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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.logic
Subject: Re: Simple enough for every reader?
Date: Fri, 30 May 2025 12:36:39 +0300
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On 2025-05-29 14:47:49 +0000, WM said:

> On 29.05.2025 12:07, Mikko wrote:
>> On 2025-05-28 15:13:54 +0000, WM said:
> 
>>>> There is no induction in plain logic.
>>> 
>>> But it is in the mathematics we apply.
>> 
>> It is in certain mathematical structures but not in all.
> 
> Anyhow a reader in sci.logic should understand it.

Everybody should understand at least arithmetic induction and its
limitations. But everybody doesn't.

>>> I have said: {1} has infinitely many (ℵo) successors.
>> 
>> But you navn't proven that this infinity is not begger than some other
>> infinity.

> I am assuming Cantor's infinity. This is expressed by ℵo.

Cantor did not use ℵo for infinity in general but only for a particular
kind of infinity. He also used other symbols for other kinds of infinity.
There are also kinds of infinity Cantor did not discuss at all.

>>>> and P[n] -> P[n+1] before it can infer
>>> 
>>> I did not expect that you need this explanation:
>>> 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.

But P[n] -> P[n+1] is not there.

> Do you doubt ℵo - 1 = ℵo?

That can't be said in Peano arithmetic.

> It is basic mathematics as you learn it in the first semester.

No, it is not that basic. There are no infinities there, and no
induction, either.

>> 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.

You have shown no proof that shows that.

>>>>> 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.

Which "here"? With or without non-set collections?

> All Cantor's natural numbers can be manipulated collectively, for 
> instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have 
> disappeared.
> Could all Cantor's natural numbers be distinguished, then this 
> subtraction could also happen but, caused by the well-order, a last one 
> would disappear. Contradiction.

It is not a contradiction that at least one disappears when all disappear.

-- 
Mikko