An uncountable countable set
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From: Daryl McCullough on 24 Jun 2006 10:07 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. >> 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. Right. If f is a function from N to P(N), then the image of f is a countable set. Adding one element to the image leaves it still countable. >> 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. Yes. >It means that there is a paradox There is no paradox here. To say that f is no a surjection from N to P(N) isn't a paradox, any more than to say that 2 is an even number. It's just a provable fact. To say that there is no surjection from N to P(N) isn't a paradox, any more than saying that there is no natural number x such that x+x is odd. It's just a provable fact. What is it that you think is paradoxical? >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. So you think that there is some function f such that *every* real is in the image of f? Then show it to me. Show me why the general proof that there is no such function breaks down by giving an *example*. A real example. Do you think that there is such an f? If so, what is it? Give me a formula for it. >> 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! 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". -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 24 Jun 2006 10:14 mueckenh(a)rz.fh-augsburg.de says... >Such a function exists from |N --> |R for all real numbers which can be >individualized, i.e. which are really real numbers. If such a function exists, then demonstrate it. Give us a function f from N to R such that every real number that is "individualized" is in the image of f. -- Daryl McCullough Ithaca, NY
From: mueckenh on 24 Jun 2006 11:01 mueckenh(a)rz.fh-augsburg.de schrieb: Correction: If |N and P(|N) are potentially infinite, then the definition of a mapping doesn't cause any problem, but it can *not* be shown bijective, because there is no completeness. Regards, WM
From: Virgil on 24 Jun 2006 13:54 In article <1151135731.658456.259240(a)g10g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Cantor said: A well-ordered set remains well-ordered, if finitely many > > >or infinitely many transpositions are executed. > > > > I don't believe that Cantor ever said that. But whether he did or not, > > it's a false statement. > > As false as yours about a well-ordering of *all* rationals. In that case, " mueckenh" is declaring it true, as there are many explicit well-orderings of the set of all rationals. And, in fact, if any set, such as the rationals, can be demonstrated to be countable it is immediately well-orderable by that very demonstration of countability. So either "mueckenh" must be declaring the set of rationals uncountable or he must admit that they can be well-ordered.
From: Virgil on 24 Jun 2006 13:56
In article <1151153250.304679.39720(a)c74g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > And why do you think Galilei was punished by he pope for arguing in > favour of Copernicus? Because Galileo and Copernicus contradicted Aristotles. |