Cantor Confusion
in [Math]
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From: Virgil on 9 May 2007 14:21 In article <1178716585.114658.276830(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 00:12, Virgil <vir...(a)comcast.net> wrote: > > In article <1178659617.233805.68...(a)e51g2000hsg.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 7 Mai, 21:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 7, 3:19 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > That is not the point here. The point is that there are subsets of > > > > > countable sets which do not allow for a bijection with N. > > > > > > No. > > > > > Why "no"??? > > > > Because subsets of countable sets are countable. > > What means "countable"? 'Countability' is an attribute of some sets. A set is countable when there is a surjective function (in the strictly set theoretic meaning of function) from N to that set/ > > > > > > > > > > There are subsets of a finitely definable set which are not > > > > finitely definitable. > > > > > And they do allow for a bijection with N? > > > > Yup, but those bijections are not finitely defineable. > > There does not exist anyting in mathematics, unless it isfinitely > definable. Perhaps not in WM's mathematics, but we have all seen how lame that is. > In particular, any infinite definition is not a definition, > because there is a definition of "definition" which says: Definition: > A definition has an end, i.e., a last word and a point. Where does that definition of 'definition' occur? Certainly not in any English language nor mathematical dictionary.
From: Virgil on 9 May 2007 14:25 In article <1178716861.173911.234320(a)y5g2000hsa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 00:56, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 8, 5:26 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 7 Mai, 21:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On May 7, 3:19 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > That is not the point here. The point is that there are subsets of > > > > > countable sets which do not allow for a bijection with N. > > > > > > No. > > > > > Why "no"??? > > > > Because you made the statement " there are subsets of > > countable sets which do not allow for a bijection with N." > > > > The "No" was meant to indicate that this statement is false. > > What is true is that there are subsets of a finitely definable > > set which are not finitely definable. The difference is... > > > > WAHH! WAHH! WAHH! I'm not listening. > > > > > > > > > > There are subsets of a finitely definable set which are not > > > > finitely definitable. > > > > > And they do allow for a bijection with N? > > > > Yes, but not a finitely definable one. The difference is... > > > > WAHH! WAHH! WAHH! I'm not listening. > > If you can recover for some time, please take notice: > Either: In mathematics there is nothing existing, unless it is > finitely definable. > Or: There exists a bijection between paths and nodes but it is not > finitely definable. WM has no power to constrain mathematics to what he thinks it should be. That it is not what WM thinks it should be is by now apparent, but that it is quite productive anyway is equally apparent.
From: Virgil on 9 May 2007 14:31 In article <1178717972.186188.186060(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 9 Mai, 01:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > But why do you write down these paths: > > > > > > > Let p'(1) be the path 0111... > > > > > Let p'(2) be the path 00111... > > > > > Let p'(3) be the path 000111... > > > > > > > ? > > > > > > Any set p(1) up to p(n) can be substitued by a single path. > > > > > In fact? Why then did you write down so many useless paths? > > > > > > Look! Over there! A pink elephant! > > > > > > The set P' can be substitued by a single path. > > > > > Oh, does the set P' consist of any paths which are not paths p(n) with > > > a finite argument n? > > > > i: Any set of paths {p(1) ... p(n)) can be replaced by a single > > path. > > Why then did you write the unnecessary paths? > > > > ii: The set P' is the union of sets {p(1) ... p(n)} > > and as such a set of nodes. > > In order to avoid the clumsy expression of path bunches, I call a path > bunch now a finite path. A finite path ends at some node. > > The infinite path {0.000...} is the union of all finite paths with > only 0's, namely {0.} U { 0.0} U {0.00} U ... > The union is one infinite path and, therefore, has not more paths than > were unified. The union of infinitely many path bunches is a single path? > > All nodes of the tree are last nodes of finite paths. > The union over all finite paths is the set of all infinite paths. This > set has not more elements than were unified. Then it is NOT the same 'tree' as a CIBT in ZF or NBG. In any CIBT of ZF or NBG, the set of paths is easily shown to equinumerous with P(N). And WM cannot refute this bijection. And Card(P(N)) > card(N) in ZF or NBG, which also WM cannot refute.
