An uncountable countable set
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From: mueckenh on 13 Aug 2006 12:27 Dik T. Winter schrieb: > > But an omega-th line, because there is a line omega + 1. > > Neither are true. > Look here: {1, 3, 5,... , 2, 4, 6, ...} There is an element number omega and an element number omega + 1 etc. > > > > The lines o Cantor's list can be enmerated. We do not remain in the > > > > finite there. The edges of the infinite tree can also be enumerated. > > > > See the scheme above. > > > > > > That is proof by example. I do not see how the infinite tree can also > > > be enumerated. Pray, again, explain. > > > > Do you know how the infinite list of Cantor can be enumerated? > > No, of course not. It is not the list of Cantor, but the list provided > to Cantor by somebody else. I hope that the person that provided the > list did know how the list was enumerated. Nonsense. Cantor invented the list and constructed the first one. > > > Do you know how the infinite set of algebraic numbers can be > > enumerated? > > Yes. And why did you believe that the set of edges of my tree was uncountable? Regards, WM
From: mueckenh on 13 Aug 2006 12:29 Dik T. Winter schrieb: > > > > Every n in N is in the list. K is not in the list. > > That is not an answer. Can the number K be indexed or not? No. If it could, then it was in the list. > > > > To be clear, what do you mean with "can be indexed". You > > > are using two different, conflicting, definitions in your articles. > > > > I mean that a natural number can be attributed to every digit position. > > Yes, so can K be indexed or not? And if not, why not? Please use your > definition of indexing. I specifically stated that K was that representation > where each natural number points to a 1, and there are no other digit > positions. So by your *own* definition it can be indexed. But indeed, > it is not in the list (as you admit). Therefore it cannot be indexed. Every digit position which can be indexed is in the true list. Regards, WM
From: mueckenh on 13 Aug 2006 12:34 Dik T. Winter schrieb: > In article <1155314988.698343.137140(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1155067333.259873.193700(a)p79g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > The v. Neumann model of natural numbers is just the model of segments. > > > > n = {0, 1, 2, ..., n-1} > > > > Every natural number is finite. > > > > Hence every union of finite segments is a finite segment. > > > > > > Every finite union of finite segments is a finite segment. Your hence > > > does not follow. Again, a misinterpretation. > > > > Every union of finite segments is a finite segment. "finite" does not > > apply here because there are *infinitely* many finite numbers. > > Just because there are infinitely many finite numbers, the word "finite" > *does* apply. An infinite union of finite segments is *not* necessarily > finite. Where do you find that an infinite union of finite segments is > finite? Prove it. By definition each natural number is finite. As v. Neumann's model gives every natural number but nothing else, every member of the set is finite. > > > > You know that there are not only finitely many numbers, but infinitely > > many. Hence we can unify infinitely many sequences. Nevertheless every > > natural number is finite. > > Yes. But if we unify infinitely many sequences the result is *not* > necessarily finite and *not* necessarily a finite number. Consider the > sequence of finite sets {}, {0}, {0,1}, {0,1,2}, ... ; there are infinitely > many of them. Relevance? Hint: You keep on intermingling number of numbers and sizes of numbers. Every number of the infinitely many numbers is a finite number. That alone is important. Regards, WM
From: mueckenh on 13 Aug 2006 12:37 Virgil schrieb: > In article <1155314675.845888.174190(a)i3g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > So we have infinitely many finite triangles, but without that "final > > > edge", no infinite triangle. > > > > Cantor's list has no final line. Is it finite? > > Since a triangle requires 3 edges, without all 3 it is not a triangle. > An endless list does not require an end, so is "complete" without one. A symmetric rectangular triangle is completely determined by one edge next to the right angle. Regards, WM
From: mueckenh on 13 Aug 2006 12:39
Virgil schrieb: > In article <1155314605.714007.167990(a)i3g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > The line number is infinite by the axiom of infinity. > > Where in the axiom of infinity are infinite number lines mentioned? > > > If no line is numbered as omega, where can we find the line omega + 1? > > Who says there is one? Set theory. > > > > > Thus "mueckenh" is claiming to be able to inject the power set of the > > > naturals into the set of naturals. > > > > > > We should be interested in seeing his attempts to perform this > > > impossibility. > > > > Where does *my arguing* fail? > > Consider the binary tree which has (no finite paths but only) infinite > > paths representing the real numbers between 0 and 1. The edges (like a, > > b, and c below) connect the nodes, i.e., the binary digits. The set of > > edges is countable, because we can enumerate them > > > > 0. > > /a\ > > 0 1 > > /b\c /\ > > 0 1 0 1 > > ............. > > > > Now we set up a relation between paths and edges. Relate edge a to all > > paths which begin with 0.0. Relate edge b to all paths which begin with > > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > > a is inherited by all paths which begin with 0.00, the other half of > > edge a is inherited by all paths which begin with 0.01. Continuing in > > this manner in infinity, we see that every single infinite path is > > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > > other path. > > >The set of paths is uncountable, but as we have seen, it > > contains less elements than the set of edges. > > " What is > true only for finite cases need not be true for infinite ones. What is true up to every level n of the tree is true for the whole tree. Else: What is true for a finite segment of Cantor's list need not be true for the whole infinite list. What is the difference (in concluding from finity to infinity) between list and tree? Regards, WM |