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

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 $\{\omega_1\}$ (equivalently, the base of open sets generated by cocountable final segments $[\alpha,\omega_1]$ as well as $[0,\omega_1)$). 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.

Answer 2

To extend the answer by Tyrone, I'll explicitly discuss the modification of Isbell's space to be $T_1$ (finite sets are closed) described in that same paper, establishing several of its properties for $\pi$-Base.

Let $X=\omega_1+1$ have the basis $\{[0,\omega_1)\setminus F:F\text{ finite}\}\cup\{[\alpha,\omega_1]\setminus F:\alpha<\omega_1,F\text{ finite}\}$.

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We still have that this space is Noetherian, a.k.a. hereditarily compact (P208 of $\pi$-Base): given an open cover of a non-empty subspace, the least point of the subspace is contained in a member of the cover that misses only finitely-many points of the subspace. It follows that the space is locally compact.

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This also means that the space is compactly generated in the sense of compact subspaces (P140 of $\pi$-Base). However, it is not compactly generated in the sense of continuous maps from compact Hausdorff spaces (P141 of $\pi$-Base). We will see this by showing that $A=[0,\omega_1)$ is a non-closed set for which $f^\leftarrow[A]$ is closed for every $f:K\to X$ with $K$ compact Hausdorff. (This is in fact Tyrone's proof, just with the details written slightly differently to my personal preference; the $T_1$ modification does not affect things.)

Let $\alpha_0=\omega$, and suppose $\alpha_n<\omega_1$ is defined with $\alpha_m<\alpha_n$ for all $m<n$. Note that $A_n=[0,\alpha_n)$ is closed in $X$, so $f^\leftarrow[A_n]$ is closed and compact in $K$. Since $A$ is open, $f^\leftarrow[A]$ is open in $K$, so by normality of $K$ we may choose $U_n$ open in $K$ with $f^\leftarrow[A_n]\subseteq U_n\subseteq \overline{U_n}\subseteq f^\leftarrow[A]$.

If $\overline{U_n}=f^\leftarrow[A]$, $f^\leftarrow[A]$ is closed and we are done. If not, we may choose $\alpha_n<\alpha_{n+1}<\omega_1$ with $A_{n+1}=[0,\alpha_{n+1})$ and $\overline{U_n}\subsetneq f^\leftarrow[A_{n+1}]$.

But if $\alpha_n$ is defined for all $n<\omega$, we'd have $\alpha_\omega=\lim_{n<\omega}\alpha_n$, $A_\omega=[0,\alpha_\omega)$ closed in $X$, and $f^\leftarrow[A_\omega]$ closed and compact in $K$. But then $\{U_n:n<\omega\}$ is an open cover of $f^\leftarrow[A_\omega]$ with no proper subcover, a contradiction.

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Finally, while the space is $T_1$, it is certainly not $T_2$; in fact, the space is hyperconnected, a.k.a. anti-Hausdorff (P39 of $\pi$-Base), as every pair of non-empty open sets has uncountable intersection.

The space also fails to be US (Unique Sequential limits) (P99 of $\pi$-Base): both $0$ and $1$ are limits of the sequence $x_n=n$.

The space also fails to have closed retracts (P101 of $\pi$-Base). $f:X\to X$ defined by $f(0)=1$ and $f(x)=x$ otherwise is continuous: open sets either miss $1$ and equal their preimage, or contain $1$ and contain either $[0,\omega_1)\setminus F$ or $[1,\omega_1]\setminus F$, and thus the preimage contains $[0,\omega_1)\setminus F$ and is open. However, its retract $[1,\omega_1]$ is not closed.

Answer 3

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 4

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