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On 10/3/2024 12:49 PM, Ross Finlayson wrote:Well it's pretty simple that in mathematicsOn 10/03/2024 09:14 AM, Jim Burns wrote:>On 10/2/2024 10:01 PM, Ross Finlayson wrote:>Such casual extensionality as>
"these Q's are those Q's"
is certainly usual,
Since we aren't being lax here,
you should say
⎛ Such casual isomorphistry as
⎜ "these Q's are those Q's"
⎝ is certainly usual,The "up to isomorphism" like "the reals are>
the complete ordered field 'up to isomorphism'",
does _not_ have that R = C, and indeed it is
so that R =/= C, so,
no, we do not say "isomorphistry".
ℝ is not isomorphic to ℂ
ℚ₀ is isomorphic to ℚₛ
>
The casual isomorphistry of
ℚ₀ is (implicitly.isomorphic.to) ℚₛ
is usual.
>Having the properties of a complete ordered field,>
and, being analytic under continuous functions,
then, maintaining analyticity under
transforms of continuous functions,
is _not_ something that R and C both have,
Yes. "not".
>so, the usual idea that>
"the reals R are unique up to isomorphism
the complete ordered field",
gets broken
The usual idea of ℝ unique.up.to.isomorphism
does not involve ℂ
because ℂ is not the complete ordered field.
>
( ℝ is unique.up.to.isomorphism
==
⎛ each two models M M′ of
⎜ the axioms for the complete ordered field (ℝ)
⎝ have a '+-×÷<'.preserving bijection.
>
Map identities to identities′.
Map integers to integers′.
Map rationals map to rationals′.
Map least.upper.bounds to least.upper.bounds′.
_Because_ M and M′ are both complete ordered fields,
there exists an isomorphism between them.
>and is not considered thorough.>
Who is it who is doing this considering?
Are they unaware of the proofs?
Do they consider proofs not thorough?
>The extensionality is>
the term of model theory with regards to that
a) it's assumed that models are faithful, and
b) it's assumed that models are equivalent
in all interpretations, thusly
c) model Q_1 and Q_2 each a, b are
extensionally equivalent and considered equals,
while of course
no properties of their structure that isn't
equi-interpretable and bi-relatable
is considered true for both, and
such aspects of their consideration are
with regards to a different theoretical object.
>
About "isomorphistry", getting into stuff
like "equals, 'almost' everywhere", has
that: "equals, a.e., is _not_, 'equals'".
And "equals everywhere" _is_ "equals".
>And R, or R^2, and C, "up to isomorphism",>
are _not_ equals.
ℚ₀ from which ℝₛ is constructed and
ℚₛ a subset of ℝₛ
_are_ equals.
>
How does someone make a claim to you (RF) that
is _not_ subject to being carried off and
put to work in foreign contexts?
>
>
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