# Is there a “natural” topology on powersets?

Let's say a topology T on a set X is natural if the definition of T refers to properties of (or relationships on) the members of X, and artificial otherwise. For example, the order topology on the set of real numbers is natural, while the discrete topology is artificial.

Suppose X is the powerset of some set Y. We know some things about X, such as that some members of X are subsets of other members of X. This defines an order on the members of X in terms of the subset relationship. But the order is not linear so it does not define an order topology on X.

I haven't found a topology on powersets that I would call natural. Is there one?

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Cofinite topology: the proper closed sets are exactly finite sets. – user31373 Jul 18 '12 at 5:05
Have you seen this part of the powerset Wikipedia page? – Zev Chonoles Jul 18 '12 at 5:06
@LeonidKovalev The cofinite topology on what? $X$ or the $\mathscr{P}(X)$. You can define the cofinite topology on $\mathscr{P}(X)$, but it does not really require use anything about $X$. MikeC wants a topology on $\mathscr{P}(X)$ that uses some property of $X$. – William Jul 18 '12 at 5:09
Perhaps instead of "natural" I should have said "induced". That sounds more precise -- but I don't know how to define "induced"! – MikeC Jul 18 '12 at 19:45

Natural is far from well-defined. For example, I don’t see why the discrete topology on $\Bbb R$ is any less natural than the order topology; one just happens to make use of less of the structure of $\Bbb R$ than the other.

That said, the Alexandrov topology on $\wp(X)$ seems pretty natural:

$$\tau=\left\{\mathscr{U}\subseteq\wp(X):\forall A,B\subseteq X\big(B\supseteq A\in\mathscr{U}\to B\in\mathscr{U}\big)\right\}\;.$$

In other words, the open sets are the upper sets in the partial order $\langle\wp(X),\subseteq\rangle$.

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Yes, a topology defined in terms of the partial order is exactly the sort of thing I was looking for. Thanks! – MikeC Jul 18 '12 at 16:47
@BrianM.Scott can we embed $(X,\mathcal{T})$ in $( \wp(X), \tau)$ ? For a given topological space $(X,\mathcal{T})$ Is there exists a topology $\mathcal {T'}$ on $\wp(X)$ so that we can embed $(X,\mathcal{T})$ in $(\wp(X) ,\mathcal {T'})$ ? – M.A. Dec 13 '15 at 7:47
@M.A.: Yes, though it’s customary to embed it in $\wp^*(X)$, the family of non-empty subsets of $X$. For each finite $\{U_1,\ldots,U_n\}\subseteq\mathscr{T}$ let $$B(U_1,\ldots,U_n)=\left\{A\in\wp^*(X):A\subseteq\bigcup_{k=1}^n U_k\text{ and }A\cap U_k\ne \varnothing\text{ for }k=1,\ldots,n\right\}\;;$$ the family of all such sets $B(U_1,\ldots,U_n)$ is a base for a topology on $\wp^*(X)$, and $X$ embeds in this via the map $x\mapsto\{x\}$. – Brian M. Scott Dec 13 '15 at 18:20

Note that $\mathcal P(X)$ is in some sense "naturally" equivalent to (and certainly in bijection with) the product space $\prod\limits_{x\in X}\{0,1\}$, since we can identify a subset $S\subseteq X$ with the element of the product space that has a $1$ in the $x$ place if $x\in S$ and a $0$ in the $x$ term otherwise. For example, the set $\{1,3\}\subseteq \{1,2,3\}$ would be identified with the element $(1,0,1)\in \prod\limits_{i=1}^3 \{0,1\}$. Using this, we can give $\mathcal P(X)$ the product topology, taking the discrete topology on $\{0,1\}$.

However, this may not be what you're looking for, since it depends only on the underlying set $X$ and not on its topology.

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"Natural" very much depends on context. The product topology of discrete spaces described in this answer is natural in descriptive set theory, even if the factors (or exponents) carry a topology, other than the discrete one. (For example in the context of determinacy, where products of the form $\kappa^{\aleph_0}$ are common for large uncountable cardinals $\kappa$, or in the context of Borel equivalence relations, where one favors Polish topologies.) – Andrés E. Caicedo Jul 18 '12 at 21:21

Your question is vague. You want a topology on $\mathscr{P}(X)$ that uses some property of $X$. If $X$ is any arbitrary set, you don't exactly know if $X$ satisfies any particular property.

However, there is a possible topology you can put on $\mathscr{P}(X)$ if $X$ is a topological space. I am not entirely sure that this is what you are looking for since a topology on $X$ is not a property of $X$, although often times a property of $X$ (such order as mentioned) can be used to define topologies.

Anyway, suppose $X$ has a topology. You can define the topological space generated by the sub basis

$\{K \in \mathscr{P}(X) : K \subseteq U\}$

$\{K \in \mathscr{P}(X) : K \cap U \neq \emptyset\}$

where $U$ ranges over all open subsets of $X$.

This is something artificial I made up, but if you replace $\mathscr{P}(X)$ with the set of all compact subset of a topological space $X$, then you obtain the Vietoris topology of the set of all compact subsets of a topological space $X$. The Vietoris Topology is actually a useful topology.

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In E. Michael, TOPOLOGIES ON SPACES OF SUBSETS says that, the topologies on $\mathcal{P}^* (X)=\mathcal{A}(X)$ generated by $\{K \in \mathscr{P}(X) : K \subseteq U\}$ and $\{K \in \mathscr{P}(X) : K \cap U \neq \emptyset\}$ are admissible with respect to the generating topology on $X$. what does they are admissible with respect to the generating topology on $X$ means? – M.A. Dec 13 '15 at 7:20

What about the following: A set $S\subset P(X)$ is open if for any $x\in S$ it also contains all $y\subset x$.

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I wonder if this is the same as the Alexandrov topology mentioned by Brian M. Scott. It's not obvious to me one way or the other. – MikeC Jul 18 '12 at 19:43
@MikeC: If I understand it correctly, in the example of Brian M. Scott for every set in $S$ all the supersets are also in $S$. Which if I'm not mistaken would mean the open sets in that topology are exactly the complements of the open sets in my example. However from the link I gather that the term "Alexandrov topology" is more general, and I think my example should also be one. – celtschk Jul 18 '12 at 20:19

The question may be more relevant to your motivation if you suppose $Y$ is a topological space and you ask for a "reasonable" topology on the power set of $Y$. This is quite tricky but there are topologies on the family of closed subsets, or of open subsets, see for example

R. Brown and A.M. Abd-Allah, A compact-open topology on partial maps with open domain'', J. London Math Soc. (2) 21 (1980) 480-486.

The idea is to use the Sierpinski space $S= \{0,1\}$ in which $1$ is open but not closed. Then the set $\mathcal O(Y)$ of open sets of $Y$ is given by a corresponding set $\mathcal{CO}(Y)$ of characteristic maps $Y \to S$; this set can be given the compact-open topology. This topology is non-Hausdorff of course.

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