Pre-calculus

Pre-calculus

[Peter Freyd]

Starting with the closed interval, regarded as an ordered midpoint
algebra with duality, what's the best way to get the reals? My
previous answer was first to reflect the midpoint algebra in the
subcategory of abelian groups and then define the multiplication via
monotonic endomorphisms. I think there's a more natural way.

First, a re-cap of the definitions. A midpoint operation is a binary 
operation, with values here denoted as  x|y,  satisfying:
idempotence, x|x = x; commutativity, x|y = y|x; 
middle-two-interchange, (u|v)|(x|y) = (u|x)|(v|y); and cancelation,
x|y = x|z  =>  y = z.  The closed interval is a totally ordered
midpoint algebra with anti-involution, with values here denoted as  
 -x.  It follows that there's a unique element of the form  (-x)|x.  
We'll call it the center.

The middle-two-interchange law is just what is needed to see that the
monoid of midpoint-endomorphisms is itself a midpoint algebra (where
the midpoint of  f  and  g  is defined pointwise: (f|g)(x) =
(fx)|(gx)).   If  f  and  g  are both order-preserving or both order-
reversing then it's easy to see that so is  f|g.  But we'll want all 
the _monotonic_ endomorphisms (the OED defines "monotonic" as "varying
in such a way that it either never increases or never decreases") and
there's no apparent reason to expect that monotonicity is preserved by
midpointing. The wonderful fact, though, is:

    Midpoint-preserving functions between intervals are monotonic.

(The axiom of choice yields  2^{2^N}  counterexamples if the interval
is replaced by the reals.)

It follows that a midpoint-preserving function is determined by
its values on any two points. The monoid of midpoint- and duality-
preserving endo-functions on a closed interval is naturally 
isomorphic (via evaluation at Top) to the closed interval. For the
entire reals use the stalk at the center of the germs of such 
functions.


PS
For a proof suppose that  f:[-1,1] --> [-K,K]  preserves midpoints and
duality. For any  x:[-1,1]  and dyadic rational  q  such that  
qx:[-1,1], we may uniquely identify  qx  using just midpoints and
duality, hence  f(qx) = qf(x).

Suppose that  f  is not monotonic.  We wish to reach a contradiction.
Let  u < v  and  x < y  in  [-1,1]  be such that  fu < fv  and  
fx > fy.  Define  a = (-u)|v  and  b = (-x)|y  to obtain  a,b > 0  
and  fb < 0 < fa.  Let  q > 0  be a dyadic rational such that
a/b - (fa)/(bK)  <  q  <  a/b  and define  c = (-qb)|a.  Let  r  be a 
dyadic rational such that  (2K)/(fa)  <  r  <  1/c.  Then  rc  is in 
[-1,1]  but  f(rc)  can not be in  [-K,K]  (because  fc > (fa)/2).

The theorem I stated doesn't say anything about preserving duality. So
let  g  be a midpoint-preserving function between intervals. Let  a  
be the dual of the value of  g  at the center. Then the function  
\x.a|(gx)  preserves the center and continues to preserve midpoints.
Since  -x  may be uniquely identified by  (-x)|x = 0  the preservation
of the center implies preservation of the duality and  \x.a|(gx)  is
monotonic. Since  \y.a|y  not only preserves but reflects the order,
we know that  g  is monotonic.