It is known that if $(X, d)$ is a compact metric space, then hyperspace $2^X= \{A\subseteq X: A\text{ is closed set} \}$ is a compact space with Hausdorff metric

What can say about $2^X= \{A\subseteq X: A\text{ is closed set} \}$, when $(X, \mathcal{U})$ is a compact uniform space?

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    $\begingroup$ Well, it is compact Hausdorff (in the Vietoris topology). But I suppose you're asking what uniformity adds to the picture. $\endgroup$ Dec 24, 2018 at 18:26
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    $\begingroup$ In the original paper(s) by E. Michael where he introduced the Vietoris topology, he also discusses the uniformity in $2^X$. I'd start there. $\endgroup$ Dec 26, 2018 at 22:53
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    $\begingroup$ Every Hausdorff compact space $X$ has EXACTLY one uniformity (basically, given by the neighborhoods of the diagonal of space $X^2$ in $X^2$. This induces a compact topology (and uniformity) in $2^X$. This generalizes the Hausdorff metric in the metric case. $\endgroup$
    – Wlod AA
    Dec 31, 2018 at 8:04

1 Answer 1


As a first step you may want to work problem 8.5.16 in Engelking's General Topology. Keep in mind that a compact Hausdorff space has a unique uniformity, the sets of all neighbourhoods of the diagonal, hence so does the hyperspace.

Spaces of closed subsets V (see Problems 2.7.20, 3.12.27, 4.5.23, 6.3.22 and 8.5.13(i))

8.5.16. Let $(X, \mathcal U)$ be a uniform space and $2^X$ the family of all non-empty subsets of $X$ closed with respect to the topology induced by $\mathcal U$.

(a) Show that the family $\mathcal B$ of all sets $$2^V=\{(A,A')\in 2^X\times2^X; A\subset B(A',V)\text{ and }A'\subset B(A,V)\}$$ where $V\in\mathcal U$ has properties (BU1)-(BU3); the uniformity $2^X$ generated by the base $\mathcal B$ is denoted $2^{\mathcal U}$. Verify that $(X,\mathcal U)$ is uniformly isomorphic to a subspace of the uniform space thus obtained which is closed with respect to the topology induced on $2^X$ by the uniformity $2^{\mathcal U}$.

(b) Verify that if a uniformity $\mathcal U$ on a set $X$ is induced by a metric $\rho$ on the set $X$, then the uniformity $2^{\mathcal U}_{\mathcal M}$ on the family $\mathcal M$ of all bounded, non-empty closed subsets of $(X, \rho)$ coincides with the uniformity induced by the Hausdorff metric.

(c) (Michael [1951]) Show that for every uniformity $\mathcal U$ on a topological space $X$, the topology on $\mathcal Z (X)$ induced by the uniformity $2^{\mathcal U}_{\mathcal Z(X)}$ coincides with the Vietoris topology.

(d) Verify that if the uniform space $(X, \mathcal U)$ is totally bounded, then the space $(2^X,2^{\mathcal U})$ also is totally bounded.

(e) Give an example of a complete uniform space $(X, \mathcal U)$ such that the space $(2^X,2^{\mathcal U})$ is not complete. Hint. Consider the uniformity on the real line generated by the base consisting of all sets of the form $\bigcup\{A\times A; A\in \mathcal A\}$, where $\mathcal A$ is a countable cover of the real line by pairwise disjoint sets.

(f) Show that if the uniform space $(X, \mathcal U)$ is compact, then the space $(2^X,2^{\mathcal U})$ also is compact.

Michael, E. [1951] Topologies on spaces of subsets, Trans. Amer. Math. Soc. 71 (1951), 152-182. https://doi.org/10.1090/S0002-9947-1951-0042109-4 https://www.jstor.org/stable/1990864

  • $\begingroup$ For readers who don't have the book, could you at least give the statement of the exercise in question? $\endgroup$ Dec 31, 2018 at 0:21
  • $\begingroup$ @NateEldredge I have edited the post - I hope that excerpt of this length still qualifies as fair use. The problem relies on some notation defined elsewhere - but I suppose that people experienced enough with this topic are able to guess the missing notation. $\endgroup$ Dec 31, 2018 at 6:40
  • $\begingroup$ Is 'hyperspace' (mentioned in your second sentence) the name for $2^X$? $\endgroup$
    – LSpice
    Dec 31, 2018 at 18:57
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    $\begingroup$ Yes, it's one of the more common terms for that entity $\endgroup$
    – KP Hart
    Dec 31, 2018 at 19:16

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