Re: Replacement of Cardinality

On 07/28/2024 11:17 AM, Jim Burns wrote:

On 7/28/2024 8:17 AM, WM wrote:

Le 27/07/2024 à 19:34, Jim Burns a écrit :

On 7/26/2024 12:31 PM, WM wrote:

>

_The rule of subset_ proves that
every proper subset has less elements than its superset.

>

If ℕ has fewer elements than ℕ∪{ℕ}
then
|ℕ| ∈ ℕ

>
|ℕ| = ω-1 ∈ ℕ

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⎛ Each non.{}.set A of ordinals holds min.A
⎜
⎜ Ordinal j = {i:i<j} set of ordinals before j
⎜
⎜ Finite ordinal j has fewer elements than j∪{j}
⎜
⎝ ℕⁿᵒᵗᐧᵂᴹ is the set of ALL finite ordinals.
>
No finite.ordinal is last.finite,
no visibleᵂᴹ finite.ordinal,
no darkᵂᴹ finite.ordinal.
In particular, no finite.ordinal is ω-1
>
Also, no before.first infinite.ordinal is
before the first infinite.ordinal ω
In particular, no infinite.ordinal is ω-1
>
----
Consider ordinals i j k such that
i∪{i} = j  and  j∪{j} = k
>
Obviously, their order is  i < j < k
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Either they're all finite
|i| < |j| < |k|
or they're all infinite
|i| = |j| = |k|
>
No finite.to.infinite step exists.
no visibleᵂᴹ finite.to.infinite step,
no darkᵂᴹ finite.to.infinite step.
>
Defining declares the meaning of one's words.
'Defining into existence' that which doesn't exist
makes nonsense of whatever meaning one's words have.
>
⎛ if
⎜ g: j∪{j}→i∪{i}: 1.to.1
⎜ then
⎜ f(x) := (g(x)=i ? g(j) : g(x))
⎜ (Perl ternary conditional operator)
⎜ f: j→i: 1.to.1
⎜
⎜ if
⎜ f: j→i: 1.to.1
⎜ then
⎜ g(x) := (x=j ? i : f(x))
⎝ g: j∪{j}→i∪{i}: 1.to.1
>
Therefore,
i has fewer than j  iff  j has fewer than k
>

ℕ has fewer elements than ℕ

>
ℕ has ω-1 elements.

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ℕⁿᵒᵗᐧᵂᴹ holds all finite ordinals.
>
Finite doesn't need to be small.
ℕⁿᵒᵗᐧᵂᴹ holds ordinals which
are big compared to Avogadroᴬᵛᵒᵍᵃᵈʳᵒ,
but those big ordinals have an immediate predecessor,
and each non.0.ordinal before them has
an immediate predecessor.
That makes them finite, but not necessarily small.
>

Because ℕ does not have fewer elements than ℕ
ℕ does not have fewer elements than ℕ∪{ℕ}
and the rule of subsets is broken.

>
ℕ = {1, 2, 3, ..., ω-1} = {1, 2, 3, ..., |ℕ|}

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∀j ∈ ℕⁿᵒᵗᐧᵂᴹ:
∃k ∈ ℕⁿᵒᵗᐧᵂᴹ\{0}:
k = j+1 ∧ ¬∃kₓ≠k: kₓ=j+1
>
'+1': ℕⁿᵒᵗᐧᵂᴹ→ℕⁿᵒᵗᐧᵂᴹ\{0}: 1.to.1
and the rule of subset is broken.
>
>

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That's, ..., nice and all, yet, are you,
"preaching to the choir", or,
"reaching to the higher", the higher ground.
>
I.e., here it's not saying much.
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Where's the "extra"-ordinary.
>
It's a matter of deductive inference there is one,
while the naive nicely arrives at it directly.
>
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