On 22 Dez., 15:28, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > Even a matheologian should understand that: If there is no digit at a > > finite place up to that the sequence 0.333... identifies the number > > 1/3, then there is no digit at a finite place up to that the number > > 1/3 can be identified. > > Right, there is no digit at a finite place up to that the number 1/3 can be > identified. And as there are no digits at infinite places that appears to > you to be a paradox. It is not. There is *no* finite sequence of digits > that identifies 1/3. But there is an *infinite* sequence of digits that > does so.
Why not pronounce it a bit clearer? We know that every digit is the last digit of a finite initial segment of digits. There is no other digit, is it? Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3.
If you read this sentence, perhaps you get an impression of what matheology is?
> > I said, if there is a sequence that identifies > > 1/3, then the identifying digits must be at finite places. > > Right, all identifying digits (there are infinitely many) are at finite > places.
Therefore your statement can be expressed in the following way: There is no finite initial segment that identifies 1/3. But there is an infinite finite initial segment that identifies 1/3. >
> > The union of finite initial segments cannot ield an infinite initial > > segment? > > Yes. But as you have stated that your tree contained finite paths only, > such an infinite initial segment is not (according to *your* definition) > a path.
The axiom of infinity states: There is an infinite set such that with every n the follower of n belongs to the set. This is true for the binary tree. With every finite path 1, ..., n the follower 1, ..., n+1 is in the tree. The union of all those paths of course is finite, as I said, because there is no infinite path, but according to the axiom it is infinite.
> > > Does the sequence of 1/3 not consist of a union of all finite > > initial segments? > > It is, but also (according to *your* definition) it is not a path.
There can be no question: The union of all finite initial segments is finite. But according to the axiom, it is infinite. So we can both be well pleased.