Re: Does the number of nines increase?

On 7/1/2024 11:19 PM, Ross Finlayson wrote:

On 07/01/2024 06:10 PM, Jim Burns wrote:

What that means is that
I think
theory is a strong mathematical platonism,
it matters what is _attained_ to,
or that to which we _attain_,
the "true" objects of
a universe of mathematical objects,
"a" universe, then with regards to descriptions

"descriptions"

there's that
the "extent density completeness measure" provide
"extent density completeness"
which you would agree that
"extent density completeness" makes for
satisfying the IVT.
>
Then, as I mentioned,
there's a theory,
in all the universe of theories,
all the abstract and contingent and fanciful and
practical and otherwise,
one of which is "the true theory",
that among those,
there's one where it appears that
"it is so" is an axiom.
>
So, given that
you won't accept that via inspection,

"inspection"

that a least-upper-bound is given and
also a sigma algebra is given,
given that extent and density are givens,
then,
given that it's axiomatic,

"axiomatic"

and, doesn't contradict the ordinary
because
it just makes for the "only-diagonal" contra
the "anti-diagonal",
then, how's that.
>
Good sir, ....

Do I understand you correctly?
You have declined my invitation to say
what your symbol.string  n/d: 0≤n≤d: d → ∞  means
because
you consider what you've told me to have answered me:
n/d: 0≤n≤d: d → ∞  satisfies IVT
n/d: 0≤n≤d: d → ∞  is countable
etc.
You have defined it so.
Do you realize that?
Definitions are two.edged swords.
They grant you unrestricted power,
but only inside the area of what.you.mean
and outside of that, no power.
If what.you.mean by  n/d: 0≤n≤d: d → ∞  is that
n/d: 0≤n≤d: d → ∞  satisfies IVT
n/d: 0≤n≤d: d → ∞  is countable
etc.
then, okay, you can define it so, but
defining it so doesn't mean it exists.
Those proofs which
  you think  n/d: 0≤n≤d: d → ∞  disproves
prove that  n/d: 0≤n≤d: d → ∞  doesn't exist.