Re: The non-existence of "dark numbers"

On 3/22/2025 5:19 AM, WM wrote:

On 22.03.2025 06:11, Jim Burns wrote:

On 3/21/2025 2:15 PM, WM wrote:

On 21.03.2025 18:39, Jim Burns wrote:

On 3/21/2025 3:50 AM, WM wrote:

On 20.03.2025 23:25, Jim Burns wrote:

For sets not.having a WM.size,
Bob vanishing isn't a size.change.

>
Only if reducing isn't reducing.

>
What you (WM) think is reducing
isn't reducing.

>
You confuse
the clear fact that
in the reality of
sets vanishing means reducing
with
the foolish claim that
cardinality was a meaningful notion.

>
The set of all sizes.which.WM.considers.sizes
does not have a size.which.WM.considers.a.size.

>
It is infinite
but nevertheless obeys the logic of
lossless exchanges do not suffer losses.

I recently asked you (WM) to
non.circularly say what 'finite' means to you.
You quoted the beginning of Wikipedia's "Finite Set",
which approximately says that finite can be counted to.
I accepted that quote, and why shouldn't I?
It shouldn't be earth.shaking that
you accept Wikipedia.
However, I see that I've been
counting my chickens before they've hatched.
A little after your quote, in "Finite Set":
⎛
⎝ The union of two finite sets is finite, ...
⎛
⎜ Any proper subset of a finite set S is finite
⎜ and has fewer elements than S itself.
⎜ As a consequence,
⎜ there cannot exist a bijection between
⎝ a finite set S and a proper subset of S.
I think that you (WM) DON'T accept that.
That leads to the 'logic' of
lossless exchanges do not suffer losses,
but not for infinite sets.
The set of sizes of sets which DO obey that 'logic'
is a set which isn't any of those obedient sizes,
is a set which DOESN'T obey that 'logic'.
If it's otherwise,
then a subset exists larger than its superset.

It is infinite
but nevertheless obeys the logic of
lossless exchanges do not suffer losses.

Infinite sets don't need to obey the logic of
lossless exchanges do not suffer losses.
That's the difference between
infinite sets and finite sets.
(You could use that to answer my question.)

Two sets of the same cardinality,
one a proper subset of the other,
can be swapped set.wise, one for the other,
and preserve size.

>
Of course.
The reason is that
all pairs of
the bijection proving same cardinality
have
infinitely many dark successors
which cannot be bijected.

A bijection which does not biject everything
is not a bijection,
just as a square with only three sides
is not a square.