EXPANSIONS OF BAIRE SPACES

James Foran and Paul Liebnitz University of Missouri–Kansas City

Let (X, T) be a topological space. Following Hewitt [3], if T′ is a topology on X such that T ⊂ T′ we call (X, T′) an \underline{expansion} of (X, T). Several other authors [1, 2, 3, 4, 6, 7, 8] have subsequently studied the preservation of topological properties under expansions. In this note we consider the preservation of Baire spaces under expansions. A \underline{Baire space} is a topological space in which every nonempty open set is of second category. Equivalently, a space is Baire if and only if none of its nonempty open sets is the union of countably many nowhere dense sets; this is true if and only if the intersection of every sequence of dense open sets is dense.

We show that an expansion of a Baire space need not be Baire, and that the supremum of a collection of Baire topologies need not be a Baire topology. We give some sufficient conditions under which an expansion of a Baire space is Baire. It would be interesting to have necessary conditions under which stronger topologies are Baire when the original topology is Baire.

We refer the reader to [5] for definitions of the basic topological notions used herein. In the sequel, we use the symbol A^{c} to denote the \underline{complement} of a set A. We begin with an example.

\underline{Example 1}. The topological expansion of a dense–in–itself Baire space to a dense–in–itself space need not be Baire.

\underline{Construction}. Let (X, T_{1}) be a dense–in–itself Baire space with a first category dense subset A, e.g., let X = [0, 1] with the usual topology, and let A be the subspace of rational numbers in [0, 1]. Let T_{2} be the topology generated by T_{1} ∪ {A}. Each member of T_{2} is of the form (A ∩ G_{1}) ∪ G_{2} for some G_{1}, G_{2} ∈ T_{1}. Since A is of the first category in (X, T_{1}),

A = ⋃_{n=1}^{∞} C_{n}

where each C_{n} is nowhere dense. Consider a fixed C_{n}. To see that C_{n} is nowhere dense in (X, T_{2}), it suffices to observe that if (A ∩ G_{1}) ∪ G_{2} is a T_{2}–open set with A ∩ G_{1} ≠ ∅, there exists a nonempty T_{1}–open subset G_{3} of G_{1} such that G_{3} ∩ C_{n} = ∅. It follows that the T_{2}–open set A is of the first category in (X, T_{2}).

Hence, (X, T_{2}) is not Baire. That (X, T_{2}) is dense–in–itself is evident.

We now give some sufficient conditions for expansions of Baire topologies to be Baire.

\underline{Proposition 1}. Suppose (X, T_{1}) is a space, A ⊂ X, and T_{2} is the topology on X generated by T_{1} ∪ {A, A^{c}}. Then each nowhere dense subset of (X, T_{2}) is also nowhere dense in (X, T_{1}). Thus if (X, T_{1}) is a Baire space, if A is second category in each T_{1}–open set which it meets, and if A^{c} is second category in each T_{1}–open set which it meets, then (X, T_{2}) is a Baire space.

\underline{Proof}. Suppose E is nowhere dense in (X, T_{2}). Let G be a nonempty T_{1}–open set such that G ∩ E ≠ ∅. To show that E is nowhere dense in (X, T_{1}), it suffices to show that G contains a nonempty T_{1}–open set H such that H ∩ E = ∅.

\underline{Case 1}. A ∩ G ∩ E ≠ ∅. Since E is nowhere dense in (X, T_{2}), there exist G′_{1}, G′_{2} ∈ T_{1} such that ∅ ≠ (A ∩ G′_{1}) ∪ G′_{2} ⊂ A ∩ G and [(A ∩ G′_{1}) ∪ G′_{2}] ∩ E = ∅. If G′_{2} ≠ ∅, take H = G′_{2}. Otherwise, ∅ ≠ A ∩ G′_{1} ⊂ A ∩ G and A ∩ G′_{1} ∩ E = ∅. Let G_{1} = G′_{1} ∩ G ⊂ G.

Then

∅ ≠ A ∩ G′_{1} ∩ G = A ∩ G_{1} ⊂ A ∩ G.

If A^{c} ∩ G_{1} ∩ E = ∅, then G_{1} ∩ E = ∅ and we take H = G_{1}. If A^{c} ∩ G_{1} ∩ E ≠ ∅, since E is nowhere dense in (X, T_{2}), there exist G_{2}, G_{3} ∈ T_{1} such that ∅ ≠ (A^{c} ∩ G_{2}) ∪ G_{3} ⊂ A^{c} ∩ G_{1} and [(A^{c} ∩G_{2})∪G_{3}]∩E = ∅. Again, if G_{3} ≠ ∅, set H = G_{3}. Otherwise, ∅ ≠ A^{c} ∩ G_{1} ∩ G_{2}, and A^{c} ∩ G_{1} ∩ G_{2} ∩ E = ∅. Then G_{1} ∩ G_{2} may be taken as the required subset H of G. For evidently, G_{1} ∩ G_{2} is a nonempty T_{1}–open subset of G. Moreover, G_{1} ∩ G_{2} ∩ E = ∅; otherwise G_{1} ∩ G_{2} would contain points of E but no point of A^{c}, and we would have

∅ ≠ G_{1} ∩ G_{2} ∩ E ∩ A ⊂ A ∩ G_{1} ∩ E = ∅.

\underline{Case 2}. A∩G∩E = ∅. Then A^{c} ∩G∩E ≠ ∅, and the existence of the required subset H of G follows by symmetry. The remainder of the proposition is easily established.

