k-spaces, II

A space X is a k-space if for each A \subset X, A is closed in X if and only if K \cap A is closed in K for all compact K \subset X. A space X is a sequential space if for each A \subset X, A is closed in X if and only if A is a sequentially closed set in X. A set A \subset X is sequentially closed in the space X if whenever we have x_n \in A and the sequence x_n converges to x \in X, we have x \in A. A set A \subset X is sequentially open in X if X-A is sequentially closed in X. In both of these definitions, we can replace “closed” with “open” and the “only if” part of the definition always hold. Thus in working with these definitions, we only need to be concerned with the “if” part. Every sequential space is a k-space. The converse does not hold. In this short note, we show that the converse holds if every point in the space is a G_\delta-set. This is a basic fact about k-spaces. For other basic facts on k-spaces and sequential spaces, see the following:

Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I

In a given space X, W \subset X is a G_\delta-set in X if W=\bigcap \limits_{i=1}^\infty U_i where each U_i is open in X, i.e. W is the intersection of countably many open sets. A point x \in X is a G_\delta-set in X if the singleton \left\{x\right\} is the intersection of countably many open subsets of X. It is a well known fact in general topology that in a compact Hausdorff space X, if x \in X is a G_\delta-set in X, then there is a countable local base at x. It follows that if every point of a compact Hausdorff space X is a G_\delta-set in X, then X is first countable (see The cardinality of compact first countable spaces, II).

Theorem
Let X be a space in which every point is a G_\delta-set in X. Then if X is a k-space then X is a sequential space.

Proof. Suppose A \subset X is not closed in X. We show that A is not sequentially closed in X, i.e. there is a sequence x_n \in A such that x_n \mapsto x \in X and x \notin A.

Since X is a k-space and A is not closed, there is a compact K \subset X such that K \cap A is not closed in K. Every point of K is a G_\delta-set in X and thus a G_\delta-set in K. It follows that K is first countable.

Let x \in \overline{K \cap A} such that x \notin A (the closure is taken in K). Since K is first countable, there is a sequence x_n \in K \cap A such that x_n \mapsto x. This means A is not sequentially closed in X. \blacksquare

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