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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.