An uncountable countable set
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From: Virgil on 24 Jun 2006 13:59 In article <1151153340.765203.100150(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dann Corbit schrieb: > > > "Richard Tobin" <richard(a)cogsci.ed.ac.uk> wrote in message > > > > > >>>What (exactly) is the place in the Bible that contradicts anything said > > >>>by > > >>Copernicus? > > > > > > Joshua 10:13 describes a miraculous event in which Joshua commands the > > > sun to stand still, with the result that the day continues until the > > > children of Israel had avenged themselves upon their enemies. This, > > > it was argued, shows that the cycle of day and night is normally > > > caused by the movement of the sun. > > > > It shows no such thing. The sun moves in the sky (visually). He was just > > asking for extended sunshine. > > But if the shadow on the sundial run back, he would have been flung > from the surface of the earth by centrifugal forces. If an almighty God cold not arrange to have the sun appear to stand still in the sky without disastrous side effects, such a God would hardly be almighty.
From: Virgil on 24 Jun 2006 14:03 In article <1151153501.336865.131980(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > In order to be unambiguous transpositions must be carried out > > sequentially, meaning that only countably many can be applied. > > Not more is required. > > > > Mueckenh's analysis is not unambiguously a sequence of transpositions. > > > > Also he does not show that any single transposition can re-order a set > > having no first member into one which does have a first member. > > Of course that is impossible. It is just showing a contradiction in set > theory. In "mueckenh"'s set theory , perhaps, but not in anyone else's. Until "mueckenh" can show that a SINGLE transposition can switch an ordered set with a first element with one not having a first element, his alleged proof fails. > > > > Similarly in the opposite direction, no single transposition will cause > > a set with a first element to become one without a first element. > > Of course that is impossible. It is just showing that there is nothing > like an actually infinite set. Wrong! It is just showing that "mueckenh"'s "proof" is invalid. > > Regards, WM
From: Virgil on 24 Jun 2006 14:09 In article <1151153602.449685.154410(a)u72g2000cwu.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > By definition, a set M is countable if and only if there SOME surjection > > g:N --> M exists and is uncountable if no such surjection can exist. > > > > This means that, given a set M, one must allow ANY function from N to M > > to be considered. > > > > So that given that M(f) exists at all, one is not constrained to only > > the function f which defines K(f) and M(f), but can consider whatever > > functions imaginable from N to M(f). > > > > Since it is easy to construct surjections from N to the set of all > > finite subset os P(N), it is also easy to do it from N to M(F). > > Such a function exists from |N --> |R for all real numbers which can be > individualized, i.e. which are really real numbers. What real numbers are not really real numbers? In the Dedekind model, every set of rationals that is bounded below determines a real number (or bounded above, for that matter). In the Cauchy sequence model, every Cauchy seqence of rationals determines a real number. Which of the real numbers so determined are not really real numbers?
From: Virgil on 24 Jun 2006 14:14 In article <1151153687.248658.128340(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > 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! > > > > > > Regards, WM > > > > > > Mueckenh is WRONG! > > > > If surjectivity is not required, all sorts of f's. can exist! > > > > Consider the following: > > f: N --> P(N) : f(n) = {} for all n in N. > > Then the image of f is {{}} > > Then K(f) = {x in N:x not in f(x)} = N, which shows f is not a > > surjection, but can exist if not required to be surjective. > > > > So there IS a mapping in this case, and Mueckenh is wrong again. > > > So drop this assumption [surjectivity]. There is no mapping [including > K as the image of at least one source element], even if surjectivity is > not at all in question! No one is claiming that there is any mapping, f, with K(f) in it image. What everyone but "mueckenh" is saying is that of f:N --> H is any mapping from N to the set of finite subsets of N, then there is a bijection other than f from N to M = (H union {F(k)}).
From: Virgil on 24 Jun 2006 14:20
In article <1151153737.788248.133080(a)y41g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Russell schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > [snip] > > > > > > 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! > > > > There is *no* mapping? No mapping at all? How can that > > be, if P(N) is a set with at least one member? Please tell > > us more about this interesting claim of yours. > > There is no mapping including K as the image of at least one source > element, even if surjectivity is not at all in question! That is not the issue in question. "Mueckenh" has claimed that he has produced an example of a set that is both countable and uncountable (See Subject), but his example is flawed because the sets and function he has presented do not exist as he has presented them. Further, If his restrictions are eased to the point that the function and sets CAN exist, by simply not allowing K(f) to be a value of f,there is no difficulty in showing both the sets involved are countable. |