if f itself is required to be a bijection neither it not K_f nor M_f even exist, so the question of countability of a non-set is nonsense.
From: Virgil on 20 Jun 2006 21:34 In article <1150836248.705493.88420(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > M depends on f. And when you take g, then M depends on g. That is the > > > essence of M. > > > > Wrong! Given any two large sets are many functions for one to the other > > and equality of cardinality depends on whether any one of them is a > > bijection. Given two sets to compare, you cannot restrict things to only > > one allowable function between them but must allow all possible > > functions between them to be considered. > > I allow for all functions, because I did not introduce any > restrictions. You require that (1) f be a bijection and (2) its codomain contain {x in N: x not in f(x)} = K_f These conditions are mutually contradictory, so that no such bijection and no such K_f can co-exist. However, it the function is not required t have K_f as a value, so that IT is not bijection, then both f and K_f can exist and the codomain of F, namely M_f, is countable. > > > > > > > In order to show that > > > > there is no bijection between N and M you are not allowed to change M > > > > for each and every attempted mapping. > > > > > > I do not change anything. I define K by exactly that mapping which is > > > assumed. > > > > Then there are other functions between N and M_f, one of which is a > > bijection. > > Could you prove that assertion please? > > > There are two issues here. The first is whether a function of the sort > > Meucken tries to create can exist at all. He has given no example of > > such a function, nor proof of existence. > > Of course there is no proof of existence. If it was, then M was > countable. Then the non-existent set M_f, if not countable, is not uncountable either. > > > > However if we allow the assumption that such a function f exists, it is > > not hard to show that there are bijections between |N and M_f, though f > > need not be one of them.
From: Virgil on 20 Jun 2006 21:47 In article <1150836393.730055.231800(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1150728692.949790.258040(a)h76g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Daryl McCullough schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de says... > > > > > > > > >> > There is no bijective mapping f : |N --> M, > > > > >> > where M contains the set of all finite subsets of |N > > > > >> > and, in addition, the set K = {k e |N : k /e f(k)} of all natural > > > > >> > numbers k which are mapped on subsets not containing k. > > > > > > > > It's not exactly clear what you are talking about here. Do > > > > you mean that M is the set of all finite subsets of N? > > > > > > Above I spelled out clearly that M contains the set of all finite > > > subsets of |N and one more set K. What is unclear? > > > > > > > In > > > > that case, there is definitely a bijection between M and N: > > > > > > If M was the set of all finite subsets of N, that would be easy enough > > > to see. > > > > > > > If you let K = { k in N such that k is not an element of f(k) } > > > > then we can see that > > > > > > > > K = N > > > > > > Such a mapping is possible with no doubt. There remains only one > > > question: what number is mapped on K? > > > > That raises the question of whether such a set as M_f can exist at all. > > > > If it cannot exist, then the issue of whether a non-existent set is > > countable is irrelevant. > > Of course it exists. How can a function, f, from N to any subset of P(N) exist if it must be the case that for some n in N, f(n) = {k in N: k not in f(k)} ? If n were such that f(n) = K_f = {k in N: k not in f(k)}, would it be the case that n IS a member of K_f or n IS NOT a member of K_f? Assuming either establishes its negation, so neither can be the case. > > > So that Meucken's first duty is to prove that a function having the sort > > of codomain, M_f, he describes can exist at all. > > > > If ever that is established, the construction of a bijection between it > > and N is fairly trivial. > > If |N exists, then K exists too. WRONG! Existence of N is quite simple in ZFC of NBG, but the existence of K depends on the alleged properties of function f. If f is requires to have K_f as a value then f, K_f and M_f are all non-existant. If f is allowed not to have K_f as a value, then f, though no longer a bijection, can exist, as can K_f and M_f. But in this case there are other functions between N and M_f which ARE bijections.
From: Virgil on 20 Jun 2006 21:55 In article <1150836884.833877.138440(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > The element K of M does not exist. > > Excuse me. The set {f, k, K} does not exist. > > > > That requires proof, and is false in at least one case. > > Let f(n) = {} for all n, then K_f is quite well defined, and equals N. > > You are correct. Of course K_f does exist for any f (if all natural > numbers do exist). Not so. {f,K_f,M_f} cannot exist if f is required to have K_f as a value, but can and will, though not as a bijection, exist if that requirement is voided. > Nevertheless you see that the mapping is not a > bijection beause no f(n) = K. The failure of one function to be a bijection does no establish that none exist. > > > > > > If it is proven impossible to set up a mapping between M and |N, then M > > > is uncountable. > > > > Wrong! > > M is countable means: There is an enumeration of the elements of M, > i.e., a bijective mapping |N <--> M. > > > > If it is impossible to set up an injection from N to M_F or a surjection > > from M_f to N, and the axiom of choice is assumed, only then is M_f > > uncountable. > > > > And if K does not exist, > > If |N does exist, then each of its subsets must exist, including K. K can only exist if the function, f, which defines it is not required to have it for a value. > What does not exist is the bijection |N <--> M. If M_f exists at all (and it won't with Muecken's conditions on f) , then there are uncountably many bijections between N and M
From: Virgil on 20 Jun 2006 22:08
In article <1150837129.824319.182500(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Rupert schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Rupert schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > An uncountable countable set > > > > > > > > > > There is no bijective mapping f : |N --> M, > > > > > where M contains the set of all finite subsets of |N > > > > > and, in addition, the set K = {k e |N : k /e f(k)} of all natural > > > > > numbers k which are mapped on subsets not containing k. > > > > > > > > > > > > > You're using the notation "f" in two ways. > > > > > > No. > > > > Yep. In one of the occurrences it occurs preceded by a universal > > quantifier, in the other it occurs as a constant symbol. > > Could you say what you mean? > > > > > First you're denying that a > > > > function f with certain properties exists, then you're defining M in > > > > terms of some fixed function f, > > > > > > f is not fixed by any prescription. > > > > > > > It doesn't make sense to talk about the set K={k e |N: k /e f(k)} unles > > you've specified what f is. > > f is not fixed by any specification. its domain, codomain and range have been specified, but that codomain and range have been define circularly in a way which prohibits the existence of an F, K_f and M_f satisfying those conditions. The easiest condition to relax is the one requiring that f be a bijection, in particular that it have K_f as a value. If this condition is dropped then, and only then, can an {f,K_f, M_f} as described actually exist, and then M_f is easily countable. > > So you mean M is the set of all finite subsets of |N, together with all > > sets of the form K={k e |N: k /e f(k)}, where f ranges over all > > possible mappings |N->P(|N)? > > No. K is that single set which belongs to the function f. Thus you need to have that there to exist an n in N such that f(n) = {x in N: x not in f(x)} = K_f But if that n is in K_f then n is not in f(n) = K_f and if it is not in K_f then it is automatically in f(n) = K_f so that no such n can exist such that f(n) = K_f. |