Posted by: "Jens"

jd@... <mailto:

jd@...?Subject=

> Can anyone (in this list) imagine a continuous relation on sets,

Sure. 2^R in DCPO*.

Here DCPO* is the cartesian closed category of *pointed*

directed-join-complete posets, or DCPOs with bottom, and their

directed-join-preserving functions. 2 is the "flat" or V-shaped DCPO

{_|_,0,1}, with _|_ < 0 and _|_ < 1 but with 0 and 1 incomparable. R

however is not a flat domain in that sense but rather is the linearly

ordered reals including -oo (which maps must send to bottom) and oo

(which maps can send anywhere, but any map from R to 2 that sends it to

0 or 1 must send some finite real to the same value).

Claim: 2^R is a tree-shaped DCPO whose root is its bottom (the function

that maps every real to bottom, i.e. are nowhere defined) and whose

leaves are its total elements (the functions that map no real to bottom,

i.e. are everywhere defined), namely the subsets of R.

To understand 2^R precisely it helps to consider 1^R first where 1 is

the flat singleton {_|_,0}. This is just R since it amounts to Dedekind

cuts in R ordered by inclusion of order filters (reverse inclusion of

order ideals). (R is the nearest thing to the linearly ordered poset Q

of rationals in DCPO* because Q is not directed-join-complete as a

partial order.)

Bottom in 1^R is the everywhere-undefined function. This function can't

be isolated in 1^R (i.e. 1^R is atomless) because if a function is

undefined at all finite reals it is undefined at top by

directed-join-completeness.

Top in 1^R is the function defined everywhere except at bottom This

function can't be isolated (1^R is coatomless) because a coatom would

have to be undefined at some finite x, and the function that is defined

precisely at x and all larger reals would be a strictly larger function

that is still not defined everywhere.

Viewing 2^R as a tree, each path can be understood as a copy of 1^R.

This set of paths would be a nice example of a set of Brouwerian choice

sequences for which choices are made continuously working from left to

right along the real axis, with the end product of a sequence being a

fully determined subset of R.

R being the unique complete (in the order interval topology rather than

the Scott topology) dense linear order, there is something canonical

about this example of a continuous tree. I suppose one could say it was

the unique complete dense binary tree, if that makes sense.

Vaughan Pratt