Re: Seven deadly sins of set theory

On 02/07/2024 09:59 AM, Jim Burns wrote:

On 2/6/2024 11:15 PM, Ross Finlayson wrote:

On 02/06/2024 02:40 PM, Jim Burns wrote:

On 2/6/2024 4:53 PM, Ross Finlayson wrote:

>

You do seem a bit fresher today,
I can sort of understand why you'd look at
the least-upper-bound appearing in the set
you imagine the sequences are from,

>
It's a theorem.
I can give you as much detail of the proof
as you'd like. More than you'd like, I'll bet.
>
But it seems as though
we're still talking past one another.
I say "non.trivial edgeless.foresplits of Q"
You say it's not conscientious (honest?) of
me to say that about _the rationals_
Have I misunderstood you?
>
I haven't said that about the rationals.
The non.trivial edgeless.foresplits of Q
aren't Q'

>
Are they not, any one, a partition of Q?

>
Exactly.
A partition of ℚ isn't an element of ℚ
>
----
Let's change tack.
>
We start by assuming there exists
ℝ the complete ordered field.
Then we work our way around to
no longer needing to assume that.
>
For each point x ∈ ℝ
there is a partition {Fᴿₓ {x} Hᴿₓ}
Fᴿₓ = {y∈ℝ|y<x}
Hᴿₓ = {y∈ℝ|y<x}
>
There is a bijection between points and partitions.
An order.isomorphism, in fact.
x ⟷ {Fᴿₓ {x} Hᴿₓ}
Moreover,
x ⟷ Fᴿₓ
x ⟷ {x}
x ⟷ Hᴿₓ
>
Fᴿₓ is a non.trivial edgeless.foresplit of ℝ
Because of the isomorphism, we could choose to
discuss non.trivial edgeless.foresplits
instead of points. The results will be the same.
>
The set {Fᴿ} of non.trivial edgeless.foresplits of ℝ
is defined on ℝ
but {Fᴿ} isn't the model of ℝ
>
There is also the set {Fꟴ} of
non.trivial edgeless.foresplits of ℚ
{Fꟴ} is the model of ℝ
>
{Fꟴ} can be a model of ℝ because
there is an isomorphism between {Fꟴ} and {Fᴿ}
>
For each Fᴿ ∈ {Fᴿ}
exists Fꟴ = Fᴿ∩ℚ ∈ {Fꟴ}
>
For each Fꟴ ∈ {Fꟴ}
exists Fᴿ = {x∈ℝ| ∃p∈Fꟴ: x<p} ∈ {Fᴿ}
>
That's what we want because,
for each two points x,y in ℝ
there is a rational p in ℚ between them.
>
Fᴿ is edgeless.
for each x ∈ Fᴿ
exists y ∈ Fᴿ: x < y  and
exists p ∈ Fᴿ∩ℚ: x < p < y
>
There are no points in Fᴿ
which are not in {x∈ℝ| ∃p∈Fꟴ: x<p} ∈ {Fᴿ}
>
Fꟴ ⟼ {x∈ℝ| ∃p∈Fꟴ: x<p}
is the inverse of
Fᴿ ⟼ Fᴿ∩ℚ
>
{Fᴿ} ⟷ {Fꟴ}
>
Also,
ℝ ⟷ {Fᴿ}
>
ℝ ⟷ {Fᴿ} ⟷ {Fꟴ}
>
{Fꟴ} has all the properties which
we require ℝ to have
because, up to isomorphism,
{Fꟴ} is ℝ
>
However,
we don't need to assume {Fꟴ} exists
the way I did up.post.
{Fꟴ} is a set of sets of rationals.
>
If
we assume or prove that
ℚ and 𝒫(ℚ) and {S ⊆ ℚ| F(S)} exist
F == "is an edgeless.foresplit"
then
{Fꟴ} exists and
the complete.ordered.field axioms for ℝ
are not of the contradictory kind.
>
>

>
No thanks, I start with an integer continuum,
then build the non-integer parts connecting
them as line-reals each, then already have
a continuous domain, then it can result that
as Eudoxus fills out any definable value as
of a point on a line modeling a continuous domain,
then this way there always exists that.
>
This way I wouldn't feel bad about
"assuming that which is set out to be shown".
>
Yeah when considering only the usual model of
the rationals the ordered field, that is not complete,
and besides looking at it as "non-Archimedean" and
as so sort of completing itself in the "non-Archimedean",
there are only rationals, and the Archimedean field,
if not the "complete Archimedean field" where that's a thing,
the ordered field Q does not have the least-upper-bound
property, which is about the most usual axiom in the
non-logical part of descriptive set theory that is
among usual models of reals the complete ordered field.
>
It's pretty well-known that LUB and "measure 1.0",
are axioms, in usual standard descriptive set theory.
>
>
I see you're showing something that's kind of consistent
within itself, but, if you remove the scaffold, then,
there wasn't one.
>
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