constructible = unconstructible
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From: WM on 9 Jun 2010 09:28 On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 9, 9:55 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 9 Jun., 13:58, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > Can you distinguish the path > > > > x= 111... > > > > from every path in the set of paths > > > > Y={ > > > > 000... > > > 1000... > > > 11000... > > > 111000... > > > ... > > > > } > > > Yes, we can! Wouldn't your question have been rather nonsensical if I > > could not? Of course we, me and you, can distinguish those paths > > because you used finite definitions like "xyz..." to name them. > > > Can you distinguish by "xyz..." more than countably many paths? > > > Please begin your answer yes or no. > > No. (I interpret your question as "Can you distinguish more > than countably many paths using finite descriptions") > It is well known that the set of finite descriptions > is countable (though not computable). Of course. Therefore it is impossible to distinguish more than countably many real numbers. > > Does the set of paths Y cover every node in the tree? > > Please begin your answer yes or no. No. The paths of the set Y do not even cover every node of the path 111... (I assume that was the correct meaning of your question). Otherwise ... But let me explain it in detail: Every path of the form 111...111000... has a 0 at a position where 111... has a 1. So when covering path 111... none of the paths 111...111000... has been covered, has it? That means for every path of the form 111...111000... there remains a node to be covered after all nodes of 111... have been covered. But my original covering of the binary tree covers all its nodes - by definition. Now I turn to your question: If all paths of the form 0,111...111000... have been covered, then the path 0,111... has not yet been covered. Otherwise the possible antidiagonal 111... of the list 000... 1000... 11000... 111000... .... would have been written down (covered) by writing (covering) all paths of the form 111...111000... of the set Y. Can that happen? If yes, then Cantor's proof is inconclusive. If no, my answer stands. Regards, WM
From: WM on 9 Jun 2010 09:37 On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > It is well known that the set of finite descriptions > is countable (though not computable). The set of all names which unavoidably include all nunbers is computable. Here it is: 0 1 00 01 10 11 000 001 .... It is simply wrong to maintain that it was uncomputable. Above you see a list of all possible symbols (encoded binary), all possible words and all possible symbols all possible languages including all possible dictionaries and all possible encyclopedias (they come somewhat later in the list). Nevertheless this list has no diagonal and, therefore, it has no antidiagonal - like many real lists. Regards, WM
From: William Hughes on 9 Jun 2010 09:50 On Jun 9, 10:28 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Jun 9, 9:55 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Jun., 13:58, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > Can you distinguish the path > > > > > x= 111... > > > > > from every path in the set of paths > > > > > Y={ > > > > > 000... > > > > 1000... > > > > 11000... > > > > 111000... > > > > ... > > > > > } > > > > Yes, we can! Wouldn't your question have been rather nonsensical if I > > > could not? Of course we, me and you, can distinguish those paths > > > because you used finite definitions like "xyz..." to name them. > > > > Can you distinguish by "xyz..." more than countably many paths? > > > > Please begin your answer yes or no. > > > No. (I interpret your question as "Can you distinguish more > > than countably many paths using finite descriptions") > > It is well known that the set of finite descriptions > > is countable (though not computable). > > Of course. Therefore it is impossible to distinguish more than > countably many real numbers. > > > > > Does the set of paths Y cover every node in the tree? > > > Please begin your answer yes or no. > > No. The paths of the set Y do not even cover every node of the path > 111... Please name one node in the tree that is not covered by a path from Y. - William Hughes
From: WM on 9 Jun 2010 10:00 On 9 Jun., 15:50, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 9, 10:28 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On Jun 9, 9:55 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 9 Jun., 13:58, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > Can you distinguish the path > > > > > > x= 111... > > > > > > from every path in the set of paths > > > > > > Y={ > > > > > > 000... > > > > > 1000... > > > > > 11000... > > > > > 111000... > > > > > ... > > > > > > } > > > > > Yes, we can! Wouldn't your question have been rather nonsensical if I > > > > could not? Of course we, me and you, can distinguish those paths > > > > because you used finite definitions like "xyz..." to name them. > > > > > Can you distinguish by "xyz..." more than countably many paths? > > > > > Please begin your answer yes or no. > > > > No. (I interpret your question as "Can you distinguish more > > > than countably many paths using finite descriptions") > > > It is well known that the set of finite descriptions > > > is countable (though not computable). > > > Of course. Therefore it is impossible to distinguish more than > > countably many real numbers. > > > > Does the set of paths Y cover every node in the tree? > > > > Please begin your answer yes or no. > > > No. The paths of the set Y do not even cover every node of the path > > 111... > > Please name one node in the tree that is not > covered by a path from Y. Do we play kindergarden here or do we do mathematics? I proved the existence of such a node. There are many proofs without construction of the proved. Can you give a well-ordering of the reals? Please look at the consequences of non-existence of such a node for Cantor's diagonal argument. That's decisive! But OK, lets make a compromise. Name the last number of Cantor's complete list, then I will name the last node of my complete tree. Regards, WM
From: William Hughes on 9 Jun 2010 10:11
On Jun 9, 11:00 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Jun., 15:50, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Jun 9, 10:28 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 9 Jun., 15:19, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Jun 9, 9:55 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 9 Jun., 13:58, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > Can you distinguish the path > > > > > > > x= 111... > > > > > > > from every path in the set of paths > > > > > > > Y={ > > > > > > > 000... > > > > > > 1000... > > > > > > 11000... > > > > > > 111000... > > > > > > ... > > > > > > > } > > > > > > Yes, we can! Wouldn't your question have been rather nonsensical if I > > > > > could not? Of course we, me and you, can distinguish those paths > > > > > because you used finite definitions like "xyz..." to name them. > > > > > > Can you distinguish by "xyz..." more than countably many paths? > > > > > > Please begin your answer yes or no. > > > > > No. (I interpret your question as "Can you distinguish more > > > > than countably many paths using finite descriptions") > > > > It is well known that the set of finite descriptions > > > > is countable (though not computable). > > > > Of course. Therefore it is impossible to distinguish more than > > > countably many real numbers. > > > > > Does the set of paths Y cover every node in the tree? > > > > > Please begin your answer yes or no. > > > > No. The paths of the set Y do not even cover every node of the path > > > 111... > > > Please name one node in the tree that is not > > covered by a path from Y. > > Do we play kindergarden here or do we do mathematics? I proved the > existence of such a node. Ok, you do not have a constuctive proof but you do claim a nonconstuctive proof. Does the set of nodes Z= { 1 11 111 ... } contain every node in x? Please begin your answer yes or no. > There are many proofs without construction > of the proved. Can you give a well-ordering of the reals? Please look > at the consequences of non-existence of such a node for Cantor's > diagonal argument. That's decisive! > > But OK, lets make a compromise. Name the last number of Cantor's > complete list, then I will name the last node of my complete tree. > You have been skipping your exercises, "Cantor's complete list" does not have a last number and the complete tree does not have a last node. - William Hughes > Regards, WM |