In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> > But the whole point is that in Cantor's argument, whatever n we choose, it > > is > > a last and largest member of a finite set.
Also the first and smallest member of an infinite set. > > > shown that the line numbered with that n is not in the list. This is > > true for an arbitrary n > > that is a last and largest member of a finite set. Also the first and smallest member of an infinite set. So what? > > > There is no specific line that is checked. And > > how we come to the n-th line is also irrelevant. For any n, the n-th line > > is not equal to the diagonal. > > And for any n, without counting, the next line is equal to the > diagonal in this example: > > 0.0 > 0.1 > 0.11 > 0.111 > ... > > Why do you think that this is not contradicting Cantor?
Because Cantor's diagonal proof only refers to infinite binary sequences, which what WM listed are NOT, and any infinite binary sequence is automatically different from every finite sequence such as the ones WM has listed.
> > No, I am showing that the set of paths is counatble.
Not in this world! Your set of FINITE paths is countable, but its countability is irrelevant to the countability of a set of paths that it is disjoint from. > > > > It is said so. But there is no possibility, after having completed the > > > construction, to introduce such a path. They have sneaked in. > > > > And so paths have sneaked in that were not introduced by the construction, > > and so when you use the paths given in the construction to map to nodes > > you do not show that the set of paths is countable, only that the subset of > > paths that you used in the construction is countable. There are other > > paths, > > that have sneaked in, that are not mapped to a node, and can not be mapped > > to a node to which no path has yet been mapped. So you have not proven the > > the set of paths is countable. > > If you believe in sneeking (Virgil claimed that they say write it with > double e) then you believe in ghosts. I do not.
How is it creating a ghost to note that an infinite binary string with infinitely 1's is not in any set of infinite binary strings each having at most finitely many 1's?
Wm is becoming delusional again. Or is it "still"? > > Constructing the tree from the list is nothing but (or at least can be > done by) writing some digits or bits in a more economical way. This > does not increase the number of sequences or paths.
Constructing a tree requires constructing its node set together with the appropriate "left child", "right child" and "parent of" relations.
Paths are merely a consequence of that construction.
For all finite complete binary trees, any set of paths whose union is the nodes set of that tree must have a path for each leaf node. Since the set of paths bijects in an obvious way with the set of leaf nodes, one has to have every path in order to get every node.
In maximal infinite binary trees, there are no leaf nodes, and there is no bijection between the set of all paths and any set of nodes.