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

On 11/4/2024 2:39 PM, Ross Finlayson wrote:

On 11/04/2024 03:52 AM, Jim Burns wrote:

On 11/2/2024 6:01 PM, Ross Finlayson wrote:

On 11/02/2024 12:37 PM, Jim Burns wrote:

On 11/2/2024 2:02 PM, Ross Finlayson wrote:

The delta-epsilonics of course,
or some put it "delta-epsilontics",
with little d and smaller e,
of often for induction arbitrary m and larger n,
is well-known to all students of calculus.
"The infinitesimal analysis", ....

>
The delta.epsilonics well.known to students of calculus
is not infinitesimal analysis.
For δ > 0 and ε > 0
there are _finite_ j and k such that
δ > ⅟j > 0 and ε > ⅟k > 0

Sure it is,
the delta-epsilonics is well known,

For δ > 0
there is finite j such that δ > ⅟j > 0
⎛ Assume otherwise.
⎜ Assume, for δ > 0, that
⎜ no ⅟j exists: δ > ⅟j > 0
⎜ δ is a lower bound of ⅟ℕ
⎜ ⅟ℕ = {⅟i: i ∈ ℕ⁺ ∧ i finite}
⎜
⎜ Let β be the greatest lower bound of ⅟ℕ
⎜ β ≥ δ > 0
⎜ 2⋅β > β > ½⋅β
⎜
⎜ ⅟ℕ ᵉᵃᶜʰ≥ β > ½⋅β
⎜ ⅟ℕ ᵉᵃᶜʰ≥ ½⋅β
⎜ ½⋅β is a lower bound of ⅟ℕ  [1]
⎜
⎜ If 2⋅β is a lower.bound of ⅟ℕ
⎜ then 2.β > β is a greater.than.β lower bound of ⅟ℕ
⎜ which is not a thing.
⎜ 2⋅β is not a lower bound of ⅟ℕ
⎜
⎜ 2⋅β is not a lower bound of ⅟ℕ
⎜ exists ⅟k ∈ ⅟ℕ: ⅟k > 2⋅β
⎜ exists ¼⋅⅟k ∈ ⅟ℕ: ¼⋅⅟k > ½⋅β
⎜ ½⋅β is not a lower bound of ⅟v ℕ
⎜
⎜ However,
⎜ [1]  ½⋅β is a lower bound of ⅟ℕ
⎝ Contradiction.
Therefore,
For δ > 0
there is finite j such that δ > ⅟j > 0

What I'm saying is that since antiquity,
it is known,
that there are at least two models of continuity,
and you may call it Archimedean and Democritan,
about the field of rationals versus atomism,
and that infinitesimal analysis includes both.

Calculus class.
Complete ordered field.
Delta.epsilonics.
No infinitesimals.

So, infinitesimal analysis includes delta-epsilonics,
if not the other way around.

Infinitesimal analysis without infinitesimals.
So creative!