An uncountable countable set
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From: Daryl McCullough on 28 Jun 2006 11:13 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: > >> mueckenh(a)rz.fh-augsburg.de says... >> >I have already shown that any list of all or even some rationals and >> >its diagonal number is uncountable. >> >> No, you didn't, since it is false. >> >> 1. Start with a function f that is a surjection from >> N to Q (Q is the rationals). >> 2. Compute the diagonal d(f). This is a real number that is >> guaranteed to not be in the image f. >> 3. Form a new set M(f) = Q union { d(f) }. >> 4. This new set is *countable* > >like any subset of P(N), No, not like any subset of P(N). Some subsets of P(N) are countable (like M(f)) and some subsets are uncountable. In particular, you are wrong to say that "any list of all or even some rationals and its diagonal number is uncountable" and wrong to say "...like any subset of P(N)". -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 28 Jun 2006 11:21 mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: >> You are just repeating once again that for any function f with >> any domain A, the set K(f) = { x in A | x is not in f(x) } >> is not in the image of f. Yes, we all agree. That means that >> if f is a function from A to P(A), then f is *not* a surjection. >> The impossibility of K(f) being in the image of f *implies* >> the impossibility of there being a surjection from A to P(A). > >No. It shows the impredicable definition of K, but not more. There is nothing impredicative about K. Once again, let A be the following set { 0, 1 } Then P(A) = { {}, {0}, {1}, {0,1} } Let f be a function from A to P(A), for example f(0) = {} f(1) = {1} Then K(f) = { x in A | x is not an element of f(x) }. = { 0 } There is nothing "impredicable" about K(f). It is a definite set, and K(f) is not in the image of f. If A is any set whatsoever, and f is any function whatsoever from A to P(A), then K(f) will be an element of P(A) that is not in the image of f. There is nothing impredicative going on here at all. >It does not stronger imply the impossibility of a surjection from N to >P(N) Yes, it most certainly does. >than it implies the impossibility of a surjection from R to P(N). The difference is that the claim that there is a surjection from N to P(N) is provably false, while the claim that there is a surjection from R to P(N) is provably true. Other than that, they are very similar. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 28 Jun 2006 11:27 mueckenh(a)rz.fh-augsburg.de says... >Daryl McCullough schrieb: > >> mueckenh(a)rz.fh-augsburg.de says... >> >There is a mathematical form containing all real numbers of [0, 1]. In >> >binary representation it is given here. Why not call it a function in >> >an extended sense? >> >> The question is this: does it map the naturals onto the real numbers >> in [0,1] such that every real number is in its image? >> >> In particular, with your representation, which natural number maps >> to the real 1/3? > >It is not necessary to know this natural number (which remains unknown >or does not exist) in oder to prove that the infinite branches of the >tree are countable. If you are claiming that your tree includes all real numbers in [0,1] then yes, it is necessary to know that it contains the real number 1/3. Cantor's claim is that no countable set contains all real numbers. To prove him wrong, you need to demonstrate both (1) that your tree has countably many branches, and (2) that it contains every real number. -- Daryl McCullough Ithaca, NY
From: Virgil on 28 Jun 2006 13:17 In article <1151505096.329945.87930(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Daryl McCullough schrieb: > > > > >> So the answer is no? There is no function f enumerating > > >> all the real numbers? So the reals are uncountable. > > > > > >There is a mathematical form containing all real numbers of [0, 1]. In > > >binary representation it is given here. Why not call it a function in > > >an extended sense? > > > > The question is this: does it map the naturals onto the real numbers > > in [0,1] such that every real number is in its image? > > > > In particular, with your representation, which natural number maps > > to the real 1/3? > > It is not necessary to know this natural number (which remains unknown > or does not exist) in oder to prove that the infinite branches of the > tree are countable. If no natural number maps onto 1/3,, then you do not have a surjection from N ONTO R at all, and your mapping does not show the set of reals to be countable after all. It is equally unneccessary to know which natural might map to 1/3 in order to prove that very same set of branches uncountable, and futhermore, such proofs of uncountability actually exist, whereas no valid "proofs" of countability can exist.
From: Virgil on 28 Jun 2006 13:34
In article <1151505625.358363.7610(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Virgil schrieb: > > > > > > > > >> It is not "Cantor's list" but ANY list. > > >> You are challenged to find a list which lists all reals, then Cantor > > >> presented a rule for showing any list you provide is incomplete. > > > > > >I have already shown that any list of all or even some rationals and > > >its diagonal number is uncountable. > > > > No, you didn't, since it is false. > > > > 1. Start with a function f that is a surjection from > > N to Q (Q is the rationals). > > 2. Compute the diagonal d(f). This is a real number that is > > guaranteed to not be in the image f. But d(f) is NOT guaranteed to be a rational number, and without such guarantee, your "proof" crashes. > > 3. Form a new set M(f) = Q union { d(f) }. > > 4. This new set is *countable* So that now "mueckenh" is claiming that Q is NOT countable but that Q union with some other set IS countable. How does "muecken's logic argue that a subset of an countable set is uncountable? Swallowing camels and straining at gnats is as nothing to the convolutions of "muecken"'s beliefs. |