Liste des Groupes | Revenir à s math |
On 28.02.2025 16:52, Jim Burns wrote:Except that FISONs do approach N, only your limited finite sets of them do not, but then your logic can't have N, so you are just stuck trying to talk about things you admit you don't have the tools to handle, but you still try, and your logic just blows itself up on the contradictions.On 2/28/2025 4:12 AM, WM wrote:Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch des folgenden, seinem wesentlichen Inhalte von Herrn Dedekind herrührenden AxiomsOn 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.
Therefore we use FISONs without approaching ℕ.and it is easy to prove it:>
Adding all natural numbers established the set ℕ.
We are finite beings. We do not do that.
The existence of Z is secured by induction: Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch des folgenden, seinem wesentlichen Inhalte von Herrn Dedekind herrührenden Axioms.How is Z accomplished?>
Z is NOT accomplishedᵂᴹ.
Without induction Z is not existing.>by induction. By what else?
Zermelo describes the Z in the discussion
He describes it by induction.{ } and a ==> {a}.>
True of Z because,
when Zermelo describes Z,
Zermelo describes such a set.>The proof of existence is done by induction.
Z being such a set is not induction.>The set Z and all its inductive subsets are proven by induction.
Induction proves inductive a subset of
a set which is its.own.only.inductive.subset.An inductive proof only proves aboutZ contains many inductive subsets.
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z>The set Z is not existing and not even defined without Zermelo's induction.>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?
Note that for *definable* elements we have
{1, 2, 3, 4, 5} \ {1} \ {2} \ {3} \ {4} \ {5} = { },
which is same as
{1, 2, 3, 4, 5} \ {1, 2, 3, 4, 5} = { }.
Hence because of ∀n ∈ UA: |ℕ \ {1, 2, 3, ..., n}| = ℵo
we get
ℕ \ A(1) \ A(2) \ A(3) \ ... =/= { }
which is same as
ℕ \ UA =/= { }.
Regards, WM
Les messages affichés proviennent d'usenet.