# Defining a topology in the Power Set

I have the following question:

1. Given a topological space $T$ is possible in general to give a topology to $2^T$ (the power set of $T$) such that this topology in $2^T$ is related to $T$.

2. If the answer in general is no, are there conditions over $T$ to do this?.

3. I'm interested to know if given a topological space $T$ and a topology on $2^T$ that is induced by the topology on $T$ one can know some topological properties of $2^T$ knowing that of $T$.

• Well, you could think of $2^T$ as the set, or space, of maps from $T$ to a discrete two point space, with the compact-open topology. Of course this is not very interesting if $T$ is connected, and maybe still not very interesting in general. – Dan Ramras Jun 7 '13 at 4:40
• A remark to the construction of Dan Ramras: it becomes much more interesting if you endow the two point space, let us denote it by $2=\{0,1\}$, with the connected topology, where $\{0\}$ is closed and $\{1\}$ is open. Then the pre-image of $\{0\}$ is closed in $T$, and the pre-image of $\{1\}$ is open. And the set $2^T$ of maps $f:T\to 2$ is in one-to-one correspondense with the set of all closed (/open) subsets in $T$, and you can endow $2^T$ with different interesting topologies. So actually, I think you should understand first, whether you need all subsets in $T$ or, say, just closed ones. – Sergei Akbarov Jun 7 '13 at 6:35
• Usually not all of $2^T$ is considered but some interesting subsets. You may wish to look at the monograph S.B. Nadler "Hyperspaces of Sets" (1978), 707pp. – Adam Przeździecki Jun 7 '13 at 7:20
• Do the study of continuous functions on $2^X$-spaces totaly coincide with that of continuous relations? Or will the question of domain for the previous make a difference? – Lehs Aug 27 '14 at 9:13
• Whatever topology you put on it, you should check whether the operation of union is continuous. If so, $2^T$ may well be contractible. (It'll be a topological, unital monoid where every element is idempotent.) – Hiro Lee Tanaka Aug 27 '14 at 11:08

## 5 Answers

It may be better for you to consider uniform spaces instead of simply topological spaces. If you have a uniform space, then there is a very natural topology that one may put on the power set. Uniform spaces are closely related to topological spaces since one may go back and forth between topological and uniform spaces because uniform spaces are topological spaces with some extra structure. If $(X,\mathcal{U})$ is a uniform space, then $(X,\mathcal{U})$ induces a completely regular topology on $X$ where $U$ is open iff for each $x\in U$ there is an entourage $R\in\mathcal{U}$ with $R[x]\subseteq U$. Furthermore, every completely regular space $X$ can be given a compatible uniformity. For example, if $C$ is a compactification of $X$ such as the Stone-Čech compactification, then $C$ has a unique compatible uniformity. This uniformity on $C$ induces a uniformity on $X$.

Suppose that $(X,\mathcal{U})$ is a uniform space. Let $H(X)$ be the set of closed subsets of $X$. Then we can put a uniformity on $P(X)$ as follows. If $R\in\mathcal{U}$, then let $R^{\sharp}$ be the binary relation on $P(X)$ where $(A,B)\in R$ if and only if $A\subseteq R[B]=\{R[b]|b\in B\}$ and $B\subseteq R[A]$. Let $\widehat{R}$ be the restriction of $R^{\sharp}$ to $H(X)$. Then the system $\{R^{\sharp}|R\in\mathcal{U}\}$ generates a uniformity on $P(X)$, but this uniformity generally does not separate points of $X$. However, if we restrict this uniformity to $H(X)$, we get a separated uniformity on $H(X)$ and this uniformity is generated by the set of entourages $\{\widehat{R}|R\in\mathcal{U}\}$. This uniformity on $H(X)$ inherits some of the properties of your original uniform space $X$. For example, if $(X,d)$ is a metric space, then the hyperspace uniformity on $H(X)$ is induced by a metric $d^{\sharp}$ called the Hausdorff metric. The metric $d^{\sharp}$ is defined by $$d^{\sharp}(C,D)=\max[\sup_{c\in C}d(c,D),\sup_{d\in D}d(d,C)]$$ $$=Max[\sup_{c\in C}\inf_{d\in D}d(c,d),\sup_{d\in D}\sup_{c\in C}d(c,d)].$$ Furthermore, the Hausdorff metric $d^{\sharp}$ is complete whenever the original metric $d$ is complete. Now, if $C$ is a compact space, then $C$ can be given a unique uniform structure. With this uniform structure, the hyperspace $H(C)$ of $C$ remains compact. We say that a uniform space $(X,\mathcal{U})$ is non-Archimedean if it is generated by equivalence relations. The hyperspace of a non-Archimedean uniform space is always non-Archimedean.

The hyperspace uniformity is closely related to the Vietoris topology on a topological space which Steven Landsburg referred to in his answer. The reader is referred to Isbell's book on uniform spaces for more information about hyperspaces of uniform spaces.

It's more customary to consider topologies on the set of closed subsets. There is a well-known construction due to Hausdorff which does this for compact metric spaces, and yields an isometric embedding (given by $x\mapsto \{x\}$ of the original space $X$ into another compact metric space. Vietoris shows that the construction can be expressed in purely topological terms. Fell (Proc AMS 01/1962) modified this construction in a way that continues to yield good results even for nonHausdorff spaces! The construction requires closer attention to fine points of general topology. In particular, since the appropriate ambient space is neither Hausdorff nor regular, the proper definition of local compactness is crucial: here it should be in the sense that every open neighborhood of a point contains a compact subneighborhood. For any topological space $X$, Fell's prescription yields a compact topological space $\mathcal{P}_{closed}(X)$ whose points are the closed subsets of $X$, and which is compact. Moreover, if $X$ is locally compact then:

(1) The space $\mathcal{P}_{closed}(X)$ is Hausdorff.

(2) $A_\eta\rightarrow A$ (convergence of nets of subsets) if and only if $\liminf A_\eta =A=\limsup A_\eta$.

(3) If $X$ is second countable then $\mathcal{P}_{closed}(X)$ is second countable and metrizable.

(4) If $X$ is Hausdorff then $X\ni x\mapsto \{x\}\in \mathcal{P}_{closed}(X)$ is an embedding.

There is an evident retraction $\mathcal{P}(X)\rightarrow\mathcal{P}_{closed}(X)$ given by $A\mapsto \overline{A}$. One might consider endowing $\mathcal{P}(X)$ with the weakest topology such that $\mathcal{P}(X)\ni A \mapsto \overline{A}\in\mathcal{P}_{closed}(X)$ is continuous. Of course this will typically yield a nonHausdorff space.

• Can you write down the definition of ${\cal P}_{closed}(X)$ (if it's not too complicated)? – Mario Carneiro Oct 21 '15 at 13:57
• It's equipped with the weakest topology such that the subset $$\{F: F\cap S \neq\varnothing\}$$ is open for every nonempty open $S\subseteq X$, and closed for every compact $S\subseteq X$. – Adam Epstein Oct 22 '15 at 19:51

See my accepted answer to this question and the comments thereon.

For a topological space $(T,\tau)$ you might also consider the topology on $\mathcal{P}(T)$ generated by the subbasis

$\mathcal{S} = \{ S \subseteq \mathcal{P}(T): (\bigcup S) \in \tau\}$.

You can embed $T$ in $\mathcal{P}(T)$ by sending $t\in T$ to $\{t\}$.

A reference to a 1951 survey: Ernest Michael, Topologies on spaces of subsets, TAMS 71(1951), 152-182.