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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 † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 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). 2