Cantor Confusion
in [Math]
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From: Virgil on 24 Feb 2007 14:22 In article <1172314973.530995.128190(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Feb., 03:08, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172252004.929693.56...(a)j27g2000cwj.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 23 Feb., 05:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1172134355.420444.86...(a)k78g2000cwa.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > ... > > > > > No? P(oo) = {p(oo), q(oo), r(oo), ...} > > > > > > > > > > p(oo) = Up(n), q(oo) = Uq(m), r(oo) = Ur(i) > > > > > > > > > > How many unions Up(n), Uq(m), Ur(i), ... can be in a union of > > > > > finite > > > > > trees? > > > > > > > > Uncountably many, > > > > > > How can the set of finite subsets of a countable set be uncountable? > > > > In what way is p(oo) a finite subset? > > Not p(oo) but the elements p(n) of which it is made are finite > subsets. Therefore all combinations form a countable set. Non sequitur, as usual. If p(oo) is to contain every path of a complete infinite binary tree, then it is easily proven uncountable. If p(oo) is anything less that the set of all such paths, then it is irrelevant. And, in particular, if WM's version of p(oo) is countable, then it is NOT the set of all paths in a complete infinite binary tree.
From: Virgil on 24 Feb 2007 14:24 In article <1172315095.706877.313610(a)8g2000cwh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > When I last looked at it in the Number > > Field Sieve for factorisation of numbers a big problem was the > > construction of the square root of an irrational algebraic number. > > What is the relation with physics? > > Irrational numbers do not exist, neither in physics nor elsewhere. > > Regards, WM They do for mathematicians. What goes on in the impenetrable jungles of physics is not relevant to what exists for civilized mathematicians.
From: William Hughes on 24 Feb 2007 21:20 On Feb 24, 12:28 pm, mueck...(a)rz.fh-augsburg.de wrote: > On 24 Feb., 00:06, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 23, 5:49 pm, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > >Otherwise you should explain how the > > > > set with n numbers > > > > {2,4,6,...,2n} can be filled up by other (namely greater) even > > > > numbers , without increasing the sizes of numbers. It is similar to > > > > create uncountably many paths in the binary tree without the presence > > > > of uncountably many nodes or edges. > > > > > Both tasks can only be managed by pure belief in these results and > > > > refraining from any attempt to understand it. > > > > As nobody has made either claim, this is not a problem. > > > You should start reacting to what people claim, not > > > what you think follows from the claim. > > > Sorry, I misread the second claim. This claim has in fact been made. > > (no problem, paths are sets of nodes and edges. Saying > > that there are only countably many nodes and edges is > > statement i, saying that a path is composed of nodes > > and edges is statement ii. Saying that there are a countable > > number of paths is statement iii. Statement iii does not follow > > from statements i and ii. Statments i and ii are true. Statement iii > > is false.) > > You misunderstand. Statement iii is not concluded from the other > statements. Reprhasing things in terms of unions of finite paths, rather than subsets of nodes, changes nothing. > Saying that there are only finitely many paths p(n) in the > finite tree T(n) is obviously correct. There are 2^n paths. These > paths are finite sequences. > > The countable union of all finite trees U(T(n)) is a tree which > contains (as subsets) only finite paths Yes. The paths in the infinite tree are the infinite paths which are the unions of finite paths. So we have Statement i There are a countable number of finite paths. Statement ii Each infinite path is made up of finite paths. Statement iii There are a countable number of infinite paths. > as the set of all natural > numbers contains only finite numbers. Don't you agree? Yes. But I don't agree that you can use this to show that the set of all natural numbers has a Wolkenmeukenheim cardinality. Statement i: Every finite number has a Wolkenmeukenheim cardinality. Statement ii: The set of all natural number contains only natural numbers. Statement iii: The set of all natural numbers has a Wolkenmueckenheim cardinality. You do not seem to know any argument other than "X in made of up Y's, so X must have the same properties as the Y's" - William Hughes > > Regards, WM
From: Dik T. Winter on 24 Feb 2007 23:26 In article <1172315095.706877.313610(a)8g2000cwh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > When I last looked at it in the Number > > Field Sieve for factorisation of numbers a big problem was the > > construction of the square root of an irrational algebraic number. > > What is the relation with physics? > > Irrational numbers do not exist, neither in physics nor elsewhere. So the Number Field Sieve does not exist? Strange that is has been able to factorise numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 25 Feb 2007 03:22
On 25 Feb., 03:20, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 24, 12:28 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 24 Feb., 00:06, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > On Feb 23, 5:49 pm, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > >Otherwise you should explain how the > > > > > set with n numbers > > > > > {2,4,6,...,2n} can be filled up by other (namely greater) even > > > > > numbers , without increasing the sizes of numbers. It is similar to > > > > > create uncountably many paths in the binary tree without the presence > > > > > of uncountably many nodes or edges. > > > > > > Both tasks can only be managed by pure belief in these results and > > > > > refraining from any attempt to understand it. > > > > > As nobody has made either claim, this is not a problem. > > > > You should start reacting to what people claim, not > > > > what you think follows from the claim. > > > > Sorry, I misread the second claim. This claim has in fact been made. > > > (no problem, paths are sets of nodes and edges. Saying > > > that there are only countably many nodes and edges is > > > statement i, saying that a path is composed of nodes > > > and edges is statement ii. Saying that there are a countable > > > number of paths is statement iii. Statement iii does not follow > > > from statements i and ii. Statments i and ii are true. Statement iii > > > is false.) > > > You misunderstand. Statement iii is not concluded from the other > > statements. > > Reprhasing things in terms of unions of finite paths, rather > than subsets of nodes, changes nothing. You are in error. I did not rephrase some claim about the set of natural numbers, but here I showed that he set of all real numbers is countable. > > > Saying that there are only finitely many paths p(n) in the > > finite tree T(n) is obviously correct. There are 2^n paths. These > > paths are finite sequences. > > > The countable union of all finite trees U(T(n)) is a tree which > > contains (as subsets) only finite paths > > Yes. > > The paths in the infinite tree are the infinite > paths which are the unions of finite paths. So we have > > Statement i > > There are a countable number of finite paths. > > Statement ii > > Each infinite path is made up of finite paths. You forget: Each infinite path is the union of finite paths. And all these unions of finite paths belong to a countable set of unions. > > Statement iii > > There are a countable number of infinite paths. > > > as the set of all natural > > numbers contains only finite numbers. Don't you agree? > > Yes. But I don't agree that you can use this to > show that the set of all natural numbers has a > cardinality. That is a gross misunderstanding by you. I do not intend to show that the set of natural numbers has a cardinality. And for the last time: If you repeat your personal insults then this was my last answer to you. No further warning will be given! Regards, WM |