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On 10/28/2024 5:45 PM, Ross Finlayson wrote:That ran(EF) is complete is rather simple, i.e. it's trivial,On 10/28/2024 01:15 PM, Jim Burns wrote:>>[...]>
Of course
you can keep in mind that "continuum limit"
...AKA letting lattice spacing approach 0...
>was not admitted to mathematics>
since modern mathematics
since at least for the past few decades,
since when for example
"everybody's favorite non-standard
not-a-real-function with standard analytical character,
function,
Dirac's delta the unit impulse function",
has of course that EF is a
non-standard and not-a-real-function,
in the usual sense,
yet has real analytical character itself.
I thought that I at least knew what it was
which you are talking about. Not so much, now.
>
Note, on the Dirac delta not.a.function:
It does not have a single non.0 value
such that it integrates to 1
>
That is an informal gloss of the not.a.function.
A useful intuition.builder, but
the existence of the Dirac delta doesn't prove
that a thick point like that exists.
>
----Well, the property that>
"a set is complete
if it contains each of its least-upper-bounds",
...if it contains,
for each of its bounded nonempty subsets,
a least upper bound...
>is completeness,>
is the one ascribed to line-reals,
ran(EF).
It just clicked with me that
you (RF) have been saying "range" and
I (JB) have been thinking "image".
My mistake.
>
Perhaps you should be saying "image".
>
It is the image EF(N) which is countable,
assuming a single.valued ("Cartesian"?) function.
>
The range is a superset of the image.
It might or might not also be countable.
>
You can't use the range to argue against
Cantor's uncountability proof.
Having a countable subset doesn't prove
countability,
>It's an upper-bound, it's least, rather trivially>
as either a finite set contains its upper-bound,
or, a finite set has a next-greater upper-bound,
one or the other of those is discernible and
one or the other of those exists and
one or the other of those is an upper-bound and
one or the other of those is least,
thusly the least-upper-bound "LUB" property holds,
completeness.
You have completely lost me.
>as either a finite set contains its upper-bound,>
or, a finite set has a next-greater upper-bound,
A finite nonempty set contains its upper bound.
So, yes,
a finite set has the least.upper.bound property.
Was it finite sets you've been talking about?
>
A finite set has the LUB property, yes,
but a finite set isn't a superset of Q
The smallest superset of Q with the LUB property
is R, which is what I've been talking about.
>
>
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