Re: Replacement of Cardinality

On 8/15/2024 9:59 AM, WM wrote:

Le 14/08/2024 à 22:04, Jim Burns a écrit :

On 8/14/2024 8:34 AM, WM wrote:

What is smaller than every positive x
is smaller than the interval (0, oo).

>
That which is not.in the interval (0,∞)
is not an end of (0,∞).

>
The smallest point of (0, oo) belongs to (0, oo).
It cannot be seen. It is dark.

A definition isn't a claim something exists.
A definition is a claim how a word is used.
Defining ⅟𝔊 to be the smallest unit fraction
isn't a claim that ⅟𝔊 exists.
'⅟𝔊' can be used to derive contradictions,
which prove that ⅟𝔊 does not exist.
for each n ∈ ℕᵈᵉᶠ:
n+1 disproves by counter.example that n is largest
⇒
max.ℕᵈᵉᶠ doesn't exist,
even though we can define max.ℕᵈᵉᶠ
In mathematics,
we _propose_ the existence of abstract objects:
numbers, sets, etc.
This is different from defining them into existence.
You are not free to reject my definitions.
That is what I know I mean by what I say.
But that doesn't make what I mean true.
You are free to reject my proposals,
but the usual practice is to frame a proposal
in maximally difficult.to.reject terms.
That's what I do when I propose Boolos's ST
⎛ ∃{}
⎜ ∀x∀y∃z=x∪{y}
⎝ extensionality
You can reject those existential claims.
But, if you do, the discussion ends and
you look more than a little unreasonable.
Not logic, but forceful in its own way.
If you accept those existential claims,
I can follow them with definitions --
which you _aren't_ free to reject --
of what.I.mean.by the natural numbers,
(spoiler: the usual naturals, '+', and '×').
Iron.clad support of our usual arithmetic,
but _not_ by definition, instead by
maximally.acceptable.proposal, Boolos's ST.

The smallest point of (0, oo) belongs to (0, oo).
It cannot be seen. It is dark.

The rationals in (0,∞) aren't ends of (0,∞).
For each p/q ∈ ℚ⁺
there is disproof by counter.example p/(q+1)
to the claim p/q is smallest in (0,∞),  and
there is disproof by counter.example (p+1)/q
to the claim p/q is largest in (0,∞).
Rational p/q is not largest or smallest in (0,∞)
The split.situaters in (0,∞) aren't ends of (0,∞).
For each x ∈ ℝ⁺\ℚ⁺
x situates the non.{} split Fₓ,Hₓ of ℚ⁺
Fₓ ᵉᵃᶜʰ< x <ᵉᵃᶜʰ Hₓ
non.{} Fₓ ∋ r₋
non.{} Hₓ ∋ s₊
r₋ < x < s₊
Irrational x is not largest or smallest in (0,∞)
By definition of the real numbers,
only rationals and split.situaters are in (0,∞)
No ends of (0,∞) exist in (0,∞)
No ends of (0,∞) exist not.in (0,∞)
No ends of (0,∞) exist.