Re: Cantor Confusion

In article <1166008449.817555.138170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

William Hughes schrieb:

A_1 = {1}
A_2 = {1,2}
A_3 = {1,2,3}

B = {1,2,3}

then B is contained in the last A_i. If there is no last A_I, then
there is
no A_i that contains B

That has nothing to do with "last".

If A_i contains B, then A_i contains any A_j.
Therefore A_i is "last".

It has all to do with "every A is finite".

No. From the statement "every A is finite" we cannot conclude
that there exists and A_i that contains B.

That is correct. But this conclusion is derived from:
1) N is a linear set (every finite initial segment (= line) includes
all preceding segments)
2) There is no element of the complete segment (= diagonal) outside of
every finite segment (= line).


But conclusion (2) requires the assumption that every set is finite, so
the whole argument is circular and therefore invalid.

Which of the following three do you claim
(note that none of them require assuming that
the set of natural numbers actually exists).

- The potentially infinite sequence of
natural numers is not an initial segment

The potentially infinite sequence of natural numers is an initial
segment. (If we refrain from the physical constraints of numbers.)

- there is a natural number which is
not an element of the potentially infinite
sequence of natural numbers

No. Every natural number belongs to the sequence and to at least one
finite initial segment.

- there is a line for which you cannot
find a natural number which is not an
element of the line

Not so many nots please.

For each line you can find a natural number which is not an element of
that line. And as soon as you have found (created) that number, you
have determined (created) the first line to which it belongs.

There is no line with every natural number.

Similarly you cannot find every elment of the diagonal.

So that one can find any but not every?


If you name an
element, then the diagonal is constructed up to that element at least.

And does it stop there?

As there is no line with every number there can be no diagonal with
every number.

That is an assumption which WM imposes, not a conclusion which can be
drawn from anything incontrovertible.

If WM will admit that he is taking it as an axiom, that is one thing,
but he keeps insisting the he can deduce it from something which
requires that he assume it.
.

Relevant Pages

  • Re: Cantor and the binary tree
    ... > Then WM is only potentially numerate but not actually numerate. ... n belongs to an initial segment S which does not contain n' ... n' belongs to an initial segment S' which does not contain n ... This segment is potentially infinite, ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >>> Every countable set like N is potentially infinite but not actually ... > n belongs to an initial segment S which does not contain n' ... > n' belongs to an initial segment S' which does not contain n ... Not so (unless WM insists that N is a member of itself, ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... >> Of course every set of naturals is bounded by a natural. ... n belongs to an initial segment S which does not contain n' ... n' belongs to an initial segment S' which does not contain n ... This segment is potentially infinite, ...
    (sci.math)
  • Re: Cantor Confusion
    ... N is a linear set (every finite initial segment (= line) includes ... The potentially infinite sequence of natural numers is an initial ... Every natural number belongs to the sequence and to at least one ...
    (sci.math)
  • Re: Cantor Confusion
    ... The set of lines is also the potentially infinite set of natural ... Every initial segment of the diagonal is a line. ... The potentially infinite sequence ... not an elment of ...
    (sci.math)