Re: Contradiction of bijections as a measure for infinite sets

On 4/4/2024 11:28 PM, Ross Finlayson wrote:

On 04/04/2024 12:01 PM, Jim Burns wrote:

On 4/3/2024 10:13 PM, Ross Finlayson wrote:

Iota-values:
the word "iota" means "smallest non-zero value".

>
That which you use iota to describe
is not the continuum.

Real-values:
 all the values between negative infinity and infinity.

 There are several ambiguities in that description.
 How about instead
Real.values:
least.upper.bounds of
bounded.non.empty.sets of
differences.of.ratios of
ordinals not.fitting.predecessors.

It's simple that the continuum limit, a limit of functions,
as we used to say modeling a function as a limit of a family
of functions, like for Dirac delta, these days it's often
called a "generalized distribution", such a function, with
its real analytical character, here the Equivalency Function
is unlike the Dirac delta in that it's not just an infinite
spike at the origin only under which is area one, while,
it is integrable, which is particularly unique for a function
from a discrete domain, and under it is area one, thus that
it's a generalized distribution if you will, while also it's
a continuum limit with the usual meaning of the words.
 Then, that its range has extent, density, completeness, measure,
particularly completeness and measure, in [0,1], establishes
its range is a continuous domain, that instead of one or
the other of line-reals or field-reals, there are both.

If I recall correctly, your claim is that
lim[n→∞, n∈ℕ] {d/n: 0≤d≤n, d∈ℕ} =
[0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ = {x∈ℝ: 0≤x≤1}
Define
[0,1]ᴿꟳ =
lim[n→∞, n∈ℕ] {d/n: 0≤d≤n, d∈ℕ}
How do we know ⅟√​̅2 ∈ [0,1]ᴿꟳ ?
Using my upthread definition of ℝ
define
[0,1]ⁿᵒᵗᐧᴿꟳ =
{ x = lub bnes ⊆ ℚ: 0≤x≤1}
lub least.upper.bound
bnes bounded.non.empty.set
How we know ⅟√​̅2 ∈ [0,1]ⁿᵒᵗᐧᴿꟳ is that
⅟√​̅2 = lub bnes {p: p² < ⅟2 } ⊆ ℚ
By the upthread definition,
[0,1]ⁿᵒᵗᐧᴿꟳ = [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
Define
[0,1]ˡⁱᵐᐧˢᵘᵖ =
⋂{ ⋃{{d/q: 0≤d≤q, d∈ℕ}: q≥n, q∈ℕ}: n∈ℕ}
For each n ∈ ℕ,
each s ∈ ℕ⁺ has a multiple q > n
∀n∈ℕ, ∀s∈ℕ⁺, ∃q∈ℕ: s | q > n
∀n∈ℕ:
⋃{{d/q: 0≤d≤q, d∈ℕ}: q≥n, q∈ℕ} =
{d/s: 0≤(d/s)≤1, (d/s)∈ℚ} =
[0,1]ʳᵃᵗⁱᵒⁿᵃˡ
[0,1]ˡⁱᵐᐧˢᵘᵖ =
⋂{[0,1]ʳᵃᵗⁱᵒⁿᵃˡ: n∈ℕ} =
[0,1]ʳᵃᵗⁱᵒⁿᵃˡ
We know ⅟√​̅2 ∉ [0,1]ʳᵃᵗⁱᵒⁿᵃˡ = [0,1]ˡⁱᵐᐧˢᵘᵖ
Thus, [0,1]ˡⁱᵐᐧˢᵘᵖ ≠ [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
However,
[0,1]ˡⁱᵐᐧˢᵘᵖ uses a very.widely.used, very.sensible
definition of a limit of a set.sequence.
https://en.wikipedia.org/wiki/Set-theoretic_limit
Your definition
lim[n→∞, n∈ℕ] {d/n: 0≤d≤n, d∈ℕ}
would need some other definition,
a definition which I have not seen you (RF) give.
What is the definition of limit which you use?
Or, is it instead that
[0,1]ᴿꟳ ≠ [0,1]ᶜᵒⁿᵗⁱⁿᵘᵘᵐ
?