Exercise 5.9 (Scott 1981, PRG-19, Lecture V) #
Suppose f, g : 𝒟 → 𝒟 are approximable and f ∘ g = g ∘ f. Then:
fandghave a least common fixed pointx = f(x) = g(x)(commuting_least_common_fixed);- if in addition
f(⊥) = g(⊥), thenfix(f) = fix(g)(fixElement_eq_of_commuting_bot); - in particular
fix(f) = fix(f²)(fixElement_iterTwice).
Construction. Let a = fix(f). Commutation makes g(a) a fixed point of f
(f(g(a)) = g(f(a)) = g(a)), so a ⊑ g(a), and Exercise 4.7 produces the least
fixed point of g
above a, namely x = ⊔ₙ gⁿ(a). Each gⁿ(a) is again a fixed point of f
(induction +
commutation), and f preserves the directed union, so f(x) = x: x is a common
fixed point. Any
common fixed point w satisfies a = fix(f) ⊑ w and g(w) ⊑ w, so x ⊑ w by
Exercise 4.7's
leastness.
For the second part, commutation gives g(fⁿ(⊥)) = fⁿ⁺¹(⊥) once g(⊥) = f(⊥), so
g(fix f) = ⊔ₙ fⁿ⁺¹(⊥) = fix f; symmetrically f(fix g) = fix g, whence fix f = fix g by
leastness.
All maps and elements are choice-free; the equalities use only the order on |𝒟|.
Commutation f ∘ g = g ∘ f read elementwise: f(g(x)) = g(f(x)).
Exercise 5.9 (Scott 1981, PRG-19). Commuting approximable maps have a
least common fixed
point x = f(x) = g(x).
f.iterElem (n+1) = f(fⁿ(⊥)).
Exercise 5.9 (Scott 1981, PRG-19). If f, g commute and f(⊥) = g(⊥),
then
fix(f) = fix(g).
fⁿ(⊥) ⊑ fix(f).
(f²)ⁿ(⊥) = f^{2n}(⊥).
Exercise 5.9 (Scott 1981, PRG-19). In particular fix(f) = fix(f²): the
f²-chain
f^{2n}(⊥) is cofinal in the f-chain fⁿ(⊥), so the two least fixed points
coincide.