Re: Simple enough for every reader?

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 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.

But the important fact, since it's /your/ proof, is what that means to
/you/ and I can not know that.

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.

So you can't do more than waffle about it?  OK.  Your "proof" has other
problems so that fact it presupposes something you can't formally define
is not the worst of it.

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.

You are actually prepared to state that N (defined by Peano) and N_def
(defined by your book) are the same and also that they are also not the
same?  That seems daft even by your standards.

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!

Nothing on this (of course).

I have shown you the definition Below it is again.
>

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

Nothing on this (of course).

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).

But it seems you can't prove that 1 is in N, can you?  It should be
easy, should it not?  It is simple using the correct definition, but
yours is junk.

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

>
I do not prove it

But you need to.  It's is the base case in the proof you asked everyone
about.  You can't make a proof by induction by simply asserting things.

Mind you, since you can't even prove that 1 is in N (or N_def), it is
hardly surprising that you can't prove this base case.

I'd like to see the base case proved.

>
It cannot be proved but only assumed.

Well don't ask others to accept your proof by induction if you yourself
think the base case can't be proved.

--
Ben.