Sujet : Re: The set of necessary FISONs
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 01. Mar 2025, 18:47:32
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <7ac03d2e-5d72-4186-8b33-6fdddb7b83ff@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 26
User-Agent : Mozilla Thunderbird
On 3/1/2025 7:51 AM, WM wrote:
On 28.02.2025 20:25, Jim Burns wrote:
On 2/28/2025 12:04 PM, WM wrote:
On 28.02.2025 16:52, Jim Burns wrote:
An inductive proof only proves about
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
>
Z contains many inductive subsets.
>
(We don't know whether Z=Z₀ or Z≠Z₀)
>
Z contains every set a.
Zermelo describes set a as an element of Z
The claim
"Z contains every element of Z"
is not false, but
it isn't helpful, either.
https://en.wikipedia.org/wiki/Cooperative_principleIt's a claim which suggests that
what the words you use mean
is not what you think they mean.
https://www.youtube.com/watch?v=dTRKCXC0JFgZ_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
If the subset {i:A(i)} ⊆ Z₀ of k such that A(k)
is inductive,
then the subset {i:A(i)} = Z₀
By definition,
∀k ∈ {i:A(i)}: A(k)
{i:A(i)} = Z₀
∀k ∈ Z₀: A(k)
That universal claim did not come from a supertask.
It came from narrowing our focus from Z to Z₀.
In narrower Z₀,
inductivity identifies a unique set.
Consider this WM.inductive.proof:
>
It is sufficient
It is necessary
It is sufficient
for my purpose to consider
0, {0}, {{0}}, ...
which is an infinite set produced by induction.
https://www.youtube.com/watch?v=dTRKCXC0JFg