EXTENSIONS OF TOPOLOGIES

T. L. Hicks University of Missouri–Rolla

\underline{Abstract}. The m–extension of a topology is defined, some of its properties are studied, and its relation to a simple extension is noted. Also, an associated closure operation on the lattice of all topologies on a fixed set is studied.

Let X be a set, m = {M_{ν} : ν ∈ I} a non–empty collection of subsets of X and, for each ν ∈ I, t_{ν} a topology for M_{ν}. We define a collection t(m) of subsets of X as follows: V ∈ t(m) if and only if for each ν ∈ I, V ∩ M_{ν} ∈ t_{ν}. It is clear that t(m) is a topology for X.

Unless otherwise specified we adopt the terminology of Gaal [2].

\underline{Theorem 1}. t(m) is the strongest topology for X such that t(m) ≤ t_{ν} on every M_{ν}.

\underline{Proof}. If A ⊆ M_{ν} and A is a member of the subspace topology t(m)∩M_{ν} on M_{ν}, then A = U ∩M_{ν} where U ∈ t(m). Thus A ∈ t_{ν}. Suppose t′ is a topology for X such that t′ ≤ t_{ν} on every M_{ν}. If V ∈ t′ and ν ∈ I, then V ∩ M_{ν} ∈ t′ ∩ M_{ν} ≤ t_{ν}. Hence V ∈ t(m).

\underline{Theorem 2}. If there exists a topology t′ for X such that t′ = t_{ν} on every M_{ν}, then t(m) = t_{ν} on every M_{ν} and t(m) is the strongest topology for X with this property.

\underline{Proof}. If such a topology t′ exists, then t′ ≤ t(m) by Theorem 1. For ν ∈ I, t_{ν} = t′ ∩ M_{ν} ≤ t(m) ∩ M_{ν} ≤ t_{ν}. Hence t(m) = t_{ν} on M_{ν}.

In the remainder of this paper we suppose that t is a topology for X and, for each ν ∈ I, t_{ν} is the subspace topology t ∩ M_{ν}. In this case, t(m) is called the m–\underline{extension} of t. It follows from Theorem 2, that the m–extension of t is the strongest topology on X that agrees with t on every M_{ν}. Thus t ≤ t(m).

\underline{Theorem 3}. The following properties hold.

1) t(m) is discrete on X − ⋃{M_{ν} : ν ∈ I}, and if X ∈ m, t = t(m).

2) A subset B of X is t(m) closed if and only if, for every ν ∈ I, B ∩ M_{ν} is closed in (M_{ν}, t_{ν}).

3) If (X, t) is a compact Hausdorff space, then t = t(m) if and only if t(m) is compact.

4) If m is the collection of all closed compact subsets of (X, t), then t = t(m) if and only if X is a k–space. \underline{Proof}. 1 and 2 follow easily. For 3, suppose t(m) is compact.

t ≤ t(m). t = t(m) since it is not possible to have a Hausdorff topology strictly weaker than a compact topology. For 4, recall that a k–space is one that satisfies the condition: if a subset A of X intersects each closed compact set in a closed set, then A is closed.

In [1], Levine defined a topology t^{∗} > t to be a \underline{simple extension} of t if there exists a subset A of X such that A ∉ t and t^{∗} = {U ∪ (V ∩ A) : U, V ∈ t}. We write t^{∗} = t(A). Many of his results dealt with the case X −A ∈ t. We need the following results which appear in [1].

(a) t and t(A) agree on A and also on X − A.

(b) A is closed in (X, t) if and only if A is closed in (X, t(A)). \underline{Theorem 4}.

(1) t(A) is the weakest topology on X such that t < t(A) and A ∈ t(A).

(2) Let m = {A, X − A}. t(A) = t(m) if and only if X − A ∈ t. \underline{Proof}. (1) If t′ > t and A ∈ t′, let N ∈ t(A). N = U ∪(V ∩A) where U, V ∈ t. Since A ∈ t′ and t < t′, N ∈ t′.

(2) By (a), t and t(A) agree on every set of m. Hence t(A) ≤ t(m) by Theorem 1. Suppose X − A ∈ t. If U ∈ t(m), U ∩ A ∈ t ∩ A = t(A) ∩ A. Since A ∈ t(A), U ∩ A ∈ t(A). Also, X − A ∈ t(A) implies that U ∩ (X − A) ∈ t(A). Therefore, U = (U ∩ A) ∪ (U ∩ X − A) ∈ t(A).

X − A ∈ t(m), so that if t(m) = t(A), X − A ∈ t(A).

\underline{Theorem 5}.

(1) t ≤ t(m) and, in general, t ≠ t(m). In fact, if m consists only of singleton subsets, then t(m) is the discrete topology.

(2) (t(m))(m) = t(m) and if m ⊆ n, then t(m) = (t(m))(n).

(3) If m ⊆ n, t(m) ≥ t(n).

(4) If X ∈ m, t(m) = t.

(5) If m = {M_{ν} : ν ∈ I} is a collection of t open sets such that ⋃{M_{ν} : ν ∈ I} = X, then t = t(m).

(6) If u is a topology for X and t ≤ u, then t(m) ≤ u(m). \underline{Proof}. Since the properties are easily established, we only offer

a proof for 2. t(m) ≤ (t(m))(n) by 1. If A ∈ (t(m))(n) and m ⊆ n, then A ∩ M_{ν} ∈ t(m) ∩ M_{ν} = t_{ν} for every ν ∈ I. Thus A ∈ t(m) and equality holds.

