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