From: Virgil on 9 May 2007 14:47 In article <1178720551.560480.296960(a)y80g2000hsf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > > I did not reject anything. I ask you whether you can put infinitely > many questions. Please answer. If the answer is no, then you should > recognize that you cannot ask whether all paths which accompany > 0.101010... will get separated from it. > > > > I would think there are only finitely many questions. But all this is not > > mathematics but philosphy. > > No. That is mathematics, because you imply to be able to ask > infinitely many question. Since Wm has so often proved himself so lousy a judge of what is mathematics, he is hardly in a position to set up as an expert on where mathematics ends and philosophy takes over. > > > > > > There is a subtle difference. With the diagonal proof we always > > > > remain in > > > > the finite. > > > > > > That's why it fails for numbers with infinitely many digits. > > > > No. > > Assertion, no argument. That is WM way. We have made the arguments, but WM merely ignores them, so now WM complains when we no longer bother to repeat what he has so often ignored! In fact, to establish the difference between the diagonal and any one other only requires finitely many digits, which WM knows but chooses to ignore. > > This formula is obviously only correct for finite sets f nubers WM changes his spots. When he is arguing, he claims that what holds for finite cases also hold in the limit, but when others argue similarly, WM says the opposite. > > Li(x) switch place. Nevertheless, it *is* a natural number > > Are you sure? Look at > > What evidence is there that 2^65536 is a natural number? The principle of induction in ZF or NBG proves that for m and n naturals, so is m^n, at least in ZF and NBG. What goes on in WM's world is up to him, but does not affect anyone else's world, except possibly his students'.
From: WM on 9 May 2007 14:49
On 9 Mai, 17:39, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 9, 11:23 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 9 Mai, 16:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 9, 9:39 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 9 Mai, 01:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > But why do you write down these paths: > > > > > > > > > Let p'(1) be the path 0111... > > > > > > > > Let p'(2) be the path 00111... > > > > > > > > Let p'(3) be the path 000111... > > > > > > > > > ? > > > > > > > > Any set p(1) up to p(n) can be substitued by a single path. > > > > > > > In fact? Why then did you write down so many useless paths? > > > > > > > > Look! Over there! A pink elephant! > > > > > > > > The set P' can be substitued by a single path. > > > > > > > Oh, does the set P' consist of any paths which are not paths p(n) with > > > > > > a finite argument n? > > > > > > i: Any set of paths {p(1) ... p(n)) can be replaced by a single > > > > > path. > > > > > Why then did you write the unnecessary paths? > > > Why then did you write the unnecessary paths? Why then did you write the unnecessary paths? > > > > > > ii: The set P' is the union of sets {p(1) ... p(n)} > > > > > and as such a set of nodes. > > > > > In order to avoid the clumsy expression of path bunches, I call a path > > > > bunch now a finite path. A finite path ends at some node. > > > > > The infinite path {0.000...} is the union of all finite paths with > > > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > > > The union is one infinite path and, therefore, has not more paths than > > > > were unified. > > > > There are only a countable number of finite paths in an infinite path > > > The infinite path {0.000...} is the union of all finite paths with > > only 0's, namely {0.} U { 0.0} U {0.00} U ... > > accepted? > > Yes, but don't let the pink elephants distract you. > The infinite path {0.000...} is a union of finite paths, > or a *set* of nodes. The number of these sets is 1, i.e., less than the number of sets unioned. > > > > > > > All nodes of the tree are last nodes of finite paths. > > > > The union over all finite paths is the set of all infinite paths. > > > > Not without a pink elephant. > > An infinite path is set of nodes. > > A set of infinite paths is a set of sets of nodes The number of these sets is less than the number of sets unioned. Regards, WM |