Now that we've gotten a feel for open and closed sets, we can move on to the next topic, which is compact sets. I'm going to admit before I begin that this topic is really pushing the limits of my understanding. I'm not sure if that's because of lack of prior familiarity or because of the way it's presented in the textbook. Hopefully I won't get anything seriously wrong. One thing that I'm concerned about is that my examples are mostly in Euclidean spaces, and Euclidean spaces are sufficiently well behaved that I'm concerned that I might be overlooking important examples from other types of metric spaces.

Thinking about the process of studying math, one thing that is becoming clear is that you can't study just one topic. If you really want to understand one area of mathematics, you have to be willing to study lots of other, not necessarily related, subjects as well. In my case, one of my underlying interests is probability theory. However, at a certain level, to really understand what's going on in probability theory, you have to have a good understanding of certain topics in real analysis. So I'm studying real analysis, but now I'm finding that really understanding analysis requires an understanding of topology. So now I'm trying to make sense of compact sets. In another example, I studied a bit of abstract algebra last fall just for fun, and have since come to appreciate how that provides useful background for both probability theory and real analysis. I didn't realize at the time that it would be relevant, but now I appreciate having looked at the topic.

To study compact sets, let's start in the same place that we always start. We have a metric space, X. We have a set in the metric space, E. Right now we don't know anything about the set. For ease of visualization, we can picture the space as a piece of paper and the set as a square (or some other shape) marked out on the paper.

We are now going to define an open cover of E. An open cover is a collection of open sets in X such that every point in E is in at least one of the open sets. Going back to the piece of paper, drop a bunch of circles on the page. Each circle represents an open set. (The open sets don't have to be circles, so use other shapes if you want.) The open sets can partially overlap E, but they can also include points in X which are not in E. If every point in E is covered by a circle, then the collection of circles (open sets) is an open cover of E.

It should be obvious that lots of different collections of open sets can all be open covers of the same set E. In particular, there can potentially be an infinite number of sets in the open cover. One obvious way is if E extends to infinity in some direction. We can say that E is bounded if E is a subset of some neighborhood of some point. If E is not bounded, then if each set in the open cover of E is bounded, there will need to be an infinite number of sets in the open cover in order to cover every point. Sooner or later I'm going to extend the piece of paper example too far, but imagine that the paper extends infinitely in every direction, and that E is a column on the page that also extends infinitely. If the open sets are finite circles, then each additional set can cover another piece of the column, but the column will alway extend beyond the last circle. So the circles can only be an open cover of E if there is no last circle, in other words if the collection of circles is infinite.

But you can get an infinite open cover even if E is bounded. The trick is to go in the other direction, and start making the sets in the open cover smaller and smaller. There's no rule that the sets in the open cover can't be subsets of other sets in the open cover, so you can pick a point in E and consider all neighborhoods of that point as part of the open cover. As the radius gets smaller, each neighborhood is a subset of the larger neighborhoods, but they are all part of the open cover, so the number of sets in the open cover is infinite.

A set is compact if any open cover of the set always contains a finite subcollection which is also an open cover. Now that we've buried our square under an infinite pile of circles, start removing circles which entirely overlap other circles. If we can always reduce the pile to a finite set of circles, which still covers the square, regardless of what the collection of circles originally looked like, then the square is a compact set. Of course, when I say square here, I mean any set E in any metric space, and when I say circles, I really mean open sets. E could have several disconnected pieces. In three dimensions, E could be a box and the open sets could be balls. Don't get too attached to one visualization. In fact, it's better to always try to find new visualizations.

Let's look at an example. Define our metric space as the real number line, with our distance function being the usual absolute value of the difference between the two points. Define E as the set of all points that can be written as 1/n, where n is any natural number. We've looked at this set before. It's not open since the gaps between points in the set mean that every neighborhood of every point contains lots of points not in the set. It's not closed since 0 is a limit point of the set, but 0 is not a member of the set. Today's question is: is this a compact set?

Well, let's create an open cover for the set. Take the open segment (−1,2). Since every point in E is greater than −1 and less than 2, this set by itself is an open cover of the set. Since 1 set is a finite number of sets, this is a finite open cover. But E isn't compact unless *every* open cover includes a subcollection which is a finite open cover. So let's look at another open cover.

For the point 1/n, for any natural number n, the closest point in the set is the point 1/(n+1). So, for each point, take a neighborhood of radius of half the distance to the nearest point, (for any point 1/n, the algebra works out to a radius of 1/[2n(n+1)]). This neighborhood is an open set, so we can build an open cover of E by including one neighborhood for each point in E. Each neighborhood contains exactly one point in E, and there are an infinite number of points in E, so there are an infinite number of sets in the open cover. No sets can be removed while continuing to cover the entire set, so E is not compact.

Let's look at another example. In the same space, again look at the set of all points that can be written as 1/n, where n is a natural number, but also add 0. (Call this set F to distinguish it from the last example.) Now every open cover of F must contain an open set which contains 0. The open set which contains 0 must include all points in a neighborhood of some radius r around zero (by the definition of open sets). r is some positive real number. Therefore, there exists some number N such that 1/N < r. (This fact is the archimedean principle, which we have probably used before, but perhaps not by its name.) Therefore, every point of the form 1/n, where n is greater than or equal to N, is contained in the open set which contains 0. There are only a finite number of natural numbers less than N, so there only a finite number of points in F which are outside of the set containing 0. Therefore, even if every other open set contains a only a single point in F, since there must be an open set containing 0, every open cover of F has a finite subcollection which is also an open cover of F. Therefore, F is a compact set.

Now that we've looked at an example, next time we will start looking at properties of compact sets in general.