Re: Two results of set geometry

On 26 Sep., 23:04, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Sep 26, 2:32 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

On 26 Sep., 20:03, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

Appending one to each of the *elements* of the complete
column is not adding one to omega. Appending one to the
*set* of all elements, the complete column, is adding one to omega.
Appending 1 to the complete column produces a column
of length omega+1.

Every finite segment is the set of smaller finite segments.

No, every finite segment is a *set* of indices.
This set contains the index that corresponds to the finite
segment in question. (You are
confusing finite segments with ordinals, they are not quite
the same thing).

Finite segments of the first column are representing the finite
positive ordinals. They are quite the same thing.

There is
no reason to distinguish the complete set from the others.

Correct. The complete set is the *set* of all the indices
(equivalently, the setof all the finite segments).
However, it is not a finite segment. Nor is the complete set an element of the complete set.

The complete set represents the first infinite ordinal. Unless it is
in bijection with the finite sequences in the rows, there are only
finite ordinals in bijection with the finite sequences in the rows.
Therefore the set of rows is not sufficient to fill an infinite set.

Look here:

1
11
111

In this matrix we have a set of three natural numbers in bijection
with *all* three possible non-empty segments of the first column.

If, in the complete matrix, we have not a set of natural numbers in
bijection with all possible non-empty segments of the first column,
then there are less natural numbers than non-empty segments of the
first column.

By th way, your conclusion is equivalent to the assertion [a, b] = [a,
b) which is only true if b does not exist.


There is a bijection between the *elements* of the complete column,
and the *elements* of the set of rows.

There is a bijection between the *finite* elements of the complete
column, and the elements of the set of rows. That is not in question.
But you conclude that therefore there are more than finitely many
elements which are finite. Why don't you conclude (with the same
right) that there are finitely many elements which are infinite?

Regards, WM

.

Relevant Pages

  • Re: Two results of set geometry
    ... Finite segments of the first column are representing the finite ... Sets can only be in bijection with sets. ... Therefore the set of rows is not sufficient to fill an infinite set. ... with *all* three possible non-empty segments of the first column. ...
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  • Re: Two results of set geometry
    ... And, no, there is no order-preserving bijection between the ordered set ... of initial segments and the ordered set of natural numbers. ... elements in the first column has order type omega. ...
    (sci.math)
  • Re: Two results of set geometry
    ... Every finite segment is the set of smaller finite segments. ... easily seen t be in bijection with the SET of finite initial sequences. ... finite ordinals in bijection with the finite sequences in the rows. ... with *all* three possible non-empty segments of the first column. ...
    (sci.math)
  • Re: Two results of set geometry
    ... >>> Only the fact that the first column has a lot of finite initial ... >>> segments and an infinite initial segment while the rows are all finite ... > corresponds to (is in bijection with) a natural number. ...
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  • Re: Two results of set geometry
    ... So the elements of the first column are sequences of '1's? ... how can you talk about a bijection between the elements of the first ... > and the initial segments are those of the first column. ... segments and an infinite initial segment while the rows are all finite ...
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