Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 19.11.2024 17:26, Jim Burns wrote:

On 11/19/2024 6:01 AM, WM wrote:

That implies that
our well-known intervals

 Sets with different intervals are different.
Our sets do not change.

The intervals before and after shifting are not different. Only their positions are.
Is the set {1} different from the set {1} because they have different positions? Is the set {1} in 1, 2, 3, ... different from the set {1} in -oo, ..., -1, 0, 1,... oo?

Sets of our well.known.intervals
can match some proper supersets without growing

They cannot match the rational numbers without covering the whole positive real line. That means the relative covering has increased from 1/5 to 1.

Relative covering isn't measure.

It is a measure! For every finite interval between natural numbers n and m the covered part is 1/5.

You haven't defined 'relative covering'.
Giving examples isn't a definition.

If you are really too stupid to understand relative covering for finite intervals, then I will help you. But I can't believe that it is worthwhile. Your only reason of not knowing it is to defend set theory which has been destroyed by my argument.

I claim that there are functions f:ℝ→ℝ
such that
⟨ f(⅟1) f(⅟2) f(⅟3) ... ⟩  =
⟨ ⅟5    ⅟5    ⅟5    ... ⟩
and  f(0)  =  1

Not in case of geometric shifting. All definable intervals fail in all definable positions.

 

So you deny analysis or / and geometry.

 I deny what you think analysis and geometry are.
I accept infinite sets
and discontinuous functions

Discontinuity is not acceptable in the geometry of shifting intervals.

What is it you (WM) accuse infinite sets of,
other than not being finite?

Nothing against infinite sets. I accuse matheologians to try to deceive.

 Note:
An infinite set
can match some proper supersets without growing

I have proven that this is nonsense.
Regards, WM