Re: The set of necessary FISONs

On 08.03.2025 20:27, Jim Burns wrote:

On 3/8/2025 9:09 AM, WM wrote:

On 08.03.2025 12:58, Jim Burns wrote:

On 3/8/2025 3:45 AM, WM wrote:

 

You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

>

Zermelo's Axiom of Infinity describes Z
Z ∋ {} ∧ ∀a: Z ∋ a ⇒ Z ∋ {a}
Multiple sets satisfy that unique description.
That's an indefinite description.

>
My description is definite.

 Your description is only definite because
ℕ ∋ n is definite.

No, my description is definite because every n can be obtained by addition of 1's (or of curly brackets). It is not necessary to have another definition of ℕ_def before.

 What ℕ is,
the interesting part of your description of Z₀,
must be found elsewhere.

You are wrong. ℕ_def is the potentially infinite sequence of sums
1
1+1
1+1+1
...
ℕ is its completion, a set larger than all FISONs.

 

If only
{{{...{{{ }}}...}}} with finitely.many curly brackets
each with immediate.predecessor ⋃{{{...{{{ }}}...}}}
and with {} being one of its priors
are in Z₀
then, yes, that is a definite description.
>
And, yes, that seems to be what you meant.
If you ever want to make your descriptions clearer,
you won't be getting complaints from me.

>
That is exactly what I meant.
And that's also the induction of my argument
UF = ℕ  ==> Ø = ℕ.

 A proof by induction
(a proof by this.inductive.subset.is.the.whole.set)
is only reliable for
a set which is its.own.only.inductive subset.

No. A proof by induction produces a set. If applied as I did above, the produced set is its own only inductive subset.

ℕ is its.own.only.inductive subset.
Z₀ is its.own.only.inductive subset.
⋃{F} union of the set {F} of all FISONs,
  is its.own.only.inductive subset.

Right.

 For any set W which
is its.own.only.inductive subset (ℕ,Z₀,⋃{F},...),
∀j∈W:∃k∈W: j < k

Yes, that is called potential infinity. A k larger than all j is not produced by induction. That is only produced by appointment.

{F} is its.own.only.inductive.subset.
No FISON in {F} is F(ω-1)

Right. ω and ω-1 cannot be attained by induction.
Therefore UF = ℕ  ==> Ø = ℕ.

you should have said somewhere
what a finite ordinal is,

>
That is not under discussion here.

 That has been under discussion for decades.
 I think that these decades of discussion
have been, in large part, you assigning
different meanings to 'finite', etc.
and matheologians (among whom I place myself)
trying to discern what your meanings are.
 Here's my best guess:
definableᵂᴹ  ==  finite

Of course. We know also what we mean by ω, but it has no FISON. Therefore it is not a visible or definable number.

darkᵂᴹ  ==  finite  ==  big

Look into the new thread "The truncated harmonic series diverges." to see that dark numbers are necessary. Having only definable denominators, the harmonic series could not diverge.

n is usually denoting
a natural number.

 Do we mean the same by 'natural number'?

There are two different meanings: All positive integers having FISONs or all positive integers.

 

FISONs are finite by definition.

 Do we mean the same by 'finite'?

A natural number n is finite. It is an integer between 0 and ω: 0 < n < ω. Sometimes 0 is included, never ω is included.
Regards, WM