Re: The set of necessary FISONs

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Sujet : Re: The set of necessary FISONs
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 05. Mar 2025, 20:26:50
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <1ba064b1-f215-494a-a6c4-9e85a6c4e29e@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
On 3/5/2025 4:22 AM, WM wrote:
On 04.03.2025 23:05, Jim Burns wrote:
On 3/4/2025 2:56 PM, WM wrote:
On 04.03.2025 17:43, Jim Burns wrote:
On 3/4/2025 5:49 AM, WM wrote:

Here is only *one* argument
standing for a long while.

⎜ And on the pedestal these words appear:
⎜ "My name is Ozymandias, King of Kings:
⎜ Look on my works, ye Mighty, and despair!"
⎜ No thing beside remains. Round the decay
⎜ Of that colossal wreck, boundless and bare
⎜ The lone and level sands stretch far away.

— Percy Shelley, "Ozymandias"

The union ⋃{F} of FISONs and
the intersection ⋂𝒫ⁱⁿᵈ of inductive subsets
are the same set.
>
Correct.
>
Proper supersets
>
We talk about subsets!
A subset B  ⇔  B superset A

Proper supersets
of ⋂𝒫ⁱⁿᵈ contain extra elements.
You (WM) have previously said that
your (WM's) ℕ doesn't have extra elements.
>
ℕ is a proper superset of ⋃F.
It contains all the dark natural numbers.
Your darkᵂᴹ numbers cannot.be.said to self.equal.

⋃F contains only defined natural numbers.
⋃{F} holds each definable natural number.
⎛ For each n ∈ ⋃{F} with swap.in ⟨n-1⇄n⟩
⎜ there is a later swap.out ⟨n⇄n+1⟩

⎜ Before all swaps, Bob is in 0
⎜ After all swaps,
⎜ each of which leaves Bob somewhere,
⎝ Bob isn't anywhere.

You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
>
No.
>
Ha. Caught red handed.
Yes there is no mathematical possibility
to contradict my claim.
No BECAUSE
∈ {} ∧ a ↦ {a} is an indefinite description.
but
Z₀ is a definite set.
Saying 'simplest' or 'emptiest' or
'intersection of subsets like that'
would make Z₀ definite,
but you (WM) didn't say any of those.
I would have inserted a clause to make it definite,
but you seem to be trying to say something
(about induction?) by not.saying such a clause.
I'll tell you why you're wrong
when I figure out what you're trying to say.

For Z, not Z₀, ∀a: a ∈ Z ⇒ {a} ∈ Z
Z₀ is the simplest inductive subset of Z.
>
What does Z₀ contain that is
not in my inductive description?
You're facing the wrong direction.
Turn around.
What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Bob is NaN.
https://www.youtube.com/watch?v=TjAg-8qqR3g

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