Now that we know that k-cells are compact sets and that infinite subsets of compact sets have a limit point in the compact set, we can establish some relationships between sets in Euclidean spaces in general.

Suppose E is some set in some Euclidean space, such that every infinite subset of E has a limit point in E. Last time we showed that for a compact set K, every infinite subset of K has a limit point in K, but we have not shown that this works in the opposite direction. We can't conclude yet that E must be compact. But let's see what we can prove based on this starting point.

Given that every infinite subset of E has a limit point in E, what can we say about E? First question: is E bounded? Well, assume that E is not bounded. (This assumption is a giveaway that we are about to prove the opposite by contradiction.) Then given any starting point p in E, we can choose any distance d, and there is always some point in E at a greater distance from p than d. Pick a point p_{1} in E. Pick another point p_{2} in E such that the distance from p_{1} to p_{2} is at least 1.

Then there must exist a point p_{3} in E, such that d(p_{1},p_{3}) is greater than d(p_{1},p_{2}) + 1. (We need the additional distance to ensure that the points are not close to each other.) This allows us to construct an infinite sequence of points, such that the distance from any point to any other point in the sequence is at least 1. (If d(p_{1},p_{n+1})≥d(p_{1},p_{n})+1, then d(p_{n},p_{n+1})≥1 by the triangle inequality.) This sequence is a subset of E, and contains no limit points. This contradicts our premise that every infinite subset of E has a limit point in E, so the conclusion is that E must be bounded.

Next question: is E closed? Again, assume E is not closed, so there exists a limit point of E which is not a member of E. (The limit point is a member of the Euclidean space which contains E, but is not a member of E.) Then for every natural number n, there is a point in E such that the distance from that point to the limit point is less than 1/n. Take the set of these points for every value of n, and this is an infinite set. This infinite set is an infinite subset of E, which has a limit point, but that limit point is not a member of E.

Importantly, the subset has no other limit points. Pick some other point in the Euclidean space. This point is at a distance d from the limit point. There exists an n such that 1/n<d/2. All of the points in the subset which are closer to the limit point than 1/n are farther from the other point than d/2. This implies than that there can only be a finite number of points with distance less than d/2 from the other point, so the other point can't be a limit point of the set. Therefore, the subset is infinite but does not have a limit point which is a member of E. This again contradicts our premise, so E must be closed.

So now we've shown that if E is a set in a Euclidean space such that every infinite subset of E has a limit point in E, then E is both closed and bounded. So, if a set is compact, every infinite subset of the set has a limit point in the set. But also, if a set has the property that every infinite subset of the set has a limit point in the set, it is both bounded and closed.

Continuing the chain, suppose E is some set in some Euclidean space, such that E is both closed and bounded. At this point we are not making any other assumptions about E. Then E is a subset of some k-cell in that space. A k-cell looks like a box, although it can be a multidimensional box, which can be hard to think about. (Check out the iOS app The Fourth Dimension if thinking about what a 4-dimensional box looks like appeals to you.) Since E is bounded, we can set the bounds of the k-cell to the bounds of E, and we have now contained E in the k-cell. Since E is closed, it is a closed subset of the k-cell, which is compact. Therefore, E is also compact.

And we have completed the loop. Take a set in a Euclidean space. If it has any one of these three properties, it must have the other two. The three properties are:

- The set is closed and bounded.
- The set is compact.
- Every infinite subset of the set has a limit point which is a member of the set.

The proof that a set in a Euclidean space is closed and bounded if and only if it is compact is the Heine-Borel theorem.

The proof as constructed here relies on the set being in a Euclidean space. As I previously encountered, in the space of rational numbers with d(p,q)=|p-q|, sets can be closed and bounded without being compact, so the Heine-Borel theorem does not hold in metric spaces in general. (A set of rational numbers can have a limit point equal to an irrational number. In the real space, this set is not closed. In the rational space, the irrational number does not exist, so the set is closed, but it is no longer true that the set has a limit point at all, so the set does not have the second or third listed properties.)

However, we've shown that if a set is compact, every infinite subset has a limit point in the compact set, and this is true in any metric space. I have not shown, but it is also true, that if every infinite subset of a set (in any metric space) has a limit point in that set, the set is compact.