Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:

William Hughes schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
William Hughes schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
William Hughes schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
William Hughes schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
Virgil schrieb:
(It is contained in the union of all lines, but
the union of all lines is not a line)

That is a void assertion unless you can prove it by
showing that element by which the union differes
from all the lines.

Not quite. In order to achieve that the diagoal is
not in any linem all that is required is:
Given any line there is an element of the diagonal
not in THAT line.
It is not requires that:
There is an element of the diagonal that is not in
any line.


For linear sets you cannot help yourself by stating
that the diagonal differs form line A by element b and
from line B by element a, but a is in A and b is in B.
This outcome is wrong.

Therefore your reasoning "there is an element of the
diagonal not in THAT line. It is not required that:
There is an element of the diagonal that is not in any
line." is inapplicable for linear sets. You see it best
if you try to give an example using a finite element a
or b.


In every finite example the line that contains
the diagonal is the last line.

Every example with natural numbers (finite lines) is a
finite example.

Your claim is that there is a line which contains the
diagonal.

Because a diagonal longer than any line is not a diagonal.

Call it L_D. Question: "Is L_D the last line?"

There is no last line

Then, there is a line that comes after L_D.

Therefore :L_D does not contain every element
that can be shown to exist in the diagonal.

All elements that can be shown to exist in the diagonal can be
shown to exist in one single line. [(P1)]

This proposition P1 has _not_ yet been proved (shown).



Call it L_D

L_D contains a largest element. n.

L_D is not the last line, so there is
a line with element n+1,

Element n+1 can be shown to exist in the diagonal.


Element n+1 can be shown to exist in L_D (which is obviously a line
containing n+1).

No. L_D is bounded. The largest element of L_D is n.
L_D does not contain n+1.

You misinterpret L_D. L_D is that line which contains all numbers
contained in the diagonal.

The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too.
Hence L_D cannot be a line of the list (meaning: cannot be _in_ the
list) for any line in the list is bounded (proof by induction).

Hence contradiction to P1. P1 must be dropped.

If your L_D does not contain them, then you have the wrong L_D.

Upto now we have no line L_D at all since P1 has not been proved.

F. N.
--
xyz
.

Relevant Pages

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  • Re: Cantor Confusion
    ... That is a void assertion unless you can prove it by showing that ... element by which the union differes from all the lines. ... For linear sets you cannot help yourself by stating that the diagonal ...
    (sci.math)
  • Re: Cantor Confusion
    ... That is a void assertion unless you can prove it by showing that ... element by which the union differes from all the lines. ... For linear sets you cannot help yourself by stating that the diagonal ...
    (sci.math)
  • Re: Cantor Confusion
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    (sci.math)
  • Re: Cantor Confusion
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