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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Simple enough for every reader?
Date: Sat, 31 May 2025 16:04:37 +0200
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On 31.05.2025 02:02, Ben Bacarisse wrote:
> WM <wolfgang.mueckenheim@tha.de> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
> des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische
> Hochschule Augsburg.)
> 
>> On 30.05.2025 03:08, Ben Bacarisse wrote:
>>> WM <wolfgang.mueckenheim@tha.de> writes:
>>
>>> I thought it might be something cumbersome and vague like that.  I can't
>>> even tell if this is a inductive collection,
>>
>> It is obvious and clear. Do you know a case where a natural number can be
>> in it and cannot be in it? No. You can only curse. It is the same as
>> Peano's set. If you can't understand blame it on yourself.
> 
> Can you prove it is an inductive set/collection?

See my book. The set is defined by induction. If n is in it, then also 
n+1 is in it. Pascal and Fermat used it without axioms as well as 
Cantor: "daß die Reihe
             1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
             1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]

The axiom has only been adapted because induction holds.
> 
>>> so I must decline any
>>> request to review a proof by induction based on it.
>>
>> Of course. There is no counter argument. So you must decline.
> 
> No, I decline because I don't know if it is an inductive set.  Do you?

Every mathematician knows that the definable natural numbers are an 
inductive set.
> 
> (I note you deleted the cumbersome and vague definition.

It has been given to be understood. Now you have or have not understood. 
If not, the further presence would not help, I assume.

> If it really
> were obvious and clear, I would have left it in to show the world how
> wrong I was to call it cumbersome and vague.)

I can give you a simpler and shorter definition: Every n that can be 
expressed by digits is definable.

> I see you've cut the incorrect definition and the claim that the axioms
> directly say that 1 is in N because, presumably, you now see that they
> don't.

>> As I said that requires an intelligent reader recognizing that without ℕ
>> obeying the axioms too the paragraph would be nonsense.
> 
> That's funny!  Yes, an intelligent reader will see you've written a junk
> definition

You are a dishonest liar. But that is not relevant.

>> All natural numbers of Cantor's set ℕ can be manipulated collectively, for
>> instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have
>> disappeared.
> 
> What definition of N do you want your intelligent readers to assume?

ℕ is Cantor's infinite set. Otherwise I could not use ℵo in my proof.

ℕ \ {1, 2, 3, ...} = { }

For the set ℕ_def defined in my book we have
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

Since all numbers can be reduced to the empty set by subtracting them 
collectively,
ℕ \ {1, 2, 3, ...} = { }
they could also be reduced to the empty set by subtracting them 
individually - if this was possible. But then the well-order would force 
the existence of a last one. Contradiction.

Regards, WM