constructible = unconstructible
in [Math]
|
Prev: "NO BOX CONTAINS THE BOX NUMBERS THAT DON'T CONTAIN THEIR OWN BOX NUMBER" ~ XEN
Next: manipulating spectral norm of matrix with normalized columns
From: WM on 10 Jun 2010 11:07 On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 10, 11:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 10 Jun., 16:00, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On Jun 10, 10:33 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 10 Jun., 14:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On Jun 9, 10:37 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > 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 numbers is > > > > > > computable. > > > > > > The set of all names, S is computable. > > > It is well known that the set of computable numbers > > > is countable (though not computable). > > <important context restored> > > The set of all computable numbers, R is a subset of S. > A subset of a computable set may or may not be computable. > R is not computable. > > > And what is the real part or numerical value of an uncomputable real > > number? > > Something that cannot be computed. If you take the > position that something that cannot be computed > does not exist There is no position to take. If something that does only exist by computing it, cannot be computed, then it does not exist. That is not a position but is simply the truth. > then neither an uncomputable number > nor a list of all computable numbers exist. I showed you a computed list that contains all computable numbers (and some other names). So yout assertion is false. In addition: Cantor's diagonal numbers are computed numbers. Therefore it is meaningless to defend his false conclusion by uncomputable numbers. Cantor's "proof" proves the uncountability of a countable set. Whether or not a list of those numbers can be computed is irrelevant because we know it is countable. Regards, WM
From: William Hughes on 10 Jun 2010 11:20 On Jun 10, 12:07 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: <snip> > > then neither an uncomputable number > > nor a list of all computable numbers exist. > > I showed you a computed list that contains all computable numbers (and > some other names). So yout assertion is false. You did not show me a list that contains all computable numbers and only computable numbers. Showing a computable list that contains "all computable numbers (and some other names)" is not enough to demonstrate the existence of a computable list that contains all computable numbers and only computable numbers. - William Hughes
From: WM on 10 Jun 2010 10:55 On 10 Jun., 16:41, William Hughes <wpihug...(a)hotmail.com> wrote: > On Jun 10, 11:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 10 Jun., 16:00, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On Jun 10, 10:33 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 10 Jun., 14:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On Jun 9, 10:37 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > 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 numbers is > > > > > > computable. > > > > > > The set of all names, S is computable. > > > It is well known that the set of computable numbers > > > is countable (though not computable). > > <important context restored> > > The set of all computable numbers, R is a subset of S. > A subset of a computable set may or may not be computable. > R is not computable. > > > And what is the real part or numerical value of an uncomputable real > > number? > > Something that cannot be computed. If you take the > position that something that cannot be computed > does not exist Something that can only exist as a computed entity does not exist, of course, when it cannot be computed. That is not a position, but that is simply the truth. > then neither an uncomputable number > nor a list of all computable numbers exist. I showed you the list that contains all computable numbers (and some more names). So your claim is wrong. In addition: Cantor's diagonal proof constructs computable numbers. It is not meaningful to defend this proof by means of uncomputable numbers. They, if assumed to exist, have nothing to do with the incorrect claim obtained from the diagonal argument. Regards, WM
From: Virgil on 10 Jun 2010 16:34 In article <aff416f4-4582-425c-ae0e-61c49cadf0c4(a)j4g2000yqh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > A lesson for set theorists > > The complete infinite binary tree contains, by definition, all real > numbers between 0 and 1 as infinite paths, i. e., as infinite > sequences { 0, 1 }^N of bits. And it 'contains' all the binary rationals in that interval twice. > > 0, > / \ > 0 1 > / \ / \ > 0 1 0 1 > / > 0 ... > > The set { a_k | k in N } of nodes a_k of the tree is countable. > > a0, > / \ > a1 a2 > / \ / \ > a3 a4 a5 a6 > / > a7 ... > > If every node a_k is covered by an arbitrary infinite path p_k > containing that node, then no further node is remaining to distinguish > any further path from the paths p_k of the