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

On 11.12.2024 05:06, Jim Burns wrote:

On 12/9/2024 4:41 PM, WM wrote:

The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.

 

We need only the sequence of sets with the empty set in the limit.

 

The limit set is the same
for both sequences.
(E(1)∩E(2)∩...∩E(n)) and (E(n))
In order to stop tricksters
we go without cardinality.

 What you (WM) mean by "going without cardinality"
is
pretending that
a sequence of sets
and
the sequence of the cardinalities of those sets
are the same thing.

No, I use the sequence of sets as defined here
E(1) = ℕ
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
and not sequences like
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
That's a big difference.

Which, for end.segments of final.cardinals,
they are not the same.
 #⋂{E(i):i} = #{} = 0
 ⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀
 

There is no use for # but only for
⋂{E(i):i} = {}.
The problem is that every definable endsegment is infinite. But if all natural numbers are applied in a Cantor bijection, then they all must leave the endsegments by the only possible process:
E(k+1) = E(k) \ {k}
Regards, WM