Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

On 10/30/2024 1:36 AM, Ross Finlayson wrote:

On 10/29/2024 08:15 AM, Jim Burns wrote:

I find the term "number" problematic until
the description of _which_ numbers is given.

Most people observe an extensionality in
the "real-valued", then for example somebody like
Hardy also relates real-valued values to points
on a line.

If you want "number" to refer to one of
the smallest superset of ℚ with the LUB property (ie, to ℝ)
then I'm okay with that.
I'm only asking that it be clear that's what it refers to.

When you say "finite ordinals"
what's intended is
"all the body of structural relation
with regards to the ordinal and
with regards to the finite",
and it's doens't necessarily say so much about
"the numbers" or even "the natural numbers",
as with regards to that Peano's numbers aren't
merely zero and a closure to successor,

'Finite ordinal' doesn't mean
merely zero and a closure to successor.
A finite ordinal and each prior ordinal
either has a predecessor or is 0
That makes induction necessary.
⎛ Assume  P(0) ∧ ∀ᴼᶠⁱⁿj: P(j)⇒P(j+1)
⎜
⎜ Consider the set {n∈𝕆ᶠⁱⁿ:¬P(n)} of counter.examples.
⎜
⎜⎛ Assume {n∈𝕆ᶠⁱⁿ:¬P(n)} is non.empty.
⎜⎜
⎜⎜ ∃i = first.in.{n∈𝕆ᶠⁱⁿ:¬P(n)}
⎜⎜
⎜⎜ i ≠ 0 because P(0)
⎜⎜ 0 isn't a counter.example.
⎜⎜
⎜⎜ i is finite
⎜⎜ i-1 exists and P(i-1)
⎜⎜ because i = first.in.{n∈𝕆ᶠⁱⁿ:¬P(n)}
⎜⎜
⎜⎜ However
⎜⎜ P(i-1)⇒P((i-1)+1)
⎜⎜ P(i)
⎜⎜ i isn't a counter.example.
⎜⎝ Contradiction.
⎜
⎜ Therefore {n∈𝕆ᶠⁱⁿ:¬P(n)} is empty
⎜ {n∈𝕆ᶠⁱⁿ:P(n)} = 𝕆ᶠⁱⁿ
⎝ ∀ᴼᶠⁱⁿk: P(k)
Therefore,
if P(0) ∧ ∀ᴼᶠⁱⁿj: P(j)⇒P(j+1)
then ∀ᴼᶠⁱⁿk: P(k)
  AKA induction.

there's constant monotone increase and modularity,
and some have it's only where the operations of
the addition and division, semi-groups, come together.

Given induction and the bit of set theory already used,
we can define addition and multiplication so that
we have the Peano numbers.