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

On 1/18/2025 3:41 AM, WM wrote:

On 18.01.2025 00:08, Jim Burns wrote:

The finite extends
much further than you (WM) think it does.
Infinitely further than you think it does.

>
No.
As long as
you deny Bob's existence and violate logic
you are not a reliable source.

https://en.wikipedia.org/wiki/Finite_set
⎛
⎜ Informally, a finite set is a set which
⎜ one could in principle count and finish counting.
⎜
⎜ [...]
⎜
⎜ Any proper subset of a finite set S is finite
⎜ and has fewer elements than S itself.
⎜
⎜ [...]
⎜
⎜ In Zermelo–Fraenkel set theory without
⎜ the axiom of choice (ZF),
⎜ the following conditions are all equivalent:
⎜ 1.
⎜ S is a finite set.
⎜ That is,
⎜ S can be placed into
⎜ a one-to-one correspondence with
⎜ the set of those natural numbers less than
⎜ some specific natural number.
⎜ [...]
⎜ 3. (Paul Stäckel)
⎜ S can be given a total ordering which
⎜ is well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S
⎜ has both a least and a greatest element in the subset.
⎜ [...]
⎜ 7.
⎜ S can be well-ordered and
⎜ any two well-orderings on it are order isomorphic.
⎜ In other words, the well-orderings on S
⎝ have exactly one order type.
There is no step from finite to infinite.
¬Eᵒʳᵈk:  finite ⟦0,k⦆ ∧ infinite ⟦0,k+1⦆
⎛ Assume otherwise.
⎜ Assume
⎜ finite ⟦0,k⦆ ∧ infinite ⟦0,k+1⦆
⎜
⎜ finite⁽³⁾ ⟦0,k⦆
⎜
⎜ ∀ˢᵉᵗS ⊆ ⟦0,k⦆: one of
⎜⎛ 1.
⎜⎜ S = {}
⎜⎜ 2.
⎜⎜ S ≠ {}
⎜⎝ min.S,max.S ∈ S
⎜
⎜ ⟦0,k+1⦆ = ⟦0,k⦆∪{k}
⎜ ∀ˢᵉᵗT ⊆ ⟦0,k+1⦆: one of
⎜⎛ 1.
⎜⎜ T = {}
⎜⎜ 2.
⎜⎜ T = {k}
⎜⎜ min.T = max.T = k
⎜⎜ 3.
⎜⎜ T = S ≠ {}  ∧  S ⊆ ⟦0,k⦆
⎜⎜ min.T = min.S
⎜⎜ max.T = max.S
⎜⎜ 4.
⎜⎜ T = S∪{k} ≠ {k}  ∧  S ⊆ ⟦0,k⦆
⎜⎜ min.T = min.S
⎜⎝ max.T = k > max.S
⎜
⎜ ∀ˢᵉᵗT ⊆ ⟦0,k+1⦆: one of
⎜⎛ 1.
⎜⎜ T = {}
⎜⎜ 2.
⎜⎜ T ≠ {}
⎜⎝ min.T,max.T ∈ T
⎜
⎜ finite⁽³⁾ ⟦0,k+1⦆
⎜
⎜ However,
⎜ infinite ⟦0,k+1⦆
⎝ Contradiction.
Therefore,
¬Eᵒʳᵈk:  finite ⟦0,k⦆ ∧ infinite ⟦0,k+1⦆
There is no step from finite to infinite.

No finite ordinal has
an infinite immediate successor.

>
Maybe. But then there is no infinite ordinal.

Each ordinal ⟦0,λ⦆ has
has an immediate successor ⟦0,λ⦆∪{λ}
Each finite ordinal ⟦0,k⦆
doesn't have an infinite immediate successor,
but it has a finite immediate successor ⟦0,k⦆∪{k}
and it isn't the last finite ordinal.
A last finite ordinal doesn't exist.
Observation of finite ordinals did not contribute to
the determination of that fact.