\underline{Definition}. By a \underline{closure operation} on a partially ordered set (P, ≤) we mean a mapping ψ : P → P such that:

(1) x ≤ y implies ψ(x) ≤ ψ(y).

(2) x ≤ ψ(x) for every x ∈ P.

(3) ψ(ψ(x)) = ψ(x) for every x ∈ P.

Suppose m = {M_{ν} : ν ∈ I} is fixed. If we look at properties 1, 2, and 6 in the above theorem we see that m defines a closure operation (t → t(m)) on the lattice P^{∗} of all topologies on X.

\underline{Theorem 6}. Suppose \{t_{α} : α ∈ J\} is a non–empty collection of topologies on X.

(1) (glb_{α} t_{α})(m) ≤ glb_{α} t_{α}(m) = (glb_{α} t_{α}(m))(m).

(2) (lub_{α} t_{α})(m) = (lub_{α} t_{α}(m))(m).

\underline{Proof}. (1) The equality in (1) is a special case of a Theorem of Ward [4, p. 68 or 5], but the reader can easily verify (1).

(2) For each α, t_{α} ≤ lub_{α} t_{α} so that t_{α}(m) ≤ (lub_{α} t_{α})(m). Hence lub_{α} t_{α}(m) ≤ (lub_{α} t_{α})(m) and consequently (lub_{α} t_{α}(m))(m) ≤ (lub_{α} t_{α})(m). Also, lub_{α} t_{α} ≤ lub_{α} t_{α}(m) implies that (lub_{α} t_{α})(m) ≤

(lub_{α} t_{α}(m))(m).

\underline{Example 1}. The ≤ in (1) of Theorem 6 may be <. Put X = {a, b, c, d} and m = {M} where M = {a, b, c}. Let t = {{c}, X, ∅} and u = {{d, c}, X, ∅}. Then (t ∩ u)(m) < t(m) ∩ u(m).

\underline{Corollary}.

(1) (lub_{α} t_{α})(m) = lub_{α} t_{α}(m) if and only if lub_{α} t_{α}(m) is a fixed point of m.

(2) If lub_{α} t_{α} is a fixed point of m, then so is lub_{α} t_{α}(m).

\underline{Proof}. In general, lub_{α} t_{α}(m) ≤ (lub_{α} t_{α})(m) = (lub_{α} t_{α}(m))(m). \underline{Theorem 7}. If (glb_{α} t_{α}) ∩ M_{ν} = glb_{α} (t_{α} ∩ M_{ν}) for every ν ∈ I,

then glb_{α} t_{α}(m) = (glb_{α} t_{α})(m).

\underline{Proof}. Suppose U ∈ t_{α}(m) for every α and let ν be fixed.

Since U ∩ M_{ν} ∈ t_{α} ∩ M_{ν}, U ∩ M_{ν} ∈ glb_{α} (t_{α} ∩ M_{ν}) = (glb_{α} t_{α}) ∩ M_{ν}.

Hence U ∈ (glb_{α} t_{α})(m) and glb_{α} t_{α}(m) ≤ (glb_{α} t_{α})(m).

\underline{Question}. Is every closure operation on P^{∗} of the above form? That is, given a closure operation ψ on P^{∗} does there exist a collection m = {M_{ν}} of subsets of X such that ψ(t) = t(m) for every t ∈ P^{∗}?

\underline{Theorem 8}. Suppose ψ is given. If there exists a class m = {M_{ν} : ν ∈ I} such that ψ(t) = t(m) for every t ∈ P^{∗}, then ψ(t) = t(m^{∗}) for every t ∈ P^{∗} where m^{∗} = {M ⊆ X : t and ψ(t) agree on M for every t ∈ P^{∗}}.

\underline{Proof}. Since t ≤ ψ(t) = t(m), t and ψ(t) agree on every M ∈ m. Hence m ⊆ m^{∗} and consequently t(m) ≥ t(m^{∗}). By Theorem 2, t(m^{∗}) ≥ ψ(t). We have ψ(t) = t(m) ≥ t(m^{∗}) ≥ ψ(t) for every t ∈ P^{∗}.

\underline{Example 2}. This example shows that the answer to the above question is no. Let X denote an infinite set, t^{∗} the discrete topology, t′ the indiscrete topology, and u the topology of finite complements. Define ψ as follows: ψ(t) = u if t ≤ u and ψ(t) = t^{∗} otherwise. The reader can easily verify that ψ is a closure operation on P^{∗}. We show that m^{∗}, defined in the above theorem, consists of all singleton subsets. Then since t(m^{∗}) is the strongest topology that agrees with t on every M ∈ m^{∗}, t(m^{∗}) = t^{∗} for every t ∈ P^{∗} and consequently ψ(t) ≠ t(m^{∗}) for any t ≤ u.

Suppose x ≠ y and let A = {x, y}. It suffices to prove that ψ(t′) = u and t′ do not agree on A. A ∩ (X − {y}) = {x} so that {x} ∈ u ∩ A. {x} ∉ t′ ∩ A.

\underline{Problem}. Characterize every closure operation on P^{∗}.

\underline{References}

1. N. Levine, “Simple Extensions of Topologies,” Amer. Math. Monthly, 71(1964), 22–25.

2. S. A. Gaal, Point Set Topology, Academic Press, New York, 1964.

3. J. L. Kelley, General Topology, Van Nostrand, New York, 1955.

4. G. Szasz, Introduction to Lattice Theory, Third Edition, Academic Press, 1963.

5. M. Ward, “The Closure Operators of a Lattice,” Annals Math., 43(1942) 191–196.