Re: The infintely small number b

On Nov 17, 6:33 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:
On Nov 18, 3:44 am, lwal...@xxxxxxxxx wrote:
I'm not quite sure why VR defines b / b as 1, but
leaves b - b undefined
Thanks for noticing. Thats an error and I've posted a correction
immediately. b/b is undefined too.

Now with VR's correction, there's another way to interpret
what VR is writing, using either hyperreals or surreals this time.

Here 0 is zero of course, b is a strictly positive infinitesimal,
n is a positive number that is neither infinite nor infinitesimal
(some nonstandard analysts call such values "appreciable"),
and Inf is a positive infinite value.

We now look at VR's rules again:

Operations with zero (non-existence of extent)

b + 0 = b
An infinitesimal plus zero is still infinitesimal.

b - 0 = b
An infinitesimal minus zero is still infinitesimal.

b * 0 = 0
An infinitesimal times zero is zero.

(b / 0) is undefined and not to be taken as inf.
Division by zero is undefined even for infinitesimals.

Operations with itself

b + b = b
The sum of two positive infinitesimals is still a
positive infinitesimal.

b * b = b
The product of two infinitesimals is still infinitesimal.

b/b is undefined too.
The quotient of two infinitesimals may be infinitesimal,
appreciable, or infinite. Indeed, if in the surreals, we
let x = epsilon and y = epsilon^2, then y/x is
infinitesimal, x/x is appreciable, and x/y is infinite.

(b - b) is undefined.
The difference of two positive infinitesimals may be
positive, zero, or negative. Indeed, x-y is positive,
x-x is zero, and y-x is negative.


Operations with n where b<n<inf:
n * b = b
The product of an appreciable number and an
infinitesimal is still infinitesimal.

n / b = inf.
The quotient of an appreciable number and an
infinitesimal is infinite.

(n/0 is undefined and can not be taken as inf.)
Division by zero is undefined, even in the
hyperreals and surreals.

n + b = n + b
n - b = n - b

The sum or difference of an appreciable number
and an infinitesimal is still appreciable.

But of course, we now wonder why VR wants to
distinguish among n+b, n, and n-b.

At this point, we must remind ourselves what
VR's original intent in coming up with the new
infinitesimal b in the first place. Although VR has
not stated his purpose in this thread, in other
threads, he points out that he wants to assign
measures to sets such that proper subsets of
bounded intervals have strictly smaller measures.

VR hasn't told us, but I can make the educated
guess that he wishes to assign measures to
intervals as follows:

Closed interval [x,y] has measure y-x+b.
Semiopen interval [x,y) has measure y-x.
Open interval (x,y) has measure y-x-b.

The degenerate interval [x,x], which is actually the
singleton {x}, would have measure b. The empty
set would have measure 0. And of course, an
unbounded interval would have measure inf.

Although VR achieves his purpose of assigning
strictly smaller measures to proper subintervals of
bounded intervals, it doesn't necessarily assign
strictly smaller measures to proper subsets that
are not intervals. Indeed, notice that:

[x,y] has measure y-x+b
(x,y) has measure y-x-b
[x,y] \ (x,y) = {x,y}
{x,y} has measure (y-x+b) - (y-x-b) = b+b.

By VR's rules above, b+b = b. So the set with
two elements, {x,y}, would have the same measure,
namely b, as its singleton subsets. The problem,
of course, is that VR has not demonstrated how to
find measures of sets that are not intervals.

We know that non-Lebesgue measurable sets
exist (in ZFC, including the Axiom of Choice). I
wonder whether non-VR measurable sets exist.
.

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