\underline{Proposition 2}. Suppose (X, T_{1}) is a space and A is a subset of X such that A^{c} is of the first category. Let T_{2} be the topology on X generated by T_{1} ∪ {A}. Then every first category subset of (X, T_{2}) is first category in (X, T_{1}). Thus if (X, T_{1}) is a Baire space, and if A is residual, then (X, T_{2}) is a Baire space.

\underline{Proof}. We will show that if G is T_{2}–open and dense, then G^{c} must be category 1 in (X, T_{1}). This will imply that every subset of X which is nowhere dense in (X, T_{2}), and, consequently, every subset of X which is category 1 in (X, T_{2}), is category 1 in (X, T_{1}). For if E is nowhere dense in (X, T_{2}), then X − \overline{E} is T_{2}–open and dense. Hence \overline{E} is category 1 in (X, T_{1}), and, consequently, so is E.

Assume, therefore, that G is T_{2}–open and dense. If G is T_{1}– open, then G^{c} is nowhere dense and hence category 1 in (X, T_{1}). Otherwise, there exist G_{1}, G_{2} ∈ T_{1} such that G = (A∩G_{1})∪G_{2} with A ∩ G_{1} ≠ ∅. Let H be the interior of G_{2}^{c} in (X, T_{1}) furnished with the relativized T_{1} topology. Now G_{1} ∩ H is dense in H. Otherwise, there would exist a nonempty open subset G_{3} of H such that G_{3} ∩ G_{1} ∩ H = ∅. Hence G_{3} ∩ [(A ∩ G_{1}) ∪ G_{2}] = G_{3} ∩ G = ∅, contrary to the assumption that G is T_{2}–dense. It follows that H − G_{1}, the boundary of G_{1} ∩ H in H, is nowhere dense, and hence of category 1 in H. Hence H − G_{1} is category 1 in (X, T_{1}). Since bdry G_{2} (the boundary of G_{2} in (X, T_{1})) and H − A are both category 1 subsets of (X, T_{1}), and since G^{c} = (H − A) ∪ (H − G_{1}) ∪ bdry G_{2}, it follows that G^{c} is category 1 in (X, T_{1}). The remainder of the proposition follows easily.

The following example shows that the supremum of an increasing sequence of Baire topologies need not be Baire.

\underline{Example 2}. There exists a dense–in–itself Baire space (X, T_{0}) and a sequence T_{0}, T_{1}, T_{2}, ⋯ of topologies on X such that for each i, T_{i}, ⊂ T_{i+1} and (X, T_{i}) is Baire, but (X, T) is not Baire if T is the supremum of the sequence {T_{n}}.

\underline{Construction}. Let X be the space of real numbers with the usual topology. By standard methods we decompose X into a sequence A_{0}, A_{1}, A_{2} ⋯ of pairwise disjoint sets such that for each nonempty open set G, and for each i, A_{i}∩G is nonempty and second category. We divide {A_{0}, A_{1}, A_{2}, ⋯} into classes C_{0}, C_{1}, C_{2}, ⋯ so that C_{k} = {B(0, k), B(1, k), ⋯, B(2^{k} − 1, k)} where

B(i, k) = ⋃_{j=0}^{∞} A_{i+j}(2^{k}), 0 ≤ i ≤ 2^{k} − 1 .

For each k,

X = ⋃ C_{k} = ⋃_{i=0}^{2^{k}−1} B(i, k),

B(i, k)^{c} = ⋃_{ℓ=1,ℓ≠i}^{2^{k}−1} B(ℓ, k) ,

and

B(i, k) ∩ B(ℓ, k) = ∅ if i ≠ ℓ .

Let T_{0} be the usual topology for the space X of real numbers. (X, T_{0}) is a Baire space. Having constructed Baire topologies T_{0} ⊂ T_{1} ⊂ ⋯ ⊂ T_{k} on X, let T_{k+1} be the expansion of T_{k} generated by

T_{k} ∪ ( ⋃_{i=1}^{2^{k}−1}{B(i, k), B(i, k)^{c}}) .

By Proposition 1, each of the spaces (X, T_{0}), (X, T_{1}), (X, T_{2}), ⋯ is a Baire space and T_{i} ⊂ T_{i+1} for each i. Let T be the supremum of T_{0}, T_{1}, T_{2}, · · ·. We assert that (X, T) is not Baire. To prove this we will show that A_{n} is nowhere dense in (X, T) for each n. So consider a fixed set A_{n}. Let H be a nonempty T –open set. We will show that H contains a nonempty T –open set which misses A_{n}. Note first that there is a nonempty T_{0}–open set G and a corresponding basic T –open set G∩B(i, k) such that ∅ ≠ G∩B(i, k) ⊂ H for some

nonnegative integers i and k. If n ≠ i + j(2^{k}) for some nonnegative

j, then A_{n} ∩ B(i, k) = ∅, and consequently, G ∩ B(i, k) ∩ A_{n} = ∅.

If n = i + j(2^{k}) for some j, then

B(i + (j + 1)2^{k}, k + (j + 1)) ⊂ B(i, k)

and

B(i + (j + 1)2^{k}, k + (j + 1)) ∩ A_{n} = ∅.

Hence

G ∩ B(i + (j + 1)2^{k}, k + (j + 1)) ≠ ∅,

and

G ∩ B(i + (j + 1)2^{k}, k + (j + 1)) ∩ A_{n} = ∅.

\underline{References}

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3. E. Hewitt, A problem of set–theoretic topology, Duke Math. J., 10 (1943), pp. 309–333, MR 5, 46.

4. S. Ikenaga and I. Yoshioka, Extensions of topologies, Proc. Japan Acad. 14 (1965), pp. 11–16.

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