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From: joes <noreply@example.org>
Newsgroups: sci.logic
Subject: Re: Simple enough for every reader?
Date: Tue, 24 Jun 2025 18:16:09 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 29 May 2025 16:47:49 +0200 schrieb WM:
> On 29.05.2025 12:07, Mikko wrote:
>> On 2025-05-28 15:13:54 +0000, WM said:

>>>> and P[n] -> P[n+1] before it can infer
>>>
>>> If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
>>> 3, ..., n, n+1} has infinitely many (ℵo) successors because here the
>>> number of successors has been reduced by 1, and ℵo - 1 = ℵo. There is
>>> no way to avoid this conclusion if ℵo natural numbers are assumed to
>>> exist. And that is the theory that I use.
>> 
>> To me this does not look like P[n] -> P[n+1].
> 
> P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
> P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
> Do you doubt ℵo - 1 = ℵo?
You have doubted that.

>> As I said the theory must be specified.
>> In Peano arithmetic the induction axiom is applicable to everything. If
>> you want something else you must specify some other theory, perhaps
>> some set theory.
> Induction is applied to every natural number of the Peano set. The proof
> shows that it cannot be applied to every natural number of the Cantor
> set.
Lol. Is the successor of every "definable" number always "definable"?

>>>>> The set of finite initial segments of natural numbers is potentially
>>>>> infinite but not actually infinite.
>>>> There is nothing potential in a set.
>>> Then call it a collection.
>> Things get soon complicated if we allow other than objects, first order
>> functions and first order predicates.
> Here nothing gets complicated, but all remains very simple.
Cool. Sets don't change.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.