Existence of a *really* nice topology on the powerset of a topological space

Question

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, 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.) 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 to 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 $\chi_U\circledast\chi_V=\int^{x,y\in X}\mathrm{Hom}_{X}(-,xy)\times\chi_U(x)\times\chi_V(y)$.

Answer 1

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. $\square$

Answer 2

$\newcommand{\Cld}{\mathrm{Cld}}$I have found partial answers for the case of the Vietoris topology, although they are only for the subspaces $\mathrm{Cld}(X)$ of closed subsets of $X$ and $\mathcal{P}_{0}(X)$ of nonempty subsets of $X$ (and I haven't checked whether they extend to all of $\mathcal{P}(X)$ myself).

Here are the results I found (feel free to add more or edit!):

• Condition 1 (Partial Positive): Corollary 3a of page 166 of Kuratowski–Jaworowski's Topology, Volume I states the map $\iota\colon X\to\Cld(X)$ is an embedding when $X$ is $\mathrm{T}_1$. Exercise 2.6.2 of Klein–Thompson's Theory of Correspondences states this is true (without the $\mathrm{T}_1$ assumption) for $\mathcal{P}_0(X)$ with respect to either the lower Vietoris, upper Vietoris, or Vietoris topologies.

• Condition 2 (Partial Positive): Corollary 4a of page 166 of Kuratowski–Jaworowski's Topology, Volume I states that $$\cup\colon\Cld(X)\times\Cld(X)\to\Cld(X)$$ is continuous.

• Condition 3 (Partial Positive): Corollary 1a of page 180 of Kuratowski–Jaworowski's Topology, Volume I states that if $X$ is $\mathrm{T}_4$, then $$\cap\colon\Cld(X)\times\Cld(X)\to\Cld(X)$$ is upper semicontinuous.

• Condition 4 (Partial Positive): The corollary in page 182 of Kuratowski–Jaworowski's Topology, Volume I states that if $X$ is $\mathrm{T}_3$, then $$\setminus\colon\Cld(X)\times\Cld(X)\to\Cld(X)$$ followed by taking closure is lower semicontinuous.

• Condition 5 (Partial Positive): Theorem 7.2.3 and Corollary 7.2.4 of Klein–Thompson's Theory of Correspondences state that the map $$\bigcup\colon\mathcal{P}_0(\mathcal{P}_0(X))\to\mathcal{P}_0(X)$$ is continuous with respect to either the lower Vietoris, upper Vietoris, or Vietoris topologies applied twice on $\mathcal{P}_0(\mathcal{P}_0(X))$ and once on $\mathcal{P}_0(X)$.

• Condition 6: Still no idea.

• Condition 7 (Partial Negative+Positive): Theorem 2 of page 165 of Kuratowski–Jaworowski's Topology, Volume I states that:

  • $f^{-1}\colon\Cld(Y)\to\Cld(X)$ is continuous (as a relation) iff $f$ is closed and open.
  • If $f$ is closed, then $f_*\colon\Cld(X)\to\Cld(Y)$ is continuous (as a relation).

  Propositions 7.2.9 and 7.2.11 of Klein–Thompson's Theory of Correspondences extend the above results to $\mathcal{P}_0(X)$ (and without a closedness assumption on $f$ for the second result).

• Conditions 8 and 9 (Partial Positive): Theorem 3 of page 162 of Kuratowski–Jaworowski's Topology, Volume I states that if $X$ is $\mathrm{T}_1$, then so is $\Cld(X)$. In particular, $\{\{x\}\}$ is closed in $\Cld(X)$ for any $x\in X$ and so is $\{S\}$ for $S$ closed.

• Condition 10: Still no idea.

• Condition 11 (Partial Positive): Theorem 3.1 of Carruth–Hildebrant–Koch's The Theory of Topological Semigroups, Volume II states that the map $$\circledast\colon\Cld(X)\times\Cld(X)\to\Cld(X)$$ is continuous when $X$ is compact (not sure how important compactness is)

• Condition 12: Still no idea.

• Condition 13: Still no idea.

• Condition 14: Still no idea.