In article <046c5a1d-9623-406f-88f4-b79b4447dcfc@r34g2000vba.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 19 Jun., 16:35, "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 why does that imply *checking* line n-1? 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. > > Counting is checking: how many.
Aha, now we come closer. So when you state that checking line n in the list implies checking previous lines, you are using the word "checking" with two different meanings. In the first instance you are meaning that it is checked whether it is different from the diagonal. In the second instance you are meaning that you count to that line. Why do you use the word "checking" the second time with a meaning that is not related to the standard meaning of checking?
> > > 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.
Irrelevant.
> > 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.
Irrelevant.
> > 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 > ...
Eh? The diagonal is 0.111111111... I see no line that is equal to that diagonal.
> Why do you think that this is not contradicting Cantor?
Because it is nonsense. Or are you actually stating that when we check 0.1, 0.11 is equal to 0.111111111...?
> > > > > > 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.
No, you are asuming that you can exhaust the paths by exhausting the nodes, this implies that you are assuming that the set of paths is countable, because the two statements are equivalent.
> > > 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.
It was *you* who wrote about that. See above, quoted here again: > > > It is said so. But there is no possibility, after having completed the > > > construction, to introduce such a path. They have sneaked in. so by your own statement you are believing in ghosts.
> 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.
In that case 1/3 is not in your tree. And your tree is not a set of nodes, but a set of paths. That are two different things.
> > > > 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.
What are you rambling about? I am not talking about binary expansion. I am talking about a special series of sums.
> > 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?
I am not talking about a bit sequence, I am talking about an infinite sum. To recap:
*You* state: union{i = 0..n} FISON(i) is a FISON and so: union{i = 0..oo} FISON(i) is a FISON
With exactly the same reasoning I get: sum{i = 0..n} 1/(i!) is a rational number and so: sum{i = 0..oo} 1/(i!) is a rational number.
So why is the reasoning correct in the first case and wrong in the second case (according to you)? A mathematician will tell you that the reasoning is wrong in both cases. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
