In article <55abc2e2-0ae1-4e8d-b05d-79761a6de727@m19g2000yqk.googlegroups.com> WM <mueckenh@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?
> > > 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. So you assertion that you need to check all lines in order is false...
> > So you agree now that you can check the n-th element of the list before > > checking any previous elements of the list? (How you get at n is irrel= > evant > > here, because getting at n does not involve checking lines preceding the > > n-th line.) > > 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 shown that the line numbered with that n is not in the list. This is true for an arbitrary n. 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.
> > > > 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.
> > Suppose we have the set of paths where each > > path goes to the right after some specific (for that path) node. > > That would imply all paths with tails 111... No problem. > > > Is there > > a node in your tree that is not occupied by any of the paths in that set? > > No, it isn't. Every path node is occupied by at least the head of a > path. > > > Are there possible other paths that are *not* in that set of paths (like > > a path that alternates going left and right)? > > 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.
> > > > Not in this case bacause you apply the words to different things. > > > > *Unless* you assume that what is valid in finite cases also is valid > > > > in infinite cases. But let's see: > > > > sum{i = 0 .. n} 1/(i!) > > > > is a rational number. So according to your logic: > > > > e = sum{i = 0 .. n} 1/(i!) > > > > is also a rational number. > > > > > > I don't see a difference. But I can assuer you, there is no decimal > > > expansion for irrational numbers. > > > > 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. 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?
> > > > What is the relevance? Where am I talking about binary expansions? > > > > > > Binary, decimal, whatever. It does not exist. > > > > I am not talking about whatever expansion. Pray keep to the question. > > According to your logic the complete (i.e. in the limit) sum of rational > > elements is also rational. And so by that logic, e, pi and whatever are > > rational and all numbers we do use are rational. > > No. The limit is not rational. But the limit cannot be represented by > a sequence.
We have: sum{i = 0 .. n} 1/(i!) is rational for each n. By *your* reasoning so also sum{i = 0 .. oo} 1/(i!) is rational, and so e is rational. You use that reasoning when you state union{i = 0 .. n} FISON(i) is a FISON and so union{i = 0 .. oo} FISON(i) os a FISON. I see no reason why you get at the conclusion in the second case but not in the first case. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
