**TL;DR.** Given a topological space $X$, is there a natural way to "induce" a topology on $\mathcal{P}(X)$ from the topology of $X$ in such a way that 1) all the basic operations of set theory (intersections, unions, direct images, etc.) become continuous and 2) the topologies on $X$ and $\mathcal{P}(X)$ are compatible in a certain sense?




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Given a topological space $X$, there are a couple of ways to assign a topology to its powerset $\mathcal{P}(X)$ that do not depend specifically on $X$. For example:


* We can view $\mathcal{P}(X)$ as the hom $\operatorname{Hom}(X,2)$, put either the discrete, indiscrete, or Sierpiński topology on $2$, and consider the compact-open topology;
* We can take the Alexandroff topology with respect to $\subset$ or $\supset$;
* There are the so-called [hit-and-miss topologies](https://encyclopediaofmath.org/wiki/Hit-or-miss_topology), of which the Vietoris and Fell topologies are examples.
* (A non-example would be to define a topology on $\mathcal{P}(\mathbb{R})$ using the order of $\mathbb{R}$.)


Given an assignment of topologies on $\mathcal{P}(X)$ from topologies on $X$ as above, we could consider the following sets of niceness conditions:


1. The map $\iota\colon X\to\mathcal{P}(X)$ given by $x\mapsto\{x\}$ is continuous.
2. Binary union ${\cup}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
3. Binary intersection ${\cap}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
4. Difference ${\setminus}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
5. Arbitrary union ${\bigcup}\colon\mathcal{P}(\mathcal{P}(X))\to\mathcal{P}(X)$ is continuous.
6. Arbitrary intersection ${\bigcap}\colon\mathcal{P}(\mathcal{P}(X))\to\mathcal{P}(X)$ is continuous.
7. If $f\colon X\to Y$ is a continuous map of topological spaces, then so are its direct and inverse images
\begin{align\*}
f\_{\*} &{}\colon\mathcal{P}(X)\to\mathcal{P}(Y),\\
f^{-1} &{}\colon\mathcal{P}(Y)\to\mathcal{P}(X).
\end{align\*}


I'm also interested in the following two sets of additional conditions that are a bit more specific:


**More compatibility conditions between $X$ and $\mathcal{P}(X)$:**


8. For each $x\in X$, if $\{x\}$ is closed in $X$, then $\{\{x\}\}$ is closed in $\mathcal{P}(X)$.
9. If $S\subset X$ is closed in $X$, then $\{S\}$ is closed in $\mathcal{P}(X)$.
10. If $U\subset X$ is open in $X$, then $\{U\}$ is open in $\mathcal{P}(X)$.


**Another niceness requirement for the topology on $\mathcal{P}(X)$:**


11. Given any monoid structure $(\star,1\_X)$ on $X$ making it into a topological monoid, the map
$$\circledast\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$$
given by
$$U\circledast V:=\{uv\in X\ |\ u\in U,v\in V\}$$
is continuous.¹
12. (Implies 11 by Tobias's [second comment](https://mathoverflow.net/questions/441087/existence-of-a-really-nice-topology-on-the-powerset-of-a-topological-space#comment1137836_441087).) Given topological spaces $X$ and $Y$, the map
$$\mathcal{P}(X)\times\mathcal{P}(Y)\to\mathcal{P}(X\times Y)$$
given by $(U,V)\mapsto U\times V$ is continuous.
13. Given topological spaces $X$ and $Y$, the bijection
$$\mathcal{P}(X)\times\mathcal{P}(Y)\to\mathcal{P}(X\sqcup Y)$$
given by $(U,V)\mapsto U\cup V$ is a homeomorphism.
14. Given topological spaces $X$ and $Y$, the isomorphism of suplattices
$$\mathcal{P}(X)\otimes\mathcal{P}(Y)\to\mathcal{P}(X\times Y)$$
is a homeomorphism, where $\otimes$ denotes the tensor product of suplattices (see [Eric Wofsey's answer](https://math.stackexchange.com/a/3892072) to [Concrete description of the tensor product of suplattices?](https://math.stackexchange.com/questions/3891985/concrete-description-of-the-tensor-product-of-suplattices)).


**Question I.** Does there exist a powerset topology satisfying (at least but not necessarily only) conditions 1–7? What about (1–7+11), 1–8, 1–9, or 1–10?


**Question II.** Given a topological space $X$, what is the finest topology on $\mathcal{P}(X)$ satisfying conditions 1–6? What about (1–6 + 11–14), or these plus any of 8–10? Lastly, in case this topology turns out to be definable in a way that is independent of $X$ (like the Vietoris topology), do we also have 7?




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**Footnotes.**


¹The motivation for this is that $\circledast$ is the zero-categorical analogue of ordinary Day convolution, and we may compute it via a completely analogous coend formula when viewing subsets $U$ of $X$ as functions $\chi\_U\colon X\to\{\mathrm{true},\mathrm{false}\}$, namely $\chi\_U\circledast\chi\_V=\int^{x,y\in X}\mathrm{Hom}\_{X}(-,xy)\times\chi\_U(x)\times\chi\_V(y)$.



Existence of a *really* nice topology on the powerset of a topological space
1-7 together are pretty strong. You aren't going to have a procedure for doing this without most of the powerset topologies being indiscrete.


