Re: How many different unit f [actions are lessorequal than all unit fractions? (infinitary)

On 10/28/2024 5:45 PM, Ross Finlayson wrote:

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.