Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 12/16/2024 6:23 AM, WM wrote:

On 16.12.2024 11:14, Mikko wrote:

That is the fallacy of equivocation.
The limit of analysis is a different concept from
the limit of set theory.

>
The limit of analysis proves that
the limit of set theory is wrong.
The limit of analysis proves that
the relative covering is 1/10,
the relative non-covering is 9/10.
Set theory proves that
the relative non-covering is 0.
These numbers are answering exactly the same question.

For infinite sequence ⟨A₀,A₁,A₂,…⟩ = ⟨Aₙ⟩ of sets,
limit.set Limⁿ.⟨Aₙ⟩ is the set such that
(1) x is in Limⁿ.⟨Aₙ⟩ if
⎛ x is in each of
⎜ the infinitely.many sets in ⟨Aₙ⟩
⎝ -- with only finitely.many exceptions
and
(2) y is not.in Limⁿ.⟨Aₙ⟩ if
⎛ if y is not.in each of
⎜ the infinitely.many sets in ⟨Aₙ⟩
⎝ -- with only finitely.many exceptions
⎛ That makes a lot more sense if
⎝ 'infinite' isn't merely a way to say 'enormous'.
Not all sequences have limits, because
not all sequences have, for each potential element,
one of those two conditions holding.
For example, consider the sequence
⟨{0},{1},{0},{1},…⟩
There are more.than.finitely.many exceptions
to 0 being in the sequence and also
more.than.finitely.many exceptions
to 0 not.being in the sequence.
And the same as well for 1
0 and 1 are neither in nor not.in
Lim.⟨{0},{1},{0},{1},…⟩
which is not.allowed for sets.
So, Lim.⟨{0},{1},{0},{1},…⟩ can't be a set.
For decreasing sequence ⟨Bₙ⟩,  i<j ⇒ Bᵢ⊇Bⱼ
Limⁿ.⟨Bₙ⟩  =  ⋂ⁿ⟨Bₙ⟩
(for example, end.segments)
For an increasing sequence ⟨Cₙ⟩,  i<j ⇒ Cᵢ⊆Cⱼ
Limⁿ.⟨Cₙ⟩  =  ⋃ⁿ⟨Cₙ⟩
(for example, FISONs)
For a more.general sequence ⟨Aₙ⟩
Lim.Infⁿ.⟨Aₙ⟩  =  ⋂ᵐ⋃ᵐᑉⁿ⟨Aₙ⟩
is a lower.bound of the set of
common.with.finite.exceptions elements.
and
Lim.Supⁿ.⟨Aₙ⟩  =  ⋃ᵐ⋂ᵐᑉⁿ⟨Aₙ⟩
is an upper.bound of the set of
common.with.finite.exceptions elements.
⎛ Lim.Inf.⟨{0},{1},{0},{1},…⟩  =  {}
⎝ Lim.Sup.⟨{0},{1},{0},{1},…⟩  =  {0,1}
Assuming Limⁿ.⟨Aₙ⟩ exists,
Lim.Infⁿ.⟨Aₙ⟩  ⊆  Limⁿ.⟨Aₙ⟩  ⊆  Lim.Supⁿ.⟨Aₙ⟩
Assuming Lim.Infⁿ.⟨Aₙ⟩  =  Lim.Supⁿ.⟨Aₙ⟩
Lim.Infⁿ.⟨Aₙ⟩  =  Limⁿ.⟨Aₙ⟩  =  Lim.Supⁿ.⟨Aₙ⟩
and
Limⁿ.⟨Aₙ⟩ exists.