FOM: Nonstandard analysis -- reply to Percival
Joe Shipman
shipman at savera.com
Mon Jun 12 10:18:54 EDT 2000
Percival:
>>I need to know whether non-standard analysis affords a way of dealing
with infinite products of reals (in the unit interval).<<
An interesting question.
First of all, you can add a logarithmic function to your structure
before taking an ultrapower, so that you can have a nonstandard version
of the real field which includes logarithmic and exponential functions.
However, in the usual Robinson-style constructions of "nonstandard"
extensions of the real field, you don't get convergence of Cauchy
sequences so you can't define countably infinite sums straightforwardly,
therefore arbitrary countably infinite products are also problematic (if
you could do sums you could do products by exponentiating the sum of the
logs).
But you seem to need something less general than an arbitrary product,
since in your examples you only ask for an omega-power (a countably
infinite product where all the multiplicands are the same). The
two-argument exponential function x^y can be defined as exp(y log x) and
will have the expected first-order properties, so that 0.5^omega will be
< 0.6^omega.
However, in this setup the identification of the ordinal omega with a
particular element of the extended real numbers is arbitrary!
What you really ought to use is Conway's "Surreal numbers". These are
constructed without AC and don't share all the first-order properties of
the reals, but they are complete (Cauchy sequences converge and you can
define infinite products) and there is also a natural definition of the
logarithmic and exponential functions (due to Simon Norton -- you first
define integration and then integrate 1/x), and omega the ordinal is
naturally identified with omega the "surreal".
See Conway's remarkable "On Numbers and Games" (for many years, this was
one of two math books I always had with me for inspirational reading
when I had free time; the other was Cohen's "Set Theory and the
Continuum Hypothesis".)
-- Joe Shipman
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