Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

On 5/8/2024 8:27 PM, Moebius wrote:

Am 08.05.2024 um 23:55 schrieb Jim Burns:

Hint: This formula is
>

∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ). [WM]

>
nonsense.
("~∃^ℵo" is not the problem.
It just means that
"there aren't countably many" ...).

No problem other than being wrong,
I think you mean.
Both quantifiers use 'n' but
they are different 'n' with different scope.
| ∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ)
is
| ~∃^ℵo n ∈ (ℝ\ℕ)
I haven't guessed whether
WM intends them to be different or the same.
I decided I didn't need to guess:
either way it doesn't say
the sort of thing he's been saying lately.

Simplified:
>
| ∀n ∈ ℕ: ~∃ n ∈ (ℝ\ℕ).
>
Does that make any sense to you?

No, no sense.
 From my having read many, many posts from WM,
I've decided that WM wants to write
_the opposite_ of the previous statement,
but with his darkᵂᴹ numbers moved to ℝ
What he wrote doesn't actually say
anything close to that, but this is
how I've been exchanging views with WM,
for several years now.
I have found that,
in cases in which I guess wrong,
WM is quick to tell me I'm wrong.