An uncountable countable set
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From: Virgil on 24 Jun 2006 14:31 In article <1151153940.768412.95000(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > Yes. The reasoning is very similar: > > > > If f is any function from N to P(N), then > > K(f) is an element of P(N) that is not in > > the image of f. Therefore, f is not a surjection. > > > > If L is any enumeration of reals (that is, L is a > > function from N to R), then D(L) is an element of > > R that is not in the enumeration L. Therefore, L > > is not complete. > > On the other hand we know that the set of individualizable real numbers > is countable. What are "individualizable real numbers"? > > > The theorem is this: > > > > forall A:set, > > forall f: A -> P(A), > > exists K: P(A), > > K is not in the image of f > > > > which implies > > > > forall A:set, > > forall f: A -> P(A), > > f is not a surjection from A to P(A) > > But for all sets S constructed by the diagonal argument, we find a > surjective mapping g: |N --> S. The diagonal argument does not construct sets, it merely shows that every "list" of reals is incomplete. There is, however, a slight variation on the diagonal argument which will provide, for any given list of reals, as many non-members of that list as there are reals (or as there are subsets of N). > > > > which implies > > > > forall A:set, > > there does not exist a surjection from A to P(A) > > > > >and shows only what it does show in fact, namely the > > >countability of all list numbers including all diagonal > > >numbers which can be constructed. > > > > It shows that every list of reals, there is at least one > > real that is not on the list. From that, it follows that > > there is no list containing all real numbers. > > No. It means that there is a paradox which can be remedied by switching > the mapping from f to g and this is possible for every real number > constructed. The point is that the Cantor construction is a rule which simultaneously provides for EVERY function, f or g or whatever from N to R, a real not listed by THAT function. Changing functions doesn't work with a rule that applies equally well to EVERY function. > Therefore there is no uncountable set of reals. WRONG! AGAIN! > > > By definition, > > that means that the reals are uncountable. > > > > >> What it shows is that the assumption that f is a surjection > > >> from N to P(N) leads to a contradiction. If something leads > > >> to a contradiction, then it is provably false. > > > > > >So drop this assumption. There is no mapping, even if > > >surjectivity is not at all in question! > > > > Well, there certainly is a mapping from N to P(N). For > > example, let f(n) = { n }. So I don't know what you could > > possibly mean by saying that there is no mapping. > > There is no mapping including K as the image of at least one source > element, even if surjectivity is not at all in question! But there are uncountably many which do not have K in their image, so what is your point?
From: Virgil on 24 Jun 2006 14:42 In article <1151154047.409215.307660(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > >The assumption includes that |N and P(|N) do exist as complete sets. > > > > Actually, the nonexistence of a mapping from N to P(N) has nothing > > to do with whether N and P(N) exist as "complete sets". If you > > want to think of them as potentially infinite, that's fine. It's > > still possible to have a bijection between two potentially infinite > > sets. > > Correct. If |N and P(|N) are potentially infinite, then the definition > of a mapping doesn't cause any problem, but it can be shown bijective, > because there is no completeness. Doesn't such non-reasoning make you dizzy, "mueckenh"? If there is "no completeness", how do you show bijectivity? If you cannot say "for all n in N" or "for every n in N", you cannot show anything about all of N. > > "Surjective" means that all elements are in the image. To be precise, it means that all elements of the codomain are also elements in the image. "All elements" without a referent is ambiguous. > A potentially > infinite set is always finite though surpassing any given limit. > Therefore the question of all of them cannot be decided. Then in "mueckenh"'s math, issues of injection and surjection cannot be decided at all for such potential but not actual sets, as resolution of those issues requires that one CAN speak of "all of them".
From: Virgil on 24 Jun 2006 14:49 In article <1151154151.237811.168860(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Actually, even if it IS true, it would not allow transmutation of a set > > with a first element into one without a first element, at least if the > > transpositions are applied sequentially, as they must be. > > The transpositions are just as sequentially as the construction of the > diagonal number. > > > There would have to be a first transposition in the sequence which > > effects the change, and a single transposition is incapable of doing > > this. > > This arguing stems from K?nig (1905). We know that there are only > countably many constructible reals. If there is a well-ordering of the > reals, then within this order there must be a first non-constructible > real. This is a paradox. Hence, there is no well order. No one here is claiming a well-ordering of the reals. That is a straw man argument by which "mueckenh" is trying to divert attention from the fact that his claim to have an uncountable countable set has been totally destroyed. His {f,K,M} construction in which he requires K = {x: x not in f(x)} to be in the image of f, is self-contradictory, so such{f,K,M} does not exist.
