Subject: categories: Reality check

From: Peter Freyd <[email-masked]>
Date: Mon, 31 Jul 2000 11:44:45 -0400 (EDT)

In an earlier posting I showed how to define co-inductively the closed
interval, in particular I showed that its elements are named by 
sequences of  0s  and  1s  with the usual binary-expansion equivalence
relation. There is a well-known computational problem with this 
approach, already in the definition of the midpoint operation: at what
point can you determine the first digit of the midpoint of .0000...  
and  .1111...? 

At the Como meeting I learned from Andrej Bauer about a  better
approach. Take the elements of  [-1,1]  to be named by infinite
sequences of _signed_ binary digits, that is -1, 0, +1.

[Just to confuse matters, Scedrov and I once used signed _ternary_ 
digits (n Cats and Allegators for the "Freyd curve"). The signed binary
expansions  .+1 -1  and  .0 +1  describe the same number, to wit, 1/4.]

Using signed binary expansions one can compute midpoints with a little
3-state machine that takes as input the sequence of pairs of signed 
binary digits of the given numbers  x  and  y, and produces as output
a sequence of signed binary digits for the midpoint  x|y. (There may,
indeed, be momentary delays in the output, but there will not be an
indefinite delay -- indeed, the number of output digits will never be
more than one less than the number of pairs of input digits).

The challenge is to revise the co-induction so that it is this better 
version that emerges.

In the previous version I worked in the category of posets with
_distinct_ top and bottom, that is, those posets for which

                      not[(B = x) and (x = T)].

In the revised version I'll strengthen the condition by working in the
category of posets with _separated_ top and bottom:

                         [(B < x) or (x < T)].

(The conditions are equivalent in the presence of De Morgan's law. In
a topos the top and bottom of omega are always distinct but they are
separated only when De Morgan's law is satisfied throughout.)

In the previous setting I defined what I'll now call the _thin_ 
version of the ordered-wedge of  X  and  Y, to wit, the set of pairs,
<x,y>  satisfying the condition:

                          (x = T) or (B = y).

The _thick_ version is the set of pairs, <x,y>  satisfying the two 
weaker conditions:
                          (x < T) => (B = y)
                          (B < y) => (x = T)

(each of which is classically equivalent to the single condition used 
in the thin version).

Easy exercise: if top and bottom are separated in  X  and  Y, then 
they are separated in the thick version of the ordered-wedge  XvY.
(Indeed, it's enough for top and bottom to be separated in just one of
X  and  Y . No, it is not enough to assume just that they are distinct
in each.)

A map  X -> XvX  is thus given by a pair  d,u: X -> X  such that for
all  x:
                         (dx < T) => (B = ux)
                         (B < ux) => (dx = T)

The final coalgebra for  XvX  is still the closed interval, but now in 
the better computational sense. Let me explain.

Given an arbitrary coalgebra  d,u : X -> X, I need to describe a 
coalgebra homomorphism  f: X -> I. where  I  is the set of equivalence
types of infinite sequences of signed binary digits.

The first step is to work not with elements of  X  but elements of 
XvX. Consider a machine that given  <x,y>:XvX  asks in parallel the
questions:
                         "B < y?"
                         "B < ux  and  dy < T?"
                         "x < T?"

Exercise: If the top and bottom of  X  are separated and  d,u 
describe a map to the thick ordered-wedge  XvX, then at least one of
these questions has a positive answer.

Given  z:X  obtain a sequence of signed binary digits by starting with
the pair  <x,y> = <dz,uz>  and iterating the  non-deterministic 
procedure:

  If  B < y               
    then emit  +1  as output and replace  <x,y>  with  <dy,uy>;
  If  B < ux  and  dy < T  
    then emit   0  as output and replace  <x,y>  with  <ux,dy>;
  If  x < B      
    then emit  -1  as output and replace  <x,y>  with  <dx,ux>.

Not-so-easy exercises: regardless of the non-determinism, the element 
fz:[-1,+1]  named by the resulting sequence is determined. Moreover,
f(uz) = u(fz)  and  f(dz) = df(z).

PS. There is some geometry behind this stuff. Let me here mention just
this: given a pair  <x,y>  in  XvX  map it into the four-fold 
ordered-wedge  XvXvXvX. Think of each of the four copies of  X  as one
"quarter" of the whole. If  B < y  then the point lies inside the "top
half" (the 3rd and 4th quarters). If  B < ux  and  dy < T  then the 
point lies inside the "middle half" (the 2nd and 3rd quarters). If 
x < B  then the  point lies inside the bottom half" (the 1st and 2nd
quarters). Clearly any point is inside at least one of these three 
halves. The output-digit registers which of the three halves is moved
to and the corresponding pair-replacement effects that move.

PPS. On 22 Dec I gave a Dedekind-cut proof that the interval  [0,1]  
constructed in the standard fashion from (unsigned) binary sequences 
is the final coalgebra for the functor that sends  X  (with distinct 
top and bottom) to the thin version of  XvX. That proof should be 
replaced. First note that the midpoint-algebra homomorphism from  [0,1]
to  [-1,1]  can be effected simply by replacing each  0  with  -1  and
keeping each  1  as  +1. Given  d,u:X -> X  such that for all  x  it is
the case that either  dx = T  or  ux = B , consider a machine that upon
given  x:X  asks in parallel the questions  "dx = T?"  and  "ux = B?".
 
 If  dx = T  then emit  +1  as output and replace  x  with  ux, 
 If  ux = B  then emit  -1  as output and replace  x  with  dx. 

Given  x:X  one may iterate this (non-deterministic) procedure to
obtain a sequence of  +1s  and  -1s. Pretty-easy exercises: regardless
of the non-determinism, the element  fx:[-1,+1]  named by the
resulting sequence is determined; f(ux) = u(fx); f(dx) = df(x).