First note that if $X$ is a set and $\tau$ is at topology on $\mathcal{P}(X)$ satisfying 2-4, then the function $A \mathbin{\Delta} B = (A \cup B) \setminus (A \cap B)$ is continuous, which implies that for fixed $B\subseteq X$, the map $A \mapsto A \mathbin{\Delta} B$ is a homeomorphism. Therefore for any $A$ and $B$ in $\mathcal{P}(X)$, there is a homeomorphism of $(\mathcal{P}(X),\tau)$ taking $A$ to $B$ (namely the map $C \mapsto C \mathbin{\Delta} (A \mathbin{\Delta} B)$).


When I say a topology is *non-trivial*, I mean specifically that it is not indiscrete.



> 
> **Lemma.** If $X$ is a set and $(\mathcal{P}(X),\tau)$ satisfies 2-4, then the closure $F$ of $\{\varnothing\}$ is a filter (i.e., $A \subseteq B \in F \Rightarrow A \in F$ and $A\in F \wedge B \in F \Rightarrow A \cup B \in F$). In particular, $\tau$ is non-trivial if and only if the $X \notin F$.
> 
> 
> 


*Proof.* Suppose that $A \notin F$ and $B \supseteq A$. Since $\cap$ is continuous, we have that $G = \{C : C \cap A \in F\}$ is closed. Clearly $\varnothing \in G$, so $F \subseteq G$. Furthermore, $B \notin G$, so $B \notin F$.


For showing that $F$ is closed under unions, first note that it is sufficient to show it for disjoint $A,B \in F$. So assume that $A,B \in F$ and $A$ and $B$ are disjoint. Since $C \mapsto C \mathbin{\Delta} B$ is a homeomorphism, we have that $A \in F$ if and only if $A \mathbin{\Delta} B \in \overline{\{B\}}$ (since $\varnothing \mathbin{\Delta} B = B$). Therefore $A \mathbin{\Delta} B \in \overline{\{B\}} \subseteq F$, as required.


For the final statement. Since $\mathcal{P}(X)$ has a transitive homeomorphism group, $\tau$ is non-trivial if and only if $F$ is not all of $\mathcal{P}(X)$. Since $F$ is a filter, this happens if and only if $X \notin F$. $\square\_{\text{Lemma}}$



> 
> **Proposition.** For any topological space $X$, if $(\mathcal{P}(X),\tau\_1)$ and $(\mathcal{P}(\mathcal{P}(X)),\tau\_2)$ satisfy 1-5 and 7 and $\tau\_1$ is non-trivial, then $X$ is discrete.
> 
> 
> 


*Proof.* To hopefully make this proof a little bit easier to read, I'm going to denote the empty set as $\varnothing\_1$ when we're thinking about it as an element of $\mathcal{P}(X)$ and as $\varnothing\_2$ when we're thinking about it as an element of $\mathcal{P}(\mathcal{P}(X))$.


Let $F\_1$ be the $\tau\_1$-closure of $\{\varnothing\_1\}$, and let $F\_2$ be the $\tau\_2$-closure of $\{\varnothing\_2\}$. Since $\tau\_1$ is non-trivial, $X \notin F\_1$ by the lemma. Since $\bigcup: \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ is continuous, we have that $\bigcup^{-1}(F\_1)$ is a closed set in $\mathcal{P}(\mathcal{P}(X))$. Since $\varnothing\_2 \in \bigcup^{-1}(F\_1)$, we have that $F\_2 \subseteq \bigcup^{-1}(F\_1)$, i.e., if $A \in F\_2$, then $\bigcup A \in F\_1$. Therefore, in particular, $\{X\} \notin F\_2$.


By 7, every homeomorphism $f$ of $\mathcal{P}(X)$ induces a homeomorphism $f\_\ast$ of $\mathcal{P}(\mathcal{P}(X))$. Each such homeomorphism $f\_\ast$ fixes $\varnothing\_2$ and takes singletons to singletons (specifically, for any $A \in \mathcal{P}(X)$, $f\_\ast(\{A\}) = \{f(A)\}$). By the discussion before the lemma, the homeomorphism group of $\mathcal{P}(X)$ is transitive. Therefore $\{A\} \notin F\_2$ for every $A \in \mathcal{P}(X)$ and so $F\_2 = \{\varnothing\_2\}$ (because $F\_2$ does not contain one particular singleton, $\{X\}$, so it does not contain any singletons, since it needs to be invariant under any homeomorphisms of $\mathcal{P}(\mathcal{P}(X))$ fixing $\varnothing\_2$).


Consider the function $(x,y) \mapsto \{x\} \cap \{y\}$ from $\mathcal{P}(X) \times \mathcal{P}(X) \to \mathcal{P}(\mathcal{P}(X))$. By 1 and 3, this is continuous. Since $\{\varnothing\_2\}$ is closed, this implies that the off diagonal $\{(x,y) \in \mathcal{P}(X) \times \mathcal{P}(X): x \neq y\}$ is closed and therefore the diagonal is open. It's relatively straightforward to show that if the diagonal of $Y^2$ is open, then $Y$ has the discrete topology. Therefore $\tau\_1$ is discrete. Repeating the argument again (since now we know that $\{\varnothing\_1\}$ is closed) gives that the topology on $X$ is discrete. $\square$


