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partial ordering in a topological space∗
CWoo†
2013-03-21 21:50:10
Let X be a topological space. For any x, y ∈ X, we define a binary relation
≤ on X as follows:
x≤y
iff
x ∈ {y}.
Proposition 1. ≤ is a preorder.
Proof. Clearly x ≤ x. Next, suppose x ≤ y and y ≤ z. Let C be a closed set
containing z. Since y is in the closure of {z}, y ∈ C. Since x is in the closure of
{y}, x ∈ C also. So x ≤ z.
We call ≤ the specialization preorder on X. If x ≤ y, then x is called a
specialization point of y, and y a generization point of x. For any set A ⊆ X,
• the set of all specialization points of points of A is called the specialization
of A, and is denoted by Sp(A);
• the set of all generization points of points of A is called the generization
of A, and is denoted by Gen(A).
Proposition 2. . If X is T0 , then ≤ is a partial order.
Proof. Suppose next that x ≤ y and y ≤ x. If x 6= y, then there is an open set
A such that x ∈ A and y ∈
/ A. So y ∈ Ac , the complement of A, which is a
c
closed set. But then x ∈ A since it is in the closure of {y}. So x ∈ A ∩ Ac = ∅,
a contradition. Thus x = y.
This turns a T0 topological space into a poset, where ≤ here is called the
specialization order of the space.
Given a T0 space, we have the following:
Proposition 3. x ≤ y iff x ∈ U implies y ∈ U for any open set U in X.
Proof. (⇒) : if x ∈ U and y ∈
/ U , then y ∈ U c . Since x ≤ y, we have x ∈ U c , a
contradiction. (⇐) : if x ∈
/ {y}, then for some closed set C, we have y ∈ C and
x∈
/ C. But then x ∈ C c , so that y ∈ C c , a contradiction.
∗ hPartialOrderingInATopologicalSpacei
created: h2013-03-21i by: hCWooi version:
h38775i Privacy setting: h1i hDefinitioni h54F99i
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Remarks.
• {x} =↓ x, the lower set of x. (z ∈↓ x iff z ≤ x iff z ∈ {x}).
• But if X is T1 , then the partial ordering just defined is trivial (the diagonal
set), since every point is a closed point (for verification, just modify the
antisymmetry portion of the above proof).
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