Re: Replacement of Cardinality

On 8/2/2024 11:36 AM, WM wrote:

Le 01/08/2024 à 20:12, Jim Burns a écrit :

∀ᴿx > 0:  NUF(x) ≥ ℵ₀

>
This nonsense will not become true by repeating it.

It does not become false by deleting it.
(0,x] inherits from its superset (0,1] properties by which,
for ⅟ℕᶠⁱⁿ∩(0,x] finite.unit.fractions in (0,x]
each non.{}.subset is maximummed,  and
each finite.unit.fraction is down.stepped,  and
each finite.unit.fraction in is non.max.up.stepped.
Therefore,
the finite.unit.fractions in ⅟ℕᶠⁱⁿ∩(0,x] are ℵ₀.many.
∀ᴿx > 0:  NUFᶠⁱⁿ(x) = ℵ₀
NUF(x) ≥ NUFᶠⁱⁿ(x)
∀ᴿx > 0:  NUF(x) ≥ ℵ₀

ℵ₀ unit fractions need ℵ₀*2ℵ₀ points above zero.

ℵ₀ is the cardinality of all final ordinals.
Final ordinal α is smaller than α∪{α}
For each final ordinal α:  α∪{α} is also a final ordinal.
The cardinality ℵ₀ of final ordinals can't be a final ordinal.
Otherwise,
there would be at least ℵ₀∪{ℵ₀} final ordinals,
which would be more final ordinals than final ordinals.
Therefore,
ℵ₀ is NOT smaller than ℵ₀∪{ℵ₀}

For those x > 0 your claim is wrong.

Finite doesn't need to be small.
Infinite is beyond big.