Sujet : Re: The set of necessary FISONs
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 28. Feb 2025, 16:52:11
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <4b45ff34-dc3f-4e32-90a3-237f78fbd321@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 2/28/2025 4:12 AM, WM wrote:
On 28.02.2025 01:00, Jim Burns wrote:
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach
does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
>
It does.
Eppur si muove.
Zermelo says it,
Nope.
and it is easy to prove it:
Adding all natural numbers established the set ℕ.
We are finite beings. We do not do that.
Zermello's Infinity guarantees
a superset Z of Z₀
>
How is that accomplished?
>
By Zermelo's approach:
Z ⇒ 𝒫(Z)
>
How is Z accomplished?
Z is NOT accomplishedᵂᴹ.
True claims about Z are accomplishedᵂᴹ.
We know that they are true claims
which describe
that which we refer to when we refer to Z
(as, for example, Zermelo describes Z)
because we know
what we refer to when we refer to Z.
(as, for example, Zermelo knows).
For some claims, we can see that they are
not.first.false,
and
finitely.many claims which
are each true.or.not.first.false
are each true.
It's not.even.wrong.
>
And it's not Zermelo's approach,
Z
>
Where does he get is Z from?
Zermelo describes the Z in the discussion.
Your available responses are only:
"Right. Next, make not.first.false claims."
"Wrong. I am not in this discussion."
His Z is ensured by induction.
>
Nope.
>
{ } and a ==> {a}.
True of Z because,
when Zermelo describes Z,
Zermelo describes such a set.
Z being such a set is not induction.
Induction proves inductive a subset of
a set which is its.own.only.inductive.subset.
When we have shown that there is
the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.
>
We don't even need the intersection
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers
then x+1 is a natural number.
>
Consider Robinson arithmetic.
>
No.
An inductive proof only proves about
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
Proofs by induction are unreliable
in Robinson arithmetic.
>
Irrelevant.
What you (WM) think is a proof by induction
is unreliable. But you don't care?
The set of necessary FISONs
Re: The set of necessary FISONs