Re: Cantor and the binary tree

In article <1120475065.587150.124710@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > > There is a
> > > > distinction between "the set is not infinite" and "none of the
> > > > elements is infinite".
> > >
> > > And just this distinction is wrong for ordered sets where every
> > > element has an initial segment.
>
> *** T. Winter wrote:
>
> > Eh? I wonder.
>
> WM: There is a bijection of n and its initial segment for any n. N is
> both, the union of all numbers and, simultaneously, the union of all
> initial segments. Not less and not more. The numbers remain finite.
> Necessarily the segments remain finite too. Or where, do you think, an
> infinity of segments distinguishes itself from the finity of values?

Yes, both the numbers and the initial segments remain finite. This does
*not* mean that the union is finite.

> > > What does it mean, that every number is finite?
> > > It means that the belonging initial segment is finite, for *every*
> > > number.
> > > The limit n --> omega is not taken in N as 0 is not taken by any 1/n.
>
> *** T. Winter wrote:
> > What do you mean with "the limit is not taken"? By convention the
> > limit of 1/n is 0. But there is no n such that 1/n = 0. By the
> > same convention, the limit of {1,...,n} is N. But there is no n such
> > that {1,...,n} = N.
>
> WM: Correct. With only finite numbers there is no actually infinite or
> completed set N of all even numbers.

Eh? By your reasoning neither lim{1,...,n} = N, not lim{1/n} = 0. So?

> > > Sets like N are potentially infinite but never actually infinite.
>
> *** T. Winter wrote:
> > So 1/n is potentially 0, but never actually 0. What is your problem
> > with that? The limit (which is just a convention) gets you at 0 and N.
>
> WM: The limit is *never* assumed as long as you consider only finite
> numbers n.

Yes, but we have to consider the limit to get lim{1/n} = 0 or to get
lim{1,2,...,n} = N.

> > > Le q_0 be the first element of the well-ordering. Let q_1 be the
> > > first rational, which is less than q_0. Exchange the positions of
> > > q_0 and q_1. Let now q_2 be the first element which is smaller than
> > > q_1. Exchange the positions of q_1 and q_2. Continue. According to
> > > Cantor one can perform infinitely many transpositions without changing
> > > the character of a well-ordered set.
> > >
> > > At the end we have a well-ordered set the first element of which is
> > > the smallest rational number.
>
> *** T. Winter wrote:
> > Yes, but we never reach the end.
>
> WM:
> Exactly to show that was my intention. We do *never* reach the end. Now
> apply that realization to bijections with N like Cantor's diagonal.

See below.

> > > IF Cantor can construct his antidiagonal number, then we can also
> > > construct the above set (in zero time, of course). But if we do
> > > never finish, having found the smallest rational, then Cantor will
> > > never finish his antidiagonal.
>
> *** T. Winter wrote:
>
> > There is no reason to finish the antidiagonal. A simple function is
> > given such that for each natural n we can verify that the n-th number
> > in your list is not the antidiagonal.
>
> WM: A simple function is also given in my attempt to order the
> rationals by magnitude.There is as much reason to finish the ordering.
> If the antidiagonal is ready without finishing it, then the rationals
> are also put in order without finishing it. But they can't!

But you do not have to *create* the antidiagonal, you only have to prove
that it exists. And the processes are quite different. In order to get
the n-th decimal of the antidiagonal we only need the n-th number of
your list of reals, nothing more, nor less. So the creation is not an
on-going process. On the other hand, in your ordering to get n-th exchange
you need *all* previous exchanges...

> > > The statement was by Cantor. I do not consider such pathologic sets
> > > as you give below.
>
> *** T. Winter wrote:
>
> > But you consider {1, 2, 3, ..., omega}. In what sense is that set not
> > pathologic?
>
> WM: I stated that only as an example for pathological set, i.e. an
> infinite set.

Eh? So you are of the opinion that an infinite set is pathological?

> > > I do not assert that sets as defined by set theory can exist in that
> > > form. But set theory is not applicable to real mathematics.
> > > I assert that the set of natural numbers is potentially infinite.
> > > That means: Every number and every initial segment is finite, although
> > > there is no largest element to be fixed.
>
> *** T. Winter wrote:
> > Yes? That is precisely what set theory tells us. There is no largest
> > element.
>
> WM
> It says more. It says that all n do actually exist and that there are
> sets which contain even more members. And that Cantor's antidiagonal
> can be distinguished from any enumerated real.

Yes, so what?

> > > Your bijection with f(n) = n+1 would apply to the sets {0,1,2,3,...}
> > > and {1,2,3,...} although I believe with Bolzano that a bijection
> > > does not prove or express any equivalence of sets.
>
> *** T. Winter wrote:
>
> > Nothing is shown about equivalence at all. Only about the equality of
> > the cardinal number of the set.
>
> WM: Sets with equal cardinal number are called equivalent sets.
> Therefore a bjjection proves equivalence.

Ah, equivalent in that sense.

> *** T. Winter wrote:
>
> > And omega != omega + 1 != omega + 2, ... And you are wrong in stating
> > that omega + omega = omega. You are confusing omega with oo, which is
> > a concept, quite different from omega.
>
> WM: omega is Cantor's symbol of actual infinity. But I used the correct
> arithmetic of infinity, not Cantor's.

You are just using oo as if it were a number (and call it omega just to
be confusing). I now understand. When you wish to communicate it is
better to use the same (mathematical) languages as the others.

> > > Exactly. If an actually infinite set is created, then also an
> > > actually infinite number is created, simultaneously. The same is
> > > true for the initial segments. Therefore it is clear: Both, segments
> > > and values, are unbounded but always finite. That is potential
> > > infinity.
>
> *** T. Winter wrote:
>
> > You do not consider limits.
> > 1. What is lim{n -> oo} 1/n
> > 2. What is lim{n -> oo} {1, 2, 3, ..., n}
> > So in your opinion, the first limit is a potential 0, not an actual 0,
> > if I understand you. Now I wonder in what way we should distinguish that
> > potential 0 from an actual 0.
>
> WM: For natural numbers n the limit is not assumed. Actual 0 would not
> exist as a number if it was defined only by lim(1/n) (as pi does not
> exist as a number). But 0 does actually exist by other definitions.
> Actual omega, ordinal of N, does not exist.

What is "exist"? I asked you before, and again, how do you define "exist"?

> Therefore every set of even numbers contains numbers which are larger
> than the cardinal number of that set.

Every finite set of even numbers.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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