# Proving that if X is compact, X is closed

1. Jun 23, 2011

### math771

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$_{0}$ greater than p that belongs to X and a point b$_{1}$ belonging to X such that x < b$_{1}$ < b$_{0}$. In fact, the process can be continued so that there are points belonging to X such that p < ... < b$_{n}$ < ... < b$_{1}$ < b$_{0}$.
Now we construct an open cover of the points of X between b$_{0}$ and p that has no finite subcover. This open cover will be the union of the intervals described by (p + (1/n); b$_{0}$) with n going from the nearest natural number greater than 1/(b$_{0}$ - 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$_{0}$). 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$_{0}$ and p, we can add to our open cover the open sets (b$_{1}$; $\infty$) and (-$\infty$; 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!

2. Jun 23, 2011

### mathwonk

this is true if you assume also hausdorff. but in non hausdoprff spaces, all finite sets are compact, but finite sets are not always closed.

3. Jun 23, 2011

### micromass

Hi math771!

Your proof looks ok. I have some questions though. First of all, I assume that you're talking about subsets of $\mathbb{R}$? Your proof is not true outside that context.

Why?

What's the point of continuing the process?

Why is this a cover?
The rest of the proof sounds ok...

4. Jun 23, 2011

### math771

Well, I'm working in the continuum, which I think will soon be revealed to have the same properties as R (but I haven't gotten to that part yet).

A neighborhood of a limit point p of X contains an infinite amount of points of X for this reason (I think): if we suppose that there is a neighborhood of p that contains a finite amount of points of X, we can construct a neighborhood around p that contains only those points between those elements of X that are closest to p on both sides (hopefully that's not too convoluted). This neighborhood would contain no points of