When one is first learning about topologies or $\sigma$-algebras, a common (trivial) exercise is showing that they are closed under intersections, but not in general under unions, i.e., any intersection of topologies is a topology, but the union of two topologies may not be a topology. When I was first learning topology, I found this to be rather disappointing, for otherwise one could form the "topology of topologies" on a given set $X$.


Suppose we have a property $P$, such as "$\forall y,z\in x \:(y\cap z\in x)$." Now for any set $x$, we may form the set $K\_P (x)$ of all $t\subset x$ such that $P(t)$ holds. Can we say anything about $P$ if, for all $x$, $P(K\_P (x))$ holds? Of course, a trivial example of such a $P$ is the property of being closed under intersections. Is that the only such $P$ ?



Self-Satisfying Properties
It seems that there are many properties with your requested feature. 


* For example, consider the vacuously true property $P(x)$, which holds of any set. Note that for any set $x$, the set $K\_P(x)$ is simply the power set of $x$, and $P$ holds of this set, since it holds of any set. So this property has your feature.
* Consider also the silly property, $P(x)=$"$x$ is empty or contains the emptyset as an element." Note that for every set $x$, the set $K\_P(x)$ contains the emptyset as an element, and so this weird property has the requested feature.


I wasn't clear on whether you were insisting that $P(K\_P(x))$ hold only given $P(x)$, but if so, then the following examples also work:


* Consider the property $P(x)=$"$x$ is finite", which works because a finite set has only finitely many (finite) subsets.
* Similarly, $P(x)=$"$x$ is infinite" works, since every infinite set has infinitely many infinite subsets.
* Similarly, "$x$ is uncountable", etc. There are many more.


