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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 $P$ ?

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  • $\begingroup$ Are you intending to use $y$ in two different ways? You introduce it as a bound variable representing an element of $x$, and then you write that you are looking for all $y \subset x$ such that $P(y)$. $\endgroup$
    – Oliver
    Commented Aug 17, 2011 at 22:14
  • $\begingroup$ Somewhat related question mathoverflow.net/questions/33366/…. $\endgroup$
    – Joel David Hamkins
    Commented Aug 17, 2011 at 23:12
  • $\begingroup$ @Oliver: Edited to clarify $\endgroup$
    – Daniel Miller
    Commented Aug 17, 2011 at 23:15

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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.

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  • $\begingroup$ Sorry, I was a little bit vague about whether or not I wanted $P(x)$ to hold. I am interested in when $P(K_P (X))$ holds for arbitrary $x$. $\endgroup$
    – Daniel Miller
    Commented Aug 18, 2011 at 2:28

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