Re: Simple enough for every reader?

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.

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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.