From: mueckenh on 25 Jun 2006 04:27 Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >"Surjective" means that all elements are in the image. A potentially > >infinite set is always finite though surpassing any given limit. > >Therefore the question of all of them cannot be decided. > > That's not true. If A is a "potentially infinite" set, then > saying "forall x in A, Phi(x)" just means that if x is any > thing of type A, then Phi holds of x. It doesn't matter whether > A is "completed" or not. > > We know perfectly well that if x is a natural number, then x+x > is an even natural number. From this fact, it follows that > > forall x: Natural, x+x is even > > "forall" doesn't mean that anyone has *checked* every element of > the set. But it doesn't say what "all" means. The set hasn't a fixed cardinality, because for every n considered, there is an n+1 not yet considered, because "potential" means always finite but not limited by any threshold. We are never closed. A diagonal number is never ready. The set K is never ready, unless we chose it to be finite by a suitable f. Regards, WM
From: mueckenh on 25 Jun 2006 04:35
Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: > >> Yes. The reasoning is very similar: > >> > >> If f is any function from N to P(N), then > >> K(f) is an element of P(N) that is not in > >> the image of f. Therefore, f is not a surjection. > >> > >> If L is any enumeration of reals (that is, L is a > >> function from N to R), then D(L) is an element of > >> R that is not in the enumeration L. Therefore, L > >> is not complete. > > > >On the other hand we know that the set of individualizable real numbers > >is countable. > > "individualizable" is not a very mathematically precise notion, so > I don't think you can prove anything about it. Any countable alphabet has only countably many finite words (cp. the set of finite subsets of |N). Any alphabet that we can use is countable (in fact it is finite). Hence we have only countably many symbols, names, notations. We can distinguish only countably many numbers. Further the set of elementary cells of the whole universe is finite. But mathematicians in general do not like to be frustrated by physical needs. > >> It shows that every list of reals, there is at least one > >> real that is not on the list. From that, it follows that > >> there is no list containing all real numbers. > > > >No. > > Yes. The list of all names contains all we can denote as a number. > > >It means that there is a paradox > What is it that you think is paradoxical? It is a paradox that mathematicians can believe that 2 + n and 2 * n share the same level, but 2^n does not, while exponentiation is only an abbreviation for multiplication which is only an abbreviation for addition. It is a paradox that logicians can believe that the non-existence of a paradoxical set proves uncountability. It is a paradox that we know that only countably man numbers can be identified, and that numbers do exist only in our minds, hence numbers which can not be identified in our minds cannot exist. It is a paradox that we know all this but nevertheless speak of uncountably many numbers. It is a paradox how this obviously selfcontradictory theory could occupy mathematics. > > >which can be remedied by switching the mapping from f to g and this > >is possible for every real number constructed. Therefore there is no > >uncountable set of reals. > > I don't know what you are talking about. Look at some examples: > > Here's an example: f(x) = x/2. That's a function from N to R. > It generates the decimal expansions > > 0.00 > 0.50 > 1.00 > 1.50 > 2.00 > ... > > The diagonal procedure produces a new real by adding one to > each diagonal digit: > > D(f) = 1.61111111.... > > D(f) is definitely *not* in the image of f. But it is definitely in the image of a suitable g. This can be done with all diagonal numbers D ever constructed. Hence all diagonal numbers ever constructed are countable. > So you think that there is some function f such that *every* > real is in the image of f? Then show it to me. I'll show it to you after you have constructed all numbers you wish to be contained in my mapping. Wrong is only your assumption that there are more than can be constructed. > Show me why > the general proof that there is no such function breaks down > by giving an *example*. A real example. The general proof breaks down by one counter example like that one given as the answer to your next posting. Here it would get too long. > > Do you think that there is such an f? If so, what is it? > Give me a formula for it. There is no such thing like "all reals". But if the diagonal proof fails in one case (see my next posting), it cannot be trusted in others. > >There is no mapping including K as the image of at least one source > >element, even if surjectivity is not at all in question! > > I can't tell exactly what that sentence means, but I think you > are just saying > > there is no mapping f that includes K(f) in its image. > > which is equivalent to the more straight-forward statement > > forall f from N to P(N), K(f) is not in the image of f > > How are you saying anything different from that? But the > latter statement immediately implies the statement > "forall f from N to P(N), f is not surjective". forall f from N to any set, K(f) is not in the image of f . Regards, WM |