Re: Freyd's couniversal characterization of [0,1]
"Martin H. Escardo" <Martin.H.Escardo@ens.fr> writes:
It would be interesting to test Freyd's couniversal characterization of the unit interval in many other categories.
Here I test it in Top, the category of topological spaces and continuous maps, and various full subcategories, where one would hope to get the unit interval with the Euclidean topology.
---------------------------------------------------------------------- Summary of the outcome of some tests:
(1) In Top, the final coalgebra for Freyd's functor exists. Its underlying object, however, is an indiscrete space (unsurprisingly).
(2) In the category of T0 spaces, it doesn't exist.
(3) In the category of normal spaces it does exist, and, as one would hope, its underlying object is indeed the unit interval with the Euclidean topology.
See remark below for weakening normality in (3) as much as possible. ---------------------------------------------------------------------- Arguments follow.
[text deleted]
---------------------------------------------------------------------- Question: What does one get in full subcategories of locales and of equilogical spaces?
I can answer this for equilogical spaces. The functor F: Bi[Equ] ---> Bi[Equ] has a final coalgebra. It is the equilogical space (C, ~) where C is the Cantor space C = 2^N = infinite sequences of 0's and 1's and ~ is the equivalence relation defined by a ~ b iff r(a) = r(b) where r: C --> [0,1] is defined by r(a) = \sum_{k=0}^\infty a_k / 2^{k+1} (This equivalence relation can easily be defined without reference to the closed interval [0,1].) Thus, the final coalgebra is the _(unsigned) binary digit representation_ of the closed interval [0,1]. The structure map d: (C,~) ---> F(C,~) is induced by the canonical isomorphism C ---> C + C = 2 x C. Unfortunately, this closed interval is not what we would hope for. Ideally, we would want the _signed_ binary digit representation of the interval [-1,1], i.e., the space ({-1,0,1}^N, :=:) where :=: is defined by a :=: b iff s(a) = s(b) s(a) = \sum_{k=0}^\infty a_k / 2^{k+1} We want this because the real numbers object in Equ is the built from the _signed_ representation of reals. I do not see how to adapt the construction so that it yields the signed representation. We would have to "glue" more than just a couple of points. But how do we say what should be glued, without reference to [0,1] or C? Let me explain briefly why (C, ~) is the final coalgebra. In Equ, in the pushout 1 -------> X | | | | | | | | V V X -------> FX the underlying space of FX is |FX| = |X| + |X|. This is a "small" but crucial difference between Top and Equ. We can first find the final coalgebra in Top_0 for the functor A --> A + A, which is the Cantor space C, and then compute the equivalence relation on C. In this case it is easy to verify that all the continuous maps obtained from the coalgebraic structure of C actually preserve the equivalence relations. By the way, the two distinguished points of (C, ~) are 0^\infty and 1^\infty, of course (and it doesn't matter, since C is homogeneous). -- Andrej Bauer
Andrej Bauer writes:
I can answer this for equilogical spaces.
The functor F: Bi[Equ] ---> Bi[Equ] has a final coalgebra. It is the equilogical space (C, ~) where C is the Cantor space
C = 2^N = infinite sequences of 0's and 1's
and ~ is the equivalence relation defined by
a ~ b iff r(a) = r(b)
where r: C --> [0,1] is defined by
r(a) = \sum_{k=0}^\infty a_k / 2^{k+1}
This is interesting in connection with some previous discussion in this list on "definability" of the mid-point operation "by coinduction". As it is well known, the mid-point operation is not continuously realizable via binary expansions with the Cantor topology. (As Andrej Bauer and other people have mentioned signed-digit binary expansions in this discussion, let me emphasize that, in contrast to the above situation, all continuous functions [-1,1]^n->[-1,1] are continuously realizable via signed-digit binary expansions with the Cantor topology. Put in another way, the space 3^N = (3^N)^n is projective over (the regular epimorphism) s:3^N->[-1,1], but the space 2^N is not projective over r:2^N->[0,1].) Martin Escardo
participants (2)
-
Andrej Bauer -
Martin H. Escardo