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

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?

• Well, it is compact Hausdorff (in the Vietoris topology). But I suppose you're asking what uniformity adds to the picture. – Forever Mozart Dec 24 '18 at 18:26
• 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. – Henno Brandsma Dec 26 '18 at 22:53
• 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. – Wlod AA Dec 31 '18 at 8:04

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

• For readers who don't have the book, could you at least give the statement of the exercise in question? – Nate Eldredge Dec 31 '18 at 0:21
• @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. – Martin Sleziak Dec 31 '18 at 6:40
• Is 'hyperspace' (mentioned in your second sentence) the name for $2^X$? – LSpice Dec 31 '18 at 18:57
• Yes, it's one of the more common terms for that entity – KP Hart Dec 31 '18 at 19:16