In article <bfcd6301-e7f0-4ced-906d-c111d378429e@j32g2000yqh.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 11 Jun., 15:46, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article <8d9c63c5-b605-4a98-81f2-815875aae...@21g2000vbk.googlegroups.= > com> WM <mueck...@rz.fh-augsburg.de> writes: ... > > > > 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?
> > You are wrong. A list is a mapping from N to the elements of the list. > > Through that list, given a number n, you find the n-th element of the list > > without referring to any previous elements of the list. > > 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.
> > To give an example, > > let's have a list of positive rational numbers through the mapping given > > a lnong time ago by David Tribble. To get the 51st element of the list, > > we calculate the mapping: write 51 in binary, create from it a continued > > fraction as described on > > <http://homepages.cwi.nl/~dik/english/mathematics/mueck/mapping.html> > > which is [0, 1, 2, 3], calculate it and we find 7/9. So we know the > > 51st element without knowing the 50th element. Where did I come at the > > 47th rational in that list (which is 7/11)? > > How can you write 51 without knowing what it is? Of course you must > count. In unary this is more difficult than in decimal or binary, but > the principle is the same. How do you obtain this number > 1111111111111111111111111111111111111 > unless you count the digits?
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 irrelevant here, because getting at n does not involve checking lines preceding the n-th line.)
Pray use tail in the future rather than end, to avoid confusion.
> > > Every end of a path contains aleph_0 nodes. > > > These nodes can be mapped on that path. Every node will eventually be > > > mapped on one path. Therefore all nodes are used up for constructing > > > a countable set of paths. > > > > 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. Suppose we have the set of paths where each path goes to the right after some specific (for that path) node. Is there a node in your tree that is not occupied by any of the paths in that set? Are there possible other paths that are *not* in that set of paths (like a path that alternates going left and right)?
> > 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?
> > > > Ok. The logic of finite complete sums of rational elements gives > > > > that the sum is rational. Using your logic we get that 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. > > > > > > In fact there are no binary expansions of irrational numbers. > > > > 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.
> > > > Ah, so 'e', 'pi', 'sqrt(2)' do not exist. Still I think you use > > > > at least > > > They exist as ideas but not as sums of series. > > > > So the equation e = sum{i = 0 .. oo} 1/(i!) is invalid in your logic? > > sum{i = 0 .. oo} 1/(i!) is a finite word with about 12 symbols. I > suffices to communicate what you mean. But what you think to have been > abbreviating is not abbreviated by this.
I do not think that you know what that abbreviation stands for. According to a statement you have posted quite some time ago it means (according to you) the infinite sum of the elements 1/(i!). But now you state it does not exist. Are you able to explain?
> > > > Apparently your logic. What are the rules of inference in your > > > > logic? > > > > > > The union of a complete set of linear sets is one of the sets. > > > > Eh? That is not a rule of inference. > > But it is true.
I asked for the rules of inference, so apparently you are not able to supply the rules of inference in your logic. Ny question again, what are the rules of inference in your logic? If you are not willing to supply them it is impossible to follow your logic (apparently it consists of some ad-hoc statements that must be true).
> > > > And, also apparently, in your logic the length of the diagonal of > > > > a square with sides with size 1 does not exist. > > > > > > It exists, but not as binary or decimal expansion. > > > > Eh? Now suddenly sqrt(2) does exist. > > Not suddenly. It accept that it exists. But it has no binary or > decimal expansion.
What is the relevance to have binary or decimal expansion? 1/3 does not have a binary and decimal expansion either (in your sense of not having it).
> > > > > That is a blatant lie. > > > > > > > > Eh? What is the lie? I just state that in ZF there are no no > > > > n-static sets. > > > > > > Then every element should be accessible. Why do you start always with > > > a diminishingly small one? > > > > Every element is accessible. And you start always at the beginning > > because there is no last one. > > But you claim that there is always a larger one than you atrt with. > Why that?
Because that is the definition of N.
> > > A potentially infinite set is a set that has no last element. > > > > Ok. Than N is a potentially infinite set. > > That is true. But as such a set is never exhausted, Cantor waits and > waits and waits ...
I wonder at the relevance.
> > > > Because in ZF a set does not need to have a last element. So > > > > whatever element you chose, there is *always* a larger one. > > > > > > That is potential infinity. > > > > Ok. Than N is a potentially infinite set that actually does exist in ZF. > > If one rejects actual infinity, then there is not "a continuous > function nowhere differentiable". The actual infinity is a necessary > condition of Cantor's diagonal proof > of the power-set theorem. If one rejects actual infinity, then the > theorem becomes unprovable and therefore there are not any cardinals > greater than aleph_0.
Great, you finally admit it. You just reject the axiom of infinity. So ZF is not inconsistent, it is inconsistent with your rejection of the axiom of infinity. That is normal: a theory is inconsistent with the rejection of one of its axioms. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
