constructible = unconstructible
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From: WM on 9 Jun 2010 10:26 On 9 Jun., 16:11, William Hughes <wpihug...(a)hotmail.com> wrote: > > Ok, you do not have a constuctive proof but you do > claim a nonconstuctive proof. Say it so: If my non-constructive proof fails, then Cantor's proof fails too. > [...] Attempt do change the topic - denied. The topic is: 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. > "Cantor's > complete list" does not have a last number and the complete > tree does not have a last node. That's the reason why I cannot show you the node that is missing in the above covering. Regards, WM
From: William Hughes on 9 Jun 2010 10:32 Does the set of nodes Z= { 1 11 111 ... } contain every node in 111...? Please begin your answer yes or no. - William Hughes
From: WM on 9 Jun 2010 10:54 On 9 Jun., 16:32, William Hughes <wpihug...(a)hotmail.com> wrote: > Does the set of nodes > Z= > { > 1 > 11 > 111 > ... > > } > > contain every node in 111...? > Please begin your answer yes or no. Of course I could answer this question by: If yes, then Cantor's proof fails by its argument. If no, then my Binary Tree proves that Cantor's proof fails by its result. But I don't want this change to a second step before I see that you have understood the first step: If all paths of the form 111...111000... have been covered, then the path 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: William Hughes on 9 Jun 2010 11:01 On Jun 9, 11:54 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 9 Jun., 16:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > > Does the set of nodes > > Z= > > { > > 1 > > 11 > > 111 > > ... > > > } > > > contain every node in 111...? > > Please begin your answer yes or no. > <snip refusal to answer question> Please begin your answer yes or no. > But I don't want this change to a second step before I see that you > have understood the first step: > > If all paths of the form 111...111000... have been covered, then the > path 111... has not yet been covered. Nope. 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? Yes. The anti diagonal is *covered* by the set Y but is not *in* the set Y. When you write down every node in the paths of Y you write down many paths that are not in Y. - William Hughes
From: WM on 9 Jun 2010 11:17
On 9 Jun., 17:01, William Hughes <wpihug...(a)hotmail.com> wrote: > > > 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? > > Yes. The anti diagonal is *covered* by the set Y but is > not *in* the set Y. That is wrong by the linearity of the set. If all 1's of 111... are in the list, then either all can be in one line or they cannot be in one line. Agreed? If they are in one line, then 111... is in the list. If they cannot be in one line, then there must be at least two lines containing all 1's. That means, there must be at leat two 1's of 111... that cannot be in one common line. And that is obviously wrong. So, if there are "all" 1's of 111... in the list, then they are in one line of this list. QED. Regards, WM |