k-spaces, I

There are many examples in general topology of defining a new topology on a set based on a given topology already defined on the set. For example, given a topology \tau on X, one can define a finer topology \tau_s consisting of all sequentially open subsets of X (based on the original topology \tau). Sequential spaces are precisely those spaces for which the original topology coincides with \tau_s (see the post Sequential spaces, II). A related concept is the notion of k-spaces. We show that the compactly generated open sets form a finer topology and that k-spaces are precisely those spaces for which the compactly generated topology coincides with the original topology. We also give an external characterization of k-spaces, namely, those spaces that are quotient images of locally compact spaces. All spaces under consideration are Hausdorff.

Let X be a space. We say A \subset X is a compactly generated closed set in X if A \cap K is closed in K for any compact K \subset X. We say B \subset X is a compactly generated open set in X if X-B is a compactly generated closed set in X. The space X is said to be a k-space if

A \subset X is a compactly generated closed set in X if and only if A is a closed set.

The direction \Leftarrow of the above statement always holds. So a space is a k-space if it satisfies the direction \Rightarrow in the above statement. We also want to mention that in the above definition, “closed” can be replaced by “open”.

Suppose \tau is the topology of the space X. Then define \tau_k as the set of all compactly generated open sets in X. It can be easily verified that \tau_k is a topology defined on the set X and that \tau_k is a finer topology than the original topology \tau, i.e. \tau \subset \tau_k. It follows that X is a k-space if and only if \tau=\tau_k.

Much of the discussion here mirrors the one in Sequential spaces, II. In a sequential space, the topology coincides with \tau_s, the open sets generated by convergent sequences (a particular type of compact sets). In a k-space, the topology conincides with \tau_k, the open sets generated by the compact sets. A sequential space is the quotient space of a topological sum of disjoint convergent sequences. Any k-space is the quotient space of a topological sum of disjoint compact sets.

We have the following theorem.

Theorem
For any space X, the following conditions are equivalent:

  1. X is a k-space.
  2. X is a quotient space of a locally compact space.

Proof. 1 \Rightarrow 2 Let X be a k-space. Let \mathcal{K} be the set of all compact subsets of X. Let Y=\oplus_{K \in \mathcal{K}}K be the topological sum of all K \in \mathcal{K} where each K \in \mathcal{K} has the relative topology inherited from the space X. Then Y is a locally compact space. There is a natural mapping we can define on Y onto X. The space Y is a disjoint union of all compact subsets of X. We can map each compact set K \in \mathcal{K} onto the corresponding compact subset K of X by the identity map. Call this mapping f where f:Y \mapsto X. We claim that the quotient topology generated by this mapping coincides with the original topology on X.

Let \tau be the given topology on X and let \tau_f be the quotient topology generated on the set X. Clearly, \tau \subset \tau_f. We need to show that \tau_f \subset \tau. Let O \in \tau_f. Since X is a k-space, if we can show that O is a compactly generated open set, then O \in \tau.

Let K \subset X be compact. We need to show that O \cap K is open in K. Since O \in \tau_f, f^{-1}(O) is open in Y. Since K is open in Y, f^{-1}(O) \cap K is open in Y. It is also the case that f^{-1}(O) \cap K is open in K. We can consider f^{-1}(O) \cap K as a subset of K \subset Y and as a subset of K \subset X. As a subset of K \subset X, we have f^{-1}(O) \cap K=O \cap K. Thus O \cap K is open in K and O \in \tau.

2 \Rightarrow 1 Let Y be locally compact and let f:Y \mapsto X be a quotient map. We show that X is a k-space. To this end, we show that if A \subset X is a compactly generated closed set in X, A is closed in X. Or equivalently, if A is not closed in X, then A is not a compactly generated closed set in X. Under the quotient map f, A is closed in X if and only if f^{-1}(A) is closed in Y.

Suppose A is not closed in X. Then f^{-1}(A) is not closed in Y. Then there is y \in \overline{f^{-1}(A)}-f^{-1}(A). Let U \subset Y be open in Y such that y \in U and \overline{U} is compact. Then f(\overline{U}) is compact. It follows that A \cap f(\overline{U}) is not closed in f(\overline{U}). Note that f(y) \in f(\overline{U}) and f(y) \notin A \cap f(\overline{U}). However, f(y) \in \overline{A \cap f(\overline{U})}. Thus A is not a compactly generated closed set in X. \blacksquare

Reference

  1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.
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