FOM: Reply to Friedman's ReplyToDavis

[Vaughan R. Pratt]

>From: martind at cs.berkeley.edu (Martin Davis)
>I always felt there was something mysterious about the "construction" of the
>real numbers. In, for example, Dedekind's version, one does technical
>fussing about the rational cuts (does the rational point belong in the left
>or right class), fusses about how to deal with negative cuts, and how to
>define the arithmetic operations, and then, lo and behold, we have an
>ordered field. Nonstandard analysis clarifes this. In an enlargement, we
>have Q* (where the rationals are Q) and then the reals are seen as the
>quotient ring of the finite elements of Q* modulo the (maximal) ideal of the
>infinitesimals. The completion of the rationals to form the reals is then
>seen as one example of the use of enlargements for sustaining limit
>operations, and the field properties as seen as just an instance of the
>(foundational!) transfer principle.

I agree with Martin's intuition that closing up the gaps that arise in
Dedekind's construction has something of a kludgy feel to it, although
I don't fully agree with his specific concerns.   (For example if you
close up the gaps by identifying their endpoints instead of removing
one of them you avoid the arbitrariness.)  The cardinality mismatch
bothers me more, that the continuum many cuts have only countably many
gaps to close up.

But there is an intermediate possibility that seems to me just as clean
as the above quotient.  Simply quotient the ring of Cauchy sequences in
Q by its (maximal) ideal of vanishing sequences, and viola, an ordered
field.

What does the NSA jackhammer have that this tackhammer doesn't?

Incidentally, for those that like constructions of the reals by
quotients, here's one I came up with that skips the rationals
altogether and goes straight from the integers to the reals via binary
arithmetic.  Take the group (not ring) of bounded integer sequences S
under pointwise addition, and equip it with an operation S/2 defined as
prefixing 0 to the sequence (so this operation applied to
3,1,4,1,5,9,... yields 0,3,1,4,1,5,9,...).  (Motivation: we want
3,1,4,1,5,9,...  to turn out to be one of many representations of the
real 3 + 1/2 + 4/4 + 1/8 + 5/16 + 9/32 + ..., somewhat bigger than \pi
of course, and its shifted version to represent half of it.)
Quotienting this group by the relator S/2 + S/2 = S turns out to
identify every sequence with a sequence all but the first digit of
which is 0 or 1, and which does not converge to 1.  For these canonical
sequences, the head is the floor of the real x it denotes while the
tail is x mod 1 in binary.

To see that this has even a prayer of working, note that

  1,0,0,0,... + 0,1,1,1,... = 1,1,1,1,...  		(group)
			    = 0,1,1,1,... + 0,1,1,1,... (relator)

	whence 1,0,0,0... = 0,1,1,1,... (subtract 0,1,1,1,... from both sides)

Similar reasoning deals with the remaining bounded sequences that
aren't canonical.  One then has to prove that it's an ordered field
(but you knew that already :).

Vaughan