On 21 Jul., 23:27, Virgil <vir...@ligriv.com> wrote: > In article > <2fccc922-5033-47e4-8ca2-8acb708ee...@t9g2000vbs.googlegroups.com>, > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 20 Jul., 21:26, MoeBlee <modem...@gmail.com> wrote: > > > On Jul 20, 1:08 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 20 Jul., 18:45, MoeBlee <modem...@gmail.com> wrote: > > > > > > On Jul 20, 2:23 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > That means: We can check every B_k for the presence of a path P_a with > > > > > > negative result. > > > > > > Please put "we can check for a negative result" in rigorous > > > > > mathematical terminology. > > > > > There exists a node P_a_j not element of B_k. > > > > WM clings to his "P" like a baby to its worn out doll. > > > > > You should be able to find such simple gacts by yourself. > > > > For any k, there exists a node that is not an element of B_k. > > > > That's a fact I figured out long before I met WM. > > > > > > But, of course, for any k>0, there is an infinite path whose range > > > > > differs from at least one member of B_k. > > > > > > > Nevertheless uncountably many P_a are in the limit. > > > > > > There's no need to say "nevertheless". > > > > > > For any k>0, there is an infinite path whose range differs from at > > > > > least one member of B_k > > > > > For any k, even k = 0, there are uncountably many paths, each of which > > > > has infinitely many nodes that are not in B_k. > > > > The set of nodes is the set of natural numbers. > > > > It is easy to see that each B_k is a set of nodes, viz. the set of > > > natural numbers less than or equal to k. > > > > It is easy to see that the set of nodes for the tree is U[k in w]B_k, > > > which is the set of natural numbers. > > > > A path p is a sequence of nodes such that, for all n, we have p(n+1) > > > is a child of p(n). > > > > For each infinite path p there corresponds a denumerable binary > > > sequence s as follows: > > > > s(0) = 0 > > > > s(n+1) = 0 if p(n+1) is a left child of p(n), and s(n+1) = 1 if p(n+1) > > > is a right child of p(n). > > > > That provides a bijection between the set of paths and the set of > > > denumerable binary sequences whose first term is 0. > > > > A path "has nodes" in the sense that the members of the range of the > > > path are nodes. > > > > But 0 is in the range of every COMPLETE path (where a complete path is > > > an infinite path p such that p(0) = 0, i.e., the path "starts" at the > > > root, which is 0). > > > > And, yes, for any k, there are uncountably many complete paths p such > > > that there are infinitely many nodes in the range of p that are not in > > > B_k. > > > > > It is not only the > > > > range that fails to be inside any B_k, but also the nodes, i. e. a > > > > value 0 or 1 with index j in N. > > > > See above. > > > > > > and > > > > > > There are uncountably many paths for the tree. > > > > > > and > > > > > > The set of nodes of the tree is U[b in w]B_k. > > > > > Certainly you mean U[k in w]B_k. > > > > Right, that was a typo. > > > > > > And the contradiction that WM claims to come from any of this > > > > > is .... ? > > > > > Cantor's list is U[k in w]L_k. > > > > What "Cantor's list"? > > > > 'L' has been used in various contexts in these discussions. What is > > > 'L' being used for by WM here? > > > > Maybe WM is referring to this argument in set theory: > > > > Let L be a countable sequence of denumerable binary sequences. > > > > Let G be the denumerable binary sequence defined by > > > > G(n) = 0 if L(n)(n) = 1, and G(n) = 1 if L(n)(n) = 0 > > > > (G is the "anti-diagonal for L".) > > > > It's easy to see that G is not in the range of L. > > > Let (B_k) be the countable sequence of finite initial segments of the > > well known Binary Tree. > > > Let P_a be any denumerable binary sequence. > > > Then P_a_k is not an element of B_k. > > But there is a k' such that P_a_k IS an element of B_k' > > In fact, once P_a_k is an element of B_k', it is also the case that > for any k" > k', P_a_k is an element of B_k' as well. > > > > > It is easy to see that P_a is not in UB_k. > > It may be easy for WM to see that, but it happens to be false, as are so > many of the things that WM finds it so easy to see. > > For all a and for all k there is some k' such that P_a_k is in UB_k'.
If your reasoning were right, then the anti-diagonal 0.111... of the list 0.0 0.1 0.11 0.111 ... was in the list. And again Cantor would fail.
