Is there a locally compact space which is not a k-space

Question

Definitions:

A locally compact space is a space where every point has a local base of compact neighborhoods.

A $k$-space $X$ has its topology generated by maps from compact Hausdorff spaces, i.e. $C$ is closed iff for every compact Hausdorff space $K$ and every continuous function $f: K \to X$, $f^{-1}[C]$ is closed in $K$. Strickland's notes call this compactly generated.

By compact I mean not necessarily Hausdorff.

The reason I ask is that standard constructions of non $k$-spaces for example the square of the one-point compactification of $\mathbb{Q}$ and the product $\mathbb{R}\setminus \{1,\frac{1}{2},\frac{1}{3}\} \times \mathbb{R}/\mathbb{Z}$ where the second quotient means identifying $\mathbb{Z}$ to one point, are usually not locally compact.

I'm looking for a locally compact space which is not a $k$-space.

Answer 1

https://en.wikipedia.org/wiki/Compactly_generated_space In the example section it is said that locally compact space are $k$-space.

Answer 2

Surely yes. For every C$^*$-algebra $A$, the primitive ideal space $Prim(A)$ is locally compact (in the sense of each point having a neighbourhood base of compact sets) in the hull-kernel topology. But $Prim(A)$ is seldom Hausdorff.

What about the C$^*$-algebra $A$ of the bounded operators on a separable Hilbert space? Then $Prim(A)$ consists of the ideal $\{0\}$ and the ideal $K(H)$ of compact operators. Topologically it is a two-point $T_0$ space with one point open and the other point closed. Is that a $k$-space?

Answer 3

The answer is that, assuming AC, there are locally compact spaces which are not compactly generated. (That is, do not have topologies which are final with respect to all maps with compact Hausdorff domains. i.e. do not belong to the monocoreflective hull of the class of compact Hausdorff spaces as computed in $Top$.)

The construction of such a space is due to J. Isbell, A Distinguishing Example in k-Spaces, Proc. Am. Math. Soc. 100 (1987), 593-594. Below is my attempt at his details.

Let $X=\omega_1+1$, where $\omega_1$ is the first uncountable ordinal, and topologise $X$ by giving it the base of closed sets generated by $(i)$ countable initial segments $[0,\alpha]$, and $(ii)$ the singleton $\{\omega\}$. The space $X$ is our counterexample.

$X$ is locally compact.

Proof: Indeed, more is true: every subspace of $A\subseteq X$ is compact. For $A$, if nonempty, contains a least element, every neighbourhood of which contains all of $A$ except for possibly $\omega_1$. $\square$

$X$ is not compactly generated.

Proof: The claim is that there is a non-closed subspace $A\subseteq X$ with the property that whenever $f:K\rightarrow X$ is a continuous function with $K$ a compact Hausdorff space, the preimage $f^{-1}(A)$ is a closed subset of $K$. We will show that the open subspace $A=[0,\omega_1)$ has this property.

So take $K$ and $f$ as above and suppose $f^{-1}[0,\omega)\subset K$ is not closed. Fix a countable ordinal $\alpha_1$ and consider the closed set $f^{-1}[0,\alpha_1]\subseteq K$. This is disjoint from the closed set $K\setminus f^{-1}[0,\omega_1)$, so by normality of $K$ there are disjoint open sets $U_1,V_1\subset K$ with $f^{-1}[0,\alpha_1]\subseteq U_1\subseteq X\setminus V_1\subseteq f^{-1}[0,\omega_1)$. Because we are assuming that $f^{-1}[0,\omega_1)$ is not closed we have $X\setminus V_1\neq f^{-1}[0,\alpha)$, so in particular the inclusion $U_1\subseteq f^{-1}[0,\omega_1)$ is proper. Thus there is a countable $\alpha_2>\alpha_1$ for which there is a proper inclusion $U_1\subseteq f^{-1}[0,\alpha_2]$.

Now iterate the process to obtain a sequence of proper inclusions $$f^{-1}[0,\alpha_1]\subseteq U_1\subseteq f^{-1}[0,\alpha_2]\subseteq U_2\subseteq\dots$$ Since $\bigcup_\mathbb{N}\alpha_i$ is countable, $\bigcup f^{-1}[0,\alpha_i]=f^{-1}(\bigcup [0,\alpha_i])$ is closed in $K$ and hence compact. On the other hand it is covered by the $U_i$, but by no finite subfamily of them. Thus there is a contradiction. It must be that $f^{-1}[0,\omega_1)$ is closed in $K$. $\square$

Remark: $(i)$ All that was needed of $K$ was its compactness and normality. A more general statement from which the proof flows is contained in Isbell's paper. $(ii)$ We quietly used the axiom of choice to ensure that $\bigcup\alpha_i$ as countable.