countable set P = { p_k | k > in N }. This proves that it is impossible to distinguish more than > countably many paths by infinite sequences of bits. It does no such thing. In the first place, every node is "covered by" as many paths as exist in the entire tree, since the subtree having that node as root is tree-isomorphic to the whole tree. Secondly, to "distinguish" any one path from all others by means of nodes alone requires a set of infinitely many nodes from the path to be distinguished, since for any finite number of nodes, either no path at all contains all of them or uncountably many paths contain all of them. > FIRST COUNTER ARGUMENT > > All nodes of a path are covered by other paths. Therefore covering all > nodes of the binary tree by a countable set of paths does not show > that all paths of the binary tree form a countable set. All nodes can be "covered" by a countable set of "finite paths", but each such finite path generates from its terminal node a tree isomorphic to the original tree, so no reduction has been achieved. > > Here is an example for the counter argument: We can cover every node > of path p = 0.111... by means of covering every path p# of the set > { > 0.000... > 0.1000... > 0.11000... > 0.111000... > ...} Covering nodes is not enough, as has been shown above. > > SECOND COUNTER ARGUMENT > > In order to save Cantor's diagonal proof, it must be claimed that > although all nodes of p are covered by the set of paths p#, there is > no path p# = p. Since and set of paths containing more than one path, if p# is such a set, it is not a path and ther is no path p# = P as desired. Not that WMs argument has any merit anyway. > Proof A. If the existence in different lines is claimed What does that mean in English? What are "lines" and what does it take for "the existence in different lines"? Invalid until "lines" are properly defined. > CAN N BE CONSTRUCTED? > > If this question is denied, then it is impossible to construct a > complete Cantor list It is Cantor's theorem that it is impossible to construct a complete "Cantor's list".
From: Virgil on 10 Jun 2010 16:49
In article <9ebeaa10-425a-480e-856f-8f09a5912d87(a)e5g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Jun., 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Jun 10, 7:34�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > <snip> > > > > > if we divide 1 by 9, then 0.111... does exist according > > > to the above definition. > > > If we write it in a single line, nobody will disagree. > > > If we write it such that every next 1 is placed in a new line, like > > > > > 0.1 > > > 0.11 > > > 0.111 > > > ... > > > > > then everybody will disagree that all 1 are in one line but all 1 > > > "must be there somehow". > > > > Yes and outside Wolkenmuekenheim, everyone is correct. > > However inside Wolenmuekenheim there is a 1 in 0.111... > > that is not in the list. > > What is flaming good for? My proof uses complete induction. One can > believe in it or not. One with any sense chooses not to. But I believe that it is valid for every line of > the list. Therefore I do not believe that your arguing "all 1's are in > the list but not all 1's are in the same line" is wrong, namely > disproved by induction. > > And if you think in logocal terms and understand that induction holds > for every natural number, here every line, then you must agree. > Sometimes people say that induction does not hold for the complete set > N. But that is not at all the question here. Here we have to prove > that every line contains all 1's of its predecessors. > > > > > This leads to the easily disprovable claim > > > that there are at least two lines required to contain all 1's. > > > > Nope, it leads to the easily provable claim that a > > set of lines, S, that contains all nodes must > > contain an infinite number of lines. > > > Nonsense. If someone claims that infinitely many are necessary, > instead a single one, then he should know at least two of them. It is WM spewing the nonsense. For the sequence .1, .11, .111, ..., at least 2 of them are required to contain all 1's(meaning a 1 in each possible position) but 2 are not nearly enough. It takes an infinite set of them ( and ANY infinite set of them suffices). > > > �Since any set > > of infinite lines is sufficient, and there exist disjoint > > sets of infinite lines, it follows immediately that > > no two specific lines are required. > > It follows immediately by induction, that no two lines are required. Actually, one may eliminate infinitely many "lines" provided one also leaves infinitely many, and still have for each "position' at least one line with a '1' in that position (in fact infinitely many such lines with a 1 in that position. However, for any finite set of such lines, there will be a position (in fact infinitely many of them) for which no line of that set has a 1 in that position. > Please excuse if I trust in complete induction more than in your shaky > assertions. Please excuse if we trust our logic more than your flakey arguments. |