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Let $X$ be a topological space and let $\mathcal{K}(X)$ denote the hyperspace of all its compact subspaces endowed with the Vietoris topology. When is $\mathcal{K}(X)$ a $k$-space?

We know that if $X$ is metrizable, then $\mathcal{K}(X)$ is also metrizable, and hence it is a $k$-space. Also, if $X$ is locally compact, then every compact $K ⊆ X$ has a compact neighborhood, and so $\mathcal{K}(X)$ is a $k$-space.

On the other hand, if $\mathcal{K}(X)$ is a $k$-space and $X$ is Hausdorff, then $X$ is closed in $\mathcal{K}(X)$, and so is itself a $k$-space.

Is $X$ being a $k$-space a sufficient condition for $\mathcal{K}(X)$ being a $k$-space? Or are there other sufficient conditions weaker than metrizability and local compactness? Is there any reference for results like this?

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In the paper "on hyperspace of compact subsets of $k$-spaces" by Momir Stanojevic, this question is discussed.

It refers to a Russian paper by Popov that gives examples of $X$ that are $k$-spaces such that $\mathcal{K}(X)$ is not a $k$-space, but the author also gives a machine for producing examples:

Let $X_1$ and $X_2$ be $k$-spaces such that $X_1 \times X_2$ is not a $k$-space. (e.g. see Engelking 3.3.29): $X_1 = \mathbb{R}\setminus\{\frac{1}{n}: n =2,3,4\ldots\}$, $X_2 = \mathbb{R}/\mathbb{N}$ (the positive integers in $\mathbb{R}$ indentified to a point, in the quotient topology). For the argument, see Engelking. We could use other examples, if you happen to know them.

Define $X = X_1 \oplus X_2$, their disjoint topological sum. Then $X$ is a $k$-space, clearly.

But the subset $C = \{\{x_1,x_2\}: x_1 \in X_1, x_2 \in X_2\}$ is closed in $\mathcal{K}(X)$ and homeomorphic to $X_1 \times X_2$, so not a $k$-space. (check the homeomorphism and the closedness!).

As a closed subspace of a $k$-space is a $k$-space, this implies that $\mathcal{K}(X)$ is not a $k$-space. So $X$ being a $k$-space is not enough.

In fact from standard facts he proves that all powers $X^n$ $n=1,2,\ldots$ have to be $k$-spaces too (here the square $X^2$ fails to be a $k$-space). This one often sees: hyperspaces of $X$ behave like $X^\omega$ in many ways.

Stanojevic also considers iterated hyperspaces $\mathcal{K}^{(1)}(X) = \mathcal{K}(X)$ and $\mathcal{K}^{(n+1)}(X)= \mathcal{K}(\mathcal{K}^{(n)(X)})$. There is a natural map of unions from $\mathcal{K}^{(n+1)}(X)$ to $\mathcal{K}^{(n)}(X)$ for all $n$ and the inverse limit of the inverse system so obtained, is called $\mathcal{K}^{(\omega)}(X)$.

Then theorem 7 in the quoted paper says that $\mathcal{K}(X)$ is a $k$-space iff for all $n \ge 2$, $\mathcal{K}^{(n)}(X)$ is a $k$-space iff $\mathcal{K}^{(\omega)}(X)$ is a $k$-space. So if such a hyperspace is a $k$-space it gets preserved by the $\mathcal{K}$-operation itself.

An interesting conjecture based on the above (Stanojevic does not ask this, but it seems reasonable in light of his results):

If $X^\mathbb{N}$ is a $k$-space does it follow that $\mathcal{K}(X)$ is one too?

We need at least all the finite powers to be a $k$-space.

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  • $\begingroup$ Thank you for your thorough answer. The connection with powers of $X$ gave me the following idea. The map $X^κ \to [X]^{≤κ} \setminus \{∅\}$ defined by $f \mapsto rng(f)$ is an open quotient with respect to product topology and lower Vietoris topology. This gives that if $X^{|X|}$ is a $k$-space, so is $Cl(X) \setminus \{∅\}$ with the lower Vietoris topology. $\endgroup$ – user87690 Aug 12 '17 at 19:16
  • $\begingroup$ @user87690 I haven't checked the argument, but it sounds plausible. But we want the full Vietoris topology, right? Maybe $X^n$ a $k$-space for all $n$ is equivalent to $X^\omega$ a $k$-space (and then maybe $\mathcal{K}(X)$ is a $k$-space?) There are such theorems for other non-productive properties (maybe countably tight spaces?). Sometimes there is a connection between the hyperspace and spaces of continuous function $C(X)$, in the compact-open topology as well. $\endgroup$ – Henno Brandsma Aug 14 '17 at 14:35

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