An uncountable countable set
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From: Gc on 3 Jul 2006 18:12 mueckenh(a)rz.fh-augsburg.de kirjoitti: > Gc schrieb: > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > > not exist. > > > > Yes it does. For every function N to P(N) there exists K in P(N). > > Not for any surjective function. Ash. I meant that if there were a surjection then K would have to exist. If you wan`t to find maximum amount of pairs of some function N ----> N(P) where there are no K you can just pair each a of N with {a}. > You will see it somewhat easier: > Map R on the set {1, 2, 3, K} with K(f) be { x in R | x is not an > element of f(x) }. What I think is that there has to be a function exactly defined before K exits and when you have defined that you already have that function, so K is not in the range of the function. The P(N) is different because it contains all possible K:s.
From: Virgil on 3 Jul 2006 18:15 In article <1151934453.948725.293220(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151843785.711371.113080(a)75g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Gc schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > > > > > > > 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. > > > > > > > > > > This shows M to be uncountable. > > > > > > > > > > Regards, WM > > > > > > > > The set M exist only if you have already defined exactly funktion f, > > > > because there is now certain set K before the funktion is exactly > > > > defined. So you have bijection g: N ---> M/K . Then you can add the set > > > > K to finite sets and get M. This is no paradox and this is very > > > > different than funktion f: N ---> P(N) because P(N) contains K by > > > > definition. > > > > > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > > not exist. > > > > > If K does not exist, there is no bar to f being bijective. > > > You can see this by the following argument: Map |R > > > (including |N) on P(|N) with the only condition that a natural number > > > has to be mapped on that set K e P(|N) which contains all natural > > > numbers which are not mapped on sets containing them. > > > > > > Here we have not at all any problem with too less elements in the > > > source. Nevertheless there is no surjection possible. > > > > Does every potential bijection from R to P(N) have to map some natural > > to K? why can't it map , say , pi to K? Then K can exist, and IS in P(N) > > and f(pi) = K with no problems, and f CAN be a bijection. > > > > There are all kinds of surjections from R to P(N), though none of them > > with the specific artificial restrictions you would impose. > > There are all kinds of surjections from N U {pi} to P(N). You cannot > prove that there is none. Theorem: There are no surjections from N U {pi} to P(N). Proof: Define g: N --> N U {pi} by g(0) = pi, and g(n) = n+1 for n != pi. This is clearly a bijection, so that fog:N --> P(N): n |--> f(g(n)) is a surjection if and only if f:N U {pi} --> P(N): x |--> f(x) is a surjection. But K(fog) = {x in N: x not in f(g(x))} cannot be in the image of fog, and prevents fog from being a surjection, so that f cannot be one either. Q.E.D. A similar argument works to show the union of N with any countable set cannot surject onto P(N).
From: Virgil on 3 Jul 2006 18:20 In article <1151934701.189982.53210(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > You have not proved - or even given any justification at all - that this > > > > process has any meaningful limit, nor that if it does, it is ordered as > > > > you wish. > > > > > > This is exactly the same with Cantor's diagonal proof which not valid > > > for an infinite list. > > > > > > That you do not choose to admit the difference is not evidence that > > there is no difference. > > There is none. Claimed but not proven, and the claim cannot be proven mathematically because it is false. > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > 0.111...1 > > > ... > > > > > > But in this list the number 0.111... is not contained. Hence not all of > > > its digits can be identified. > > > > You have just proved that your example supports Cantor's theorem by > > producing a number, 0.111..., not in your own list. > > All digits of a number must be indexed by natural numbers. Otherwise > they cannot be identified. All digits which can be identified are > pesent in the list which contains all unary representations of naturlal > numbers. That presumes that the set of naturalas is finite, which in ZFC abd NBG is false. And "mueckenh" has not presented any consistent system of axioms in which it is true. > 0.111... does not belong to this set. It does in ZFC and NBG in both of which N is infinite.
From: Virgil on 3 Jul 2006 18:31 In article <1151935261.123362.266490(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Take > (N and pi) --> P(N) is not provable non-surjective. > N --> P(N) is apparently provable non-surjective. False!!! Define g: N <--> N union {pi} by g(0) = pi and for n > 0, g(n) = n -1 Then g is a bijection. if f: N union {pi} --> P(N) is a surjection then so must be fog, where fog(n) = f(g(n)) for all n. Thus if N --> P(N) cannot be surjective then neither can N union {pi} --> P(n) be surjective. > > You do not wonder why that is so? Because it is not so, as proved above. > > Please stay with one single arguing, and do not judge with different > measures as you like to do. All our "measures" and "arguings" are only mathematics in action. If you don't like mathematics, you would do well to avoid it in future.
From: Virgil on 3 Jul 2006 18:37
In article <1151961136.639460.110410(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1151759090.073603.122700(a)m79g2000cwm.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > > > Eh? K(f) is defined regardles the kind of mapping of f. How can you > > > > come to the conclusion that it is not defined for some particular f? > > > > What part of the definition fails for some particular f? > > > > > > Remember what I said: > > > Map |R (including |N) on P(|N) with the only condition that a natural > > > number has to be mapped on that set K e P(|N) which contains all > > > natural numbers which are not mapped on sets containing them. > > > > Yes, such a map is impossible, but that proves nothing about the > > impossibility > > to have a surjective mapping from R to P(N). If a mapping is surjective > > the requirement is that *each* element of the target should be the image > > of an element in the source. There is nothing that requires that element > > in the source to be a natural number, *unless* the source is the set of > > natural numbers (or a subset of it). So what does it prove that such a > > required mapping is impossible? > > Try to map R on the set {1, 2, 3, K} and you will see it. R surjects to {1,2,3,L} with no problem, when L is not "mueckenh"'s circularly defined K. When one restricts the functions one is allowed to use to functions which cannot exist, as "mueckenh" keeps doing, one canot do much math. |