Exercise 4.24 (Scott 1981, PRG-19, Lecture IV) — Schröder–Bernstein via a fixed #
point
(For set theorists.) Let f : A → B and g : B → A be one-one (injective, not
necessarily onto).
Then there is a one-one correspondence h : A ≃ B (schroeder_bernstein).
Following Tarski's hint, work in the power-set domain P A (a complete lattice)
and take a fixed
point X of the monotone operator
T(X) = (A − g(B)) ∪ g(f(X)),
which exists by Knaster–Tarski (Exercise 4.14 / 4.13(2), here lfpSet). With X = T(X) fixed,
define
h(a) = f(a) if a ∈ X,
h(a) = g⁻¹(a) if a ∉ X
— well-defined because a ∉ X = T(X) forces a ∈ g(B) (it is not in (A − g(B))), so a has a
unique g-preimage (mem_range_of_not_mem). The map h is:
- injective (
sbFun_injective): withinX, by injectivity off; outsideX, by injectivity ofg; the mixed case is impossible —f(a₁) = g⁻¹(a₂)would puta₂ = g(f(a₁)) ∈ g(f(X)) ⊆ X; - surjective (
sbFun_surjective): forb ∈ B, ifg(b) ∉ Xthenh(g(b)) = b; ifg(b) ∈ Xtheng(b) ∈ g(f(X)), sog(b) = g(f(a))witha ∈ X, whenceb = f(a) = h(a).
Packaged as the equivalence schroederBernsteinEquiv. This is a set
theorists' exercise and is
inherently classical; the construction of h uses Classical.choice (the
g-preimage), exactly as
the statement demands.
Exercise 4.24 (Scott 1981, PRG-19). Tarski's bijection h : A → B: f on
the fixed set
X, and the g-inverse off it.
Equations
- Domain.Neighborhood.Exercise424.sbFun f g a = if ha : a ∈ Domain.Neighborhood.Exercise424.sbSet f g then f a else Classical.choose ⋯
Instances For
Exercise 4.24 (Scott 1981, PRG-19) — Schröder–Bernstein. Injections f : A → B and
g : B → A yield a bijection A → B.
Exercise 4.24 (Scott 1981, PRG-19). The one-one correspondence as an
Equiv A ≃ B.