Re: Does the number of nines increase?

On 7/13/2024 11:43 AM, WM wrote:

Le 12/07/2024 à 22:23, Jim Burns a écrit :

On 7/12/2024 1:04 PM, WM wrote:

says a person who believes that
by exchanging two elements
one of them can be lost.

>
I have never claimed ℵ₀ = 2

>
You have claimed that
by exchanging X and O an O can disappear,

Yes.
f(j)  :=  (j=Bob ? 1 : j⁺¹)
(Perl ternary conditional operator)
f: ℕ₁⁺ᴮᵒᵇ → ℕ₁: 1.to.1
Bob ∉ f(ℕ₁⁺ᴮᵒᵇ)

in fact infinitely many can disappear by pure exchange.

Yes.
g⟨j,k⟩  :=  ⟨ (j+k-1)(j+k-2)/2+j, 1 ⟩
g: ℕ₁×ℕ₁ → ℕ₁×{1}: 1.to.1
∀⟨j,k⟩ ∈ ℕ₁×(ℕ₁\{1}):  ⟨j,k⟩ ∉ g(ℕ₁×ℕ₁)

The set ℕ⁺ᴮᵒᵇ of
cardinals which, when augmented, are bigger plus Bob
has the same cardinality ℵ₀ as

>
Cardinality is nonsense as my proof shows.
There is no counting but only exchanging.

There are
f(j)  :=  (j=Bob ? 1 : j⁺¹)
and
g⟨j,k⟩  :=  ⟨ (j+k-1)(j+k-2)/2+j, 1 ⟩

You cannot consistently argue with
rules which are disproved.

The Disappearing.Bob Proposal:
⎛ Equi.membered sets are the same set.
⎜ Empty set ∅ exists.
⎝ Adjunct X∪{Y} exists for existing sets X and Y
 From the Disappearing Bob Proposal,
it follows not.first.false.ly
that familiar natural.number arithmetic
  is consistent
that functions f(j) and g⟨j,k⟩ which, respectively,
  disappear Bob and ℕ₁×(ℕ₁\{1})
  exist.
You have the definitions of f and g in front of you.
Your argument against believing your lying eyes is that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ don't act like finiteⁿᵒᵗᐧᵂᴹ sets.
You crow that I and others don't understand
your argument that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ don't act like finiteⁿᵒᵗᐧᵂᴹ sets.
I and others understand your argument that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ don't act like finiteⁿᵒᵗᐧᵂᴹ sets.
What we also which  understand and you (WM) don't is that
ℕ₁⁺ᴮᵒᵇ and ℕ₁×ℕ₁ aren't finiteⁿᵒᵗᐧᵂᴹ sets,  and
not.acting like they are is appropriate to their nature.