Re: Seven deadly sins of set theory

On 02/04/2024 02:46 PM, Jim Burns wrote:

On 2/4/2024 2:44 PM, Ross Finlayson wrote:

On 02/04/2024 11:22 AM, Jim Burns wrote:

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Consider only partitions of rationals,

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I would prefer to consider only
non.trivial edgeless.foresplits of the rationals,
{F ⊆ ℚ| F ᣔ<ᘁ F ᣔ<ᣔ ℚ\F}\{∅,ℚ}
which is what I am claiming is a model of the reals.
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The first important difference I see
is that
non.trivial edgeless.foresplits of the rationals
are totally.ordered by '⊆'  and
partitions.more.generally of the rationals
{F,H}: F∩H = ∅  ∧  F∪H = ℚ
aren't.
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If you intend to prove that
_partitions_ of ℚ don't model ℝ
then I wish you all the best in your endeavor,
but that's not my claim.
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Consider only partitions of rationals,
because of
the density of the rationals in the reals, and,
the fact that the rationals are
not gapless/complete/having-LUB,
establishing for a given element that
a given partition has a lesser and a greater side,

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That is not established for partitions.
However, that is established for
non.trivial edgeless.foresplits of the rationals
and their complement.
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each lesser lesser than each greater and vice-versa,
these partitions must be as by

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"must be as by" == ?
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these partitions must be as by
the complete ordered field or R itself already.

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I'm not sure how to regard "already" in this context.
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It is true that
non.trivial edgeless.foresplits of the rationals
"already" model ℝ
in that they are not time.bound.
They have always, do now, and will always model ℝ
Thus, they "already" model ℝ
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It is false that
ℝ existing is a prerequisite to
non.trivial edgeless.foresplits of the rationals
existing.
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ℚ yes
𝒫(ℚ)yes
{F ⊆ ℚ| F ᣔ<ᘁ F ᣔ<ᣔ ℚ\F}\{∅,ℚ} yes
ℝ no
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I understand that you wouldn't say that,
because,
the rationals are looking HUGE, but,
it seems ignorant,
then once informed, it seems hypocritical.

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| I beseech you, in the bowels of Christ,
| think it possible that you may be mistaken.
|
-- Oliver Cromwell.
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However huge the rationals are,
the sets of rationals are provably huger.
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It is _sets of rationals_ and not rationals
which are the "points" in
what is provably a model of ℝ
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A model might not hold
what you think it should hold.
However, if it satisfies the axioms,
it is a model.
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So, how to arrive at this
when the rationals are everywhere dense, in the reals,
but nowhere gapless, in the reals,
so clearly don't have least-upper-bound property,

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It is selected sets of rationals, not rationals,
which have the least.upper.bound.property.
A non.trivial edgeless.foresplit of the rationals
is a set of rationals, not a rational.
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I'll agree that
sets of rationals _have_ a least upper bound,
then immediately demonstrate that's because
the complete ordered field has least upper bound,

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That isn't the least upper.bound property.
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{1,2,3,4,5,6,7,8,9} has
the least.upper.bound property because
each of its bounded, non.empty subsets
has a least.upper.bound
in {1,2,3,4,5,6,7,8,9}
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ℝ is a much more interesting example,
but {1,2,3,4,5,6,7,8,9} is also an example.
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ℚ doesn't have the least.upper.bound property.
{F ⊆ ℚ| F ᣔ<ᘁ F ᣔ<ᣔ ℚ\F}\{∅,ℚ} has it.
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So, let's talk about this relation:  '^<^'.

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It is an abbreviation of "foresplit"
F ᣔ<ᣔ ℚ\F  :⇔
∀x ∈ F: ∀y ∈ ℚ\F:  x < y
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Also too, an abbreviation of "edgeless" is:
F ᣔ<ᘁ F  :⇔
∀x ∈ F: ∃x′ ∈ ℚ\F:  x < x′
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For what it's worth,
ᣔ is superscript AND
ᘁ is superscript OR
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Also, you mentioned the usual, ..., understanding of that
"the complete ordered field is unique up to isomorphism",
then I wonder if you recalled when I wrote field operations
as a limiting case for the interval [-1, 1], or,
"field operations equipping the unit interval with
being a field", and as with
regards to how it's a bit singularly different.

