In article <email@example.com> WM <firstname.lastname@example.org> writes: > On 19 Jun., 15:42, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > If every terminating path of the form 111...1000... is used, then > > > every node is covered by the last 1. > > > > But that still does not show that all nodes are covered by a countable set > > of paths. It only shows that every node is covered by a set of paths that > > contains a countable subset, not that the set itself is countable. > > It is easy to construct a bijection between a countable set of paths > and all nodes. These nodes can even be defined by the paths that lead > to them. That shows that all nodes of the tree can be covered by a > countable set of paths.
Yes, they can be covered by a countable set of paths. Nothing more. It does *not* show a bijection, and not a construction of a bijection.
> > Yes, they are in the tree, so you have to prove that they are also used in > > the covering, which you have not done. You do not have to show that each > > node is covered, you have to show that with your covering you use each > > path. You did show the former. > > If every node is covered, then there remains no path to be covered.
What do you mean with "covering a path"? Until now you were talking about "covering nodes". And whatever it may mean, in what way does "every node is covered" show that there remains "no path to be covered"?
> > > > Or can you find a node that is *not* covered by a path that after > > > > some node to the right? Can you find a node that is covered by a > > > > path that alternates going left and right that is not covered by a > > > > path that after some node always goes to the right? > > > > > > I see that every such alternating path with all its nodes is in the > > > tree, after construction. > > > > This is not an answer to my question. The path is in the tree, but does > > it contain a node that is not covered by a path that after some node > > always goes to the right? > > No. From that we can obtain that the number 1/3 is also in a list of > terminating rationals. There does not exist a sequence 0.010101... > that is longer than *every* finite sequence of that form.
Ah, so now you contend that 1/3 is a terminating rational. As the sequence of terminating rationals starting with that is: 1/4, 5/16, 21/64, 84/256, ... which of those is 1/3?
> > > We can state: The number of nodes is countable. The number of paths > > > required to cover all nodes is countable too. > > > > Right. But the number of required paths is the minimal number of paths > > needed to cover all the nodes. That is *not* the number of all paths. > > If all nodes are covered then all path that can be distinguished by > nodes, i.e., all reals that can be distinguished by digits, are there.
What do you *mean* with "all path that can be distinguished by nodes"? I see two meanings: (1) All paths that can be distinguished by a node from each other path (2) All paths that can be distinguished by a node from all other paths. As the covering of all nodes obviously does not imply (1), you must mean (2). But there is no path in the set of all paths that can be distinguished by a node from all other paths. So clearly that is not "all paths".
> > Again: > > consider the set of paths where each path after some point always goes to > > the right. This set of paths covers each node. Consider the path that > > alternates going right and left continuously. Can you point to a node in > > this alternating path that is not covered by one of the paths in the > > earlier set? But you did agree slightly higher up that that path was in > > the tree. > > All parts of that path that really exist, really are in the tree.
That is not an answer to my question. Moreover you have not defined "really exist".
> > > That is a ridiculous requirement in an informal discussion, in > > > particular if not mathematics is concerned. > > > > Oh. So in an informal discussion you can tell nonsense whatever you like > > and when challenged you do not need to back it up? > > I have read that there is a dogma of the catholic church stating that > the existence of God can be proved by means of scientific reasoning. I > do not remember the source. Does that turn this information into > nonsense? (Except that its contents is nonsense.)
Clearly it is not backed-up by facts, so it can be nonsense or not, I have no way of knowing. It might be true, if might be false, the author may have been wrong, you can remember wrong. But whatever, it is not an argument in a sensible discussion. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/