# 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.