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The complete ordered field satisfies
the complete.ordered.field axioms.
Something.a.bit.different which doesn't  isn't.
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We know that
the complete ordered field is
unique up to isomorphism
because,
given two models, let us say
the non.trivial edgeless.foresplits of ℚ
and the partitions of Cauchy sequences of ℚ
an isomorphism between them exists.
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_Any_ model of ℝ satisfies those axioms,
and it is from the axioms that
the isomorphism is proven to exist.
Thus, unique up to isomorphism.
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That <=, l.t.e. arithmetically and, you know, <= set-wise, subset,
have that the sub-set relation only exists because l.t.e., already.
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If it's "less than, a given rational",
or "less than, a given irrational",
or "less than, a given real",
those are various I'd imagine you'd agree.
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It's sort of funny the "edgeless" and "gapless", together.
Vis-a-vis, the edgeless and "the nowhere not having
a neighborhood with a gap".
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Then, there's that it must be the partitioning of the
predicate that indicates membership, in this structure,
is by a member of the complete ordered field already,
where when it's not a rational already that there's
an edgeless.foresplit of rationals in front of it,
as for that there's a distinct and unique one,
i.e. that it has elements that none others do.
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Each of these different edgeless.foresplits, or
partitions by reals, has pair-wise, in their disjoint,
pair-wise unique elements, what distinguishes them
as sets the model, by what distinguishes them as
assignment via extensionality to a real value.
It is well-known that ZF has pair-wise associativity,
but not the illative associativity, for then it's quite
simple to demonstrate with the illative, that, these
would be so many distinct rationals, as there are,
distinct partitions.
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Yeah, I know.  "I can't acknowledge that both the
rationals and irrationals have the same topological
properties with respect to each other, even throwing
in the algebraics, for that I would contradict myself."
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So, deal with it, and figure out how it is so that they
have (all) their own properties as mathematical objects,
so that then you'd have a better theory than that worse one.
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Because, otherwise you'd just be in denial,
and it's considered unconscientious.
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How I deal with it is a bit of book-keeping, to help explain
why there's a notion that involving so many copies of the
integers to result the rationals, makes the rationals in
a formal sense larger.  For example, imagine that all numerators
come from a set T and all denominators from a set B, and
show that both must be infinite to construct a model of
the rationals.  Yeah, I know that with usual countability
and Cartesian functions that this bit of formality
isn't usually part of the work-up, but, there's something
to be said for the apparatus, that, in terms of numerical
resources, it's up-front what it requires, to be a full
model of the rationals.
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So, anyways, for example, let S be any one of Q, P, or R.
Do its elements partition Q?
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Of course the edgeless.foresplits defined by P\Q or R\Q
do _not_ contain their least-upper-bound.
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And, edgeless.foresplits are a subset of the partitions,
while 2^Q is plenty large.
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About the field operations for [-1,1], it was like so.
(Field operations for [-1,1].)
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Consider a function taking [-1,1] to (-1,1) as so.
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f_n(x) = nx, x e [-1,1], n-> oo, ran(f_oo(x)) = (-oo, oo)
g_n(y) = ny, x e (-1, 1), n-> oo, dom(g^-1_oo(g_n(y)) = (-oo, oo)
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The properties of those functions is that they're linear.
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Then, with j = g^-1 o f (or as composed): dom(j) = [-1,1] and ran(j)
= (-1,1).
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With the composition of linear properties then j(x>y) > j(y), and
etcetera.
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Then, let h_n(a) = na, n->oo, dom(h(a)) = (-1,1), ran(h(a)) = (-oo,
oo), h is linear as above.
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Then a + b is h^-1( h(a) + h(b) ). And, a * b is h^-1( h(a) *
h(b) ).
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Here a, b e dom(h) with +, * defined s above, h(b), h(a) e ran(h),
with +, * as in R.
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c * (a + b)
= h'( h(c) * h( h'(h(a) + h(b)) ))
= h' ( h(c) * (h(a) + h(b)) )
= h' ( h(c) * h(a) + h(c) * h(b) )
= h' ( h(c) * h(a) ) + h' ( h(c) * h(b) )
= c * a + c * b
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Multiplication distributes over addition, and addition and
multiplication are closed as sums and products of h(x) are in (-oo,
oo).
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Via symmetry, additive inverses exist and there's an additive
identity, zero.
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Then, is there a multiplicative identity and inverse. There is a
mulitplicative inverse of h(a) in [0,oo) in R with the natural
definition of multiplication, and, with the continuous and order-
preserving mapping from [0,1), for any a with 0 < a < 1, there exists
1/h(a) in R+, in ran(h), and: h' ( h(a) * 1/h(a) ) e R[0,1).
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Basically: "A definition of field operations on the unit interval."
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https://groups.google.com/g/sci.math/c/4RBNLj-Q4Mo/m/Nn3T7v_rCnIJ
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