Re: Incompleteness of Cantor's enumeration of the rational numbers (doubling-spaces)

On 11/12/2024 07:36 PM, Chris M. Thomasson wrote:

On 11/12/2024 6:45 PM, Ross Finlayson wrote:

On 11/12/2024 06:22 PM, Ross Finlayson wrote:

On 11/12/2024 05:38 PM, Chris M. Thomasson wrote:

On 11/12/2024 5:24 PM, Ross Finlayson wrote:

On 11/12/2024 05:02 PM, Chris M. Thomasson wrote:

On 11/12/2024 3:13 PM, Ross Finlayson wrote:

On 11/12/2024 01:36 PM, Jim Burns wrote:

On 11/12/2024 12:40 PM, Ross Finlayson wrote:

On 11/11/2024 12:59 PM, Ross Finlayson wrote:

On 11/11/2024 12:09 PM, Jim Burns wrote:

On 11/11/2024 2:04 PM, Ross Finlayson wrote:

On 11/11/2024 11:00 AM, Ross Finlayson wrote:

On 11/11/2024 10:38 AM, Jim Burns wrote:

>

Our sets do not change.
Everybody who believes that
  intervals could grow in length or number
is deeply mistaken about
  what our whole project is.

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How about Banach-Tarski equi-decomposability?

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The parts do not change.

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any manner of partitioning said ball or its decomposition,
would result in whatever re-composition,
a volume, the same.

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So, do you reject the existence of these?

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No.
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What I mean by "The parts do not change" might be
too.obvious for you to think useful.to.state.
Keep in mind with whom I am primarily in discussion.
I am of the strong opinion that
"too obvious" is not possible, here.
>
Finitely.many pieces of the ball.before are
  associated.by.rigid.rotations.and.translations to
finitely.many pieces of two same.volumed balls.after.
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They are associated pieces.
They are not the same pieces.
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Galileo found it paradoxical that
each natural number can be associated with
its square, which is also a natural number.
But 137 is associated with 137²
137 isn't 137²
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I don't mean anything more than that.
I hope you agree.
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Mathematics doesn't, ....

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Mathematics thinks 137 ≠ 137²
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1 = 1^2
0 = 0^2

[...]
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Don't forget the i... ;^)
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sqrt(-1) = i
i^2 = -1
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?

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Nah, then the quotients according to the
definition of division don't have unique quotients.

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Do you know that any complex number has n-ary roots?
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[...]

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Consider for example holomorphic functions,
where there's complex division, thusly,
it could be a variety.
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https://en.wikipedia.org/wiki/Holomorphic_function#Definition
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People expect unique quotients being all "isomorphic"
to the complete ordered field, it isn't. Complex
numbers _have_ other quotients, real numbers from
the complete ordered field have _unique_ quotients.
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What's left after truncating a piece that exists
fits, though it's kind of amputated. Like, when
Cinderella's step-sister's slipper fit after
she cut her toes off to fit the slipper.
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That any complex-number, has, n-ary roots, ...
Well any number has n-ary roots.
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I think you mean "unity has n'th complex roots".
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There's the fundamental theorem of algebra, ...,
that that says a polynomial of n'th order has n many roots,
that though the multiplicity of roots isn't necessarily 1.
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It's so though that positive real numbers
have unique positive real roots.
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How about "roots of phi", ..., powers of phi are
pretty directly figured, yet, roots, ....
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The, "roots of zero" then is about where it is so
that for some integral equations, it would be, an,
indeterminate quantity, at zero, yet it's still
part of the domain, so, something like zero is
part of the "envelope", of the linear fractional
equation, and Clairaut's equation, and d'Alembert's equation,
and so is x = y = z = ..., "the identity dimension",
an "origin".
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"Roots of Identity"
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n-ary roots a complex number a such that any of the roots when raised
back up by a power, say, n. equal the exact same complex number a. It's
really fun. Actually, it's hyper fun, read all if you get the time:
>
https://paulbourke.org/fractals/multijulia/
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A friend of mine did a little write up on some of my work.

 Yeah you posted this before and I commented about it then.
 So, 1 + i0 ?