Re: Simple enough for every reader?

On 28.05.2025 01:54, Ben Bacarisse wrote:

WM <wolfgang.mueckenheim@tha.de> writes:
 

On 27.05.2025 01:57, Ben Bacarisse wrote:

WM <wolfgang.mueckenheim@tha.de> writes:
>

On 26.05.2025 02:52, Ben Bacarisse wrote:

WM <wolfgang.mueckenheim@tha.de> writes:

>

With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:

I can't comment on an argument that is based on a set you have not
defined.

>
Can you understand my proof by induction?

Not without knowing what the set N_def is, since the argument starts
"For all n in N_def".

>
It starts: For every n ∈ ℕ that can be defined.

 "i.e. ∀n ∈ ℕ_def:".
 

Then it is proved that not every n ∈ ℕ can be defined.

 The "proof" starts with an undefined collection.

Every n that can be expressed by digits should be known to you.

We both know that you can't define N_def so you need to find some way of
waffling about it that starts by assuming it is known.

Of course I can decide for every number whether it can be distinguished from all other numbers. If so, it belongs to ℕ_def.
If you are unable to do so, simply assume that every natural number can be defined. Then you get the following contradiction:
All natural numbers can be manipulated collectively, for instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all numbers have disappeared.
Assume that all natural numbers can be defined/distinguished, then the above subtraction could also happen but, caused by the well-order, a last number would disappear. Contradiction.

It sounds as if you are saying that it (your book) defines N_def, and
that it (the set defined in your textbook) is the set defined by Peano
and many others.

Yes.

That would make N_def and N the same.  Really?

The above proof contradicts that statement.

 I see you cut the request to prove that 1 is in N (or it is N_def?)
using your junk "definition".  Of course you cut it.  You can't do it!

I have shown you the definition Below it is again.

Can you even prove that 1 is in N using your definition?

1 ∈ M (4.1)
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
Of course no intelligent reader need be told that this ℕ = ℕ_def also satisfies the axioms (4.1) and (4.2).

How you prove that {1} "has ℵo" successors.

I do not prove it but I apply Cantor's set ℕ which has cardinality ℵo, that is an actual infinity of elements. And also by Cantor ℵo - 1 = ℵo.

I'd like to see the base case proved.

It cannot be proved but only assumed. My proof shows: If Cantor was right and there is an actual infinity of ℵo natural numbers, then most numbers are dark. If the assumption is wrong because only Peano's potentially infinite collection ℕ_def exists, then my proof is void.
Regards, WM