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

On 11.12.2024 21:58, Jim Burns wrote:

On 12/11/2024 2:57 PM, WM wrote:

On 11.12.2024 20:27, Jim Burns wrote:

 

⋂{E(i):i} = {}.

>
Of course. But
all intersections with finite contents
are invisible.

 We know about what's invisible by
assembling finite sequences holding only
claims which are true.or.not.first.false.
 We know that
each claim in the claim.sequence is true
by _looking at the claims_
independently of _looking at the invisible_
 _It doesn't matter_
whether any finite.cardinals are invisible.
Each finite cardinal is finite, and
that is enough to start
a finite sequence of claims holding only
claims which are true.or.not.first.false
-- claims about each finite.cardinal, visible or not.
 Some claims seem too dull to need verifying.
"Is a finite.cardinal finite?"
Better to ask "Is the Pope Catholic?"
But such obviously.true claims start us off.
 Other claims, the more interesting claims,
can be verified as not.first.false
_by looking at the claims_
NOT by looking at finite.cardinals
Look at q in ⟨p p⇒q q⟩
There is no way in which q can be first.false.
It doesn't matter what q means, or what p means.
We can see q is not.first.false in that sequence.
 Repeat the pattern ⟨p p⇒q q⟩ and a few others
for a whole finite sequence of claims,
and
that whole finite sequences of claims
holds no first false claim,
and thus holds no false claim.
 Which we know by _looking at the claims_
 

Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ

>
It is a smaller set.

 For each k in ℕ
there is unique k+1 in ℕ\{0}
 

Cardinalities are not useful.

 And yet, by ignoring them,
you (WM) end up wrong about
⎛ For each k in ℕ
⎝ there is unique k+1 in ℕ\{0}
 

All that waffle only in order to avoid the crucial question?
Simply answer by yes or no: Is the complete removal of natural numbers from the sequence of intersections bound by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
or not?
Regards, WM