What do we know about compact sets, besides their kind of clunky and nonintuitive definition? So far, not much. We know that if a set is compact, it is compact in any space which contains it, which is a good start. Let's look at other properties of compact sets.

Assume E is a compact set in some space X. As usual, I'm picturing E as a square drawn on a piece of paper, which represents X. Choose some point p which is outside of E. For every point q in E, we are going to draw two neighborhoods. V_{q} is a neighborhood of p, and W_{q} is a neighborhood of q. Note that both subscripts are q, because we are taking any single fixed point p which is not in E, and then we are considering lots of different q's in E, and we are drawing neighborhoods around p and q depending on the particular q we are currently looking at.

Okay, that was a horrible abstract mess. Mark p on the page, somewhere not inside the square. Pick any point in the square, mark it, and call it q_{1}. Measure the distance between p and q_{1} with a ruler. Call the distance d_{1}. Draw a circle with radius d_{1}/2 around p, and call the circle V_{q1}. Draw a circle with the same radius around q_{1}, and call that circle W_{q1}. Each circle represents an open neighborhood. The circles touch on the page, but they don't overlap. There are no points in both circles.

Next, mark a second point in the square and call it q_{2}. Measure the distance from q_{2} to p, and call the distance d_{2}. Draw a second circle around p with radius d_{2}/2, and call it V_{q2}. Draw a circle around q_{2} with the same radius, and call it W_{q2}.

Skip to the end. Assume we've marked every single point in E, measured the distance, and drawn lots of circles. p will be at the center of lots of circles. Each circle will have its own radius, but it is of course okay if two different circles have the same radius. Looking in E, every point will be the center of exactly one circle, but in general each point will be in the interior of lots of circles.

Now, every neighborhood of every q is an open set. Since every point in E is the center of one circle, the union of the collection of all the neighborhoods is an open cover of E. (Formally, E ⊂ ∪_{q}(W_{q}). Since each W_{q} is an open set, the collection {W_{q}} is an open cover of E.) Since E is compact, a finite subcollection of {W_{q}} is also an open cover.

Look at that finite subcollection. For each set in that subcollection, there is a corresponding neighborhood of p. (If W_{q1} is in the subcollection, V_{q1} is the corresponding neighborhood of p.) Since the collection is finite, there is a smallest neighborhood in the collection. (If the collection were infinite, it's possible that there would be no smallest neighborhood. We are dependent on the fact that E is compact to limit ourselves to a finite collection of sets.) Call the smallest neighborhood V_{qsmall}.

V_{qsmall} is a neighborhood of p. Furthermore, V_{qsmall} can't contain any points in E. In our finite subcollection of {W_{q}}, no W_{qn} can intersect its corresponding V_{qn}. Since we are looking at the smallest V_{q}, none of the W_{q} can intersect it. Go back to the piece of paper with the square and two points marked in the square. If each pair of circles for point q_{1} has a smaller radius than the pair for point q_{2}, then W_{q1} overlaps V_{q2}, but neither circle overlaps V_{q1}. Just extend that to a finite number of circles. (With an infinite number of circles, each circle could be smaller than the previous one, and it all breaks down.)

We've been talking about the set E. Let's turn things inside out and look at the complement of E, E^{c}. p is a member of E^{c}. V_{qsmall} is a neighborhood of p. V_{qsmall} does not intersect any set W_{q} in the finite open cover of E, and E is a subset of its open cover. Therefore, V_{qsmall} does not intersect E, so it is a subset of E^{c}. Therefore, p is an interior point of E^{c}. But p is just any arbitrary point in E^{c}, so every point in E^{c} is an interior point. Which means that E^{c} is an open set. (This chain of reasoning had a lot of individual steps, and if, like me, you're struggling to get up to speed on all these definitions, you may be wondering why each step follows from the last. Everything here comes from the definition of an open set.)

So if E is a compact set, then this means that E^{c} is an open set. Which means, in turn, that E is a closed set. Stated again, in conclusion, every compact set is closed. (The converse is not true. On the real number line, look at the set of all integers. This set is closed, but not compact.)

I keep thinking I'm going to get further talking about compact sets than I do, but I think I need to lay everything out in exhaustive detail to make sure that I understand it. I had hoped to cover more ground today, but I'm going to have to stop here and return to compact sets next week.