M6140 Topology Exercises - 11th Week (2022)
1 Pointless Topology
Definition 1. A frame is a complete lattice that satisfies the infinite distributivity law which says
that a ∧ i∈I bi = i∈I(a ∧ bi) for all a, bi ∈ A . A homomorphism of frames is a mapping preserving
finite meets and arbitrary joins.
Exercise 1. Show that each topology is a frame.
Exercise 2. Let f : (X, T1) → (Y, T2) be a continuous mapping. Prove that T2 → T1 given by
U → f−1(U) is a homomorphism of frames.
Definition 2. A point of a frame A is a frame homomorphism A → 2. The set of points of a frame
A is denoted by pt(A).
Exercise 3. How is the notion from the previous definition related to the usual notion of a point in
a topological space?
Definition 3. An element p of a frame A is called a prime element if p = 1 and for all x, y ∈ A:
x ∧ y ≤ p implies x ≤ p or y ≤ p.
Exercise 4. Show that there is a bijective correspondence between points in a frame A and prime
elements in a frame A.
Exercise 5. Define a mapping Φ: A → P(pt(A)) by
Φ(a) := {p: A → 2 | p(a) = 1} ∼= {q prime | a ≤ q}.
Show that Φ is a homomorphism of frames and conclude that the image of Φ is a topology on pt(A).
Remark 1. In this way we can relate the theory of frames with the theory of topological spaces and
vice versa. In the words of category theory, it can be easily shown that the functor pt: Frmop
→ Top
is right adjoint to the functor Ω: Top → Frmop
which assigns to each topological space its topology.
Definition 4. A frame A is said to have enough points if Φ: A → P(pt(A)) is a bijection.
Exercise 6. Prove that each topology has enough points.
Exercise 7. Let A be a frame. Show that the topological space pt(A) is sober.
Remark 2. Combining the results above allows us to conclude that there’s an equivalence of categories
between the opposite of the category of frames that have enough points and the category of sober
spaces.
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