On 19 Jun., 16:35, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <55abc2e2-0ae1-4e8d-b05d-79761a6de...@m19g2000yqk.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 12 Jun., 04:54, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > ... > > > > > > > Because you do not check the lines in order. It is always > > > > > > > your basic assumption that you first check the first line > > > > > > > and after that the next line. That is wrong. > > > > > > > > > > > > That is necessary because you cannot find the n-th line unless > > > > > > you know the line number n - 1 or some equivalent mark. > > > > > > But why is getting the number n in any way related to the checking of the > > > previous lines? > > > > Because by blind choice you cannot be sure to hit what you want. > > Why does it imply checking the previous lines? Why do you not answer that > question?
You cannot check line n without knowing line n-1. > > > > > But you cannot find number n without referring to the numbers less > > > > than n. > > > > > > That does not mean that you check the line with number n. So you > > > assertion that you need to check all lines in order is false. > > > > In unary or in von Neumann's representation there is the whole > > counting process incorporated in each number. In decimal, you need > > only count the powers of 10. Anyhow addressing a number means > > counting. > > And that *still* does not imply checking.
Counting is checking: how many.
> > Of course you can check the n-th element before checking any other. > > But in order to find the n-th element, you have to count from 1 to n, > > either in unary or at least in the powers of 10. Therefore it is not > > correct to talk about simultaneity. > > 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.
> 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.
> 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? > > > > > > Perhaps right, depends on how you actually do define things. But > > > > > are there nodes mapped to all paths? That is what you assert. > > > > > > > > Unless there is at least one node occupied by the path p_n that is not > > > > occupied by the paths p_0 to p_n-1, p_n is not a new path. > > > > > > Assuming countability again. > > > > Countability of nodes only. I am only interested to cover every node, > > i.e., to exhaust the tree. I am a follower of Eudoxos. > > You have to exhaust the paths, not the nodes. If you think that exhausting > the nodes implies exhausting the paths you are assuming that the set of paths > is countable.
No, I am showing that the set of paths is counatble.
> > 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.
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. >
> > > Where am I talking about decimal expansions? According to you the > > > in finite sum above (which equals e by definition) is rational. Is > > > that true? > > > > True is: There are only rational sums of decimal or binary expansions. > > That is (again) not an answer to my question. By your logic e is a rational > number. No, the contrary can be proven. e is not rational, but there is no binary sequence for e. There are only approximations. But they are not suitable to occur in Cantor's list.
> Because that is what your logic dictates. That is, every sum of a > finite initial set of the summands is rational, and so according to your logic > also the complete sum is rational. Do you indeed think e is rational?
No. See above. Do you indeed think that you can communicate e to some person by means of a bit sequence?
