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## Main Question or Discussion Point

Here is my attempt at the proof:

The statement is equivalent to the proposition that if X is not closed, X is not compact.

If X is not closed, there exists a limit point p that does not belong to X. Every neighborhood of p contains infinitely many points of X, which tells us also that X is infinite.

We will treat the next stage of the proof as two similar cases. We consider the case in which there is an infinite amount of points of X greater than p and the case in which there is an infinite amount of points of X less than p.

In the first case, there will exist a point b[itex]_{0}[/itex] greater than p that belongs to X and a point b[itex]_{1}[/itex] belonging to X such that x < b[itex]_{1}[/itex] < b[itex]_{0}[/itex]. In fact, the process can be continued so that there are points belonging to X such that p < ... < b[itex]_{n}[/itex] < ... < b[itex]_{1}[/itex] < b[itex]_{0}[/itex].

Now we construct an open cover of the points of X between b[itex]_{0}[/itex] and p that has no finite subcover. This open cover will be the union of the intervals described by (p + (1/n); b[itex]_{0}[/itex]) with n going from the nearest natural number greater than 1/(b[itex]_{0}[/itex] - p) to infinity. To show that there is no finite subcover, we can observe that for any finite n, there will be some neighborhood of p (that obviously contains points of X because p is a limit point of X) that does not intersect the interval (p + (1/n); b[itex]_{0}[/itex]). For example, the neighborhood that contains all points bewteen p + (1/n) and some number less than p.

To cover the points of X that are not located between b[itex]_{0}[/itex] and p, we can add to our open cover the open sets (b[itex]_{1}[/itex]; [itex]\infty[/itex]) and (-[itex]\infty[/itex]; p). (It looks like this is the point where the fact that p does not belong to X becomes important.)

A similar argument holds if there is an infinite amount of points of X less than p.

I'm unsure about this solution only because there are a few lemmas that I think I'm supposed to use. One of them states that the exterior of a neighborhood is open, and the other says that no finite subset of the exterior of a neighbohood N covers C\{x} where x is a point belonging to N. I've had trouble finding a way to incorporate these statements into a proof, but even if the above proof is sufficent, I would like to try to use these lemmas somehow. Right now, I thinking that we could try to prove that if X is compact, its complement is open. Because X is compact, for every point p of C\X, there is a open cover of X that could be described as a finite collection of neighborhoods none of which include p. But I fail to see how the lemmas could help you say that p is an interior point of C\X.

Any suggestions?

Thanks!

The statement is equivalent to the proposition that if X is not closed, X is not compact.

If X is not closed, there exists a limit point p that does not belong to X. Every neighborhood of p contains infinitely many points of X, which tells us also that X is infinite.

We will treat the next stage of the proof as two similar cases. We consider the case in which there is an infinite amount of points of X greater than p and the case in which there is an infinite amount of points of X less than p.

In the first case, there will exist a point b[itex]_{0}[/itex] greater than p that belongs to X and a point b[itex]_{1}[/itex] belonging to X such that x < b[itex]_{1}[/itex] < b[itex]_{0}[/itex]. In fact, the process can be continued so that there are points belonging to X such that p < ... < b[itex]_{n}[/itex] < ... < b[itex]_{1}[/itex] < b[itex]_{0}[/itex].

Now we construct an open cover of the points of X between b[itex]_{0}[/itex] and p that has no finite subcover. This open cover will be the union of the intervals described by (p + (1/n); b[itex]_{0}[/itex]) with n going from the nearest natural number greater than 1/(b[itex]_{0}[/itex] - p) to infinity. To show that there is no finite subcover, we can observe that for any finite n, there will be some neighborhood of p (that obviously contains points of X because p is a limit point of X) that does not intersect the interval (p + (1/n); b[itex]_{0}[/itex]). For example, the neighborhood that contains all points bewteen p + (1/n) and some number less than p.

To cover the points of X that are not located between b[itex]_{0}[/itex] and p, we can add to our open cover the open sets (b[itex]_{1}[/itex]; [itex]\infty[/itex]) and (-[itex]\infty[/itex]; p). (It looks like this is the point where the fact that p does not belong to X becomes important.)

A similar argument holds if there is an infinite amount of points of X less than p.

I'm unsure about this solution only because there are a few lemmas that I think I'm supposed to use. One of them states that the exterior of a neighborhood is open, and the other says that no finite subset of the exterior of a neighbohood N covers C\{x} where x is a point belonging to N. I've had trouble finding a way to incorporate these statements into a proof, but even if the above proof is sufficent, I would like to try to use these lemmas somehow. Right now, I thinking that we could try to prove that if X is compact, its complement is open. Because X is compact, for every point p of C\X, there is a open cover of X that could be described as a finite collection of neighborhoods none of which include p. But I fail to see how the lemmas could help you say that p is an interior point of C\X.

Any suggestions?

Thanks!