After I introduced set functions, I looked at complements, unions, and intersections of sets as applied to set functions. Now that we are looking at open and closed sets, we can do the same thing. Are unions of closed sets necessarily closed?

Let's start with complements of sets. If X is a metric space, and E is an open set in that metric space, what can we say about about E^{c}? For example, take X as the real numbers with distance function d(p,q)=|p−q|, and take E as the set of all real numbers x such that x<1. It's easy to show that E is an open set. Take any point x in E. x must be less than one. Define a radius r = (1−x)/2. Then the neighborhood of x with radius r must lie entirely within E. Since this is true for any point in E, E is an open set.

Let's look at the complement of E. E^{c} is the set of all points in X which are not in E, so in our case it's all real numbers greater than or equal to 1. Look at any limit point in E^{c}. Any neighborhood of the limit point must contain a point in E^{c}. This means that no neighborhood of that point can be entirely contained in E, so that point can't be in E. Therefore, that point is in E^{c}. Since every limit point of E^{c} is a member of E^{c}, E^{c} is closed. Again, in our example, no point less than 1 can be a limit point of E^{c}, so every limit point of E^{c} is in E^{c,} and E^{c} is closed.

We can run this the other direction as well. Assume E is closed. Choose any point in E^{c}. This point is not in E, and can't be a limit point of E since E is closed. Since it's not a limit point of E, there exists a neighborhood of the point which contains no points in E, or which is a subset of E^{c}. Since this is true for every point in E^{c}, E^{c} is open.

In other words, if any set in a metric space is open, its complement is closed. If any set is closed, its complement is open. Note that the null set is both closed and open. Also, it's possible for a set to be neither closed nor open, in which case its complement is also neither closed nor open.

Let's look at unions of open sets. If we have a collection of open sets, with each set labeled E_{λ} for some index set Λ, then we can take the union of all of the sets ∪_{λ} E_{λ}. If x is some point in the union, then x is a member of one of the individual sets. Since there exists a neighborhood of x in that individual set, and all of the points in that neighborhood are also in the union, every point in the union of sets has a neighborhood in the union, so the union is an open set.

We could probably prove that intersections of closed sets are closed directly. But let's be lazy. Last time I was looking at intersections, unions, and complements of sets, I stated that the complement of a union is equal to the intersection of the complements (De Morgan's Law). If we have a collection of closed sets, each of those is the complement of an open set. If F_{α} is closed, then E_{α} is open, where F_{α}=E_{α}^{c}. Then ∩_{λ} F_{λ} = ∩_{λ} (E_{λ}^{c}). But by De Morgan's Law, this equals (∪_{λ} E_{λ})^{c}. Since we've already shown that the union of open sets is an open set, and the complement of an open set is a closed set, it follows that the intersection of closed sets is a closed set.

This outcome isn't surprising. And you might expect that intersections of open sets are open, and unions of closed sets are closed. This is sometimes true, but there's a catch, which we'll look at next time.