Re: Simple enough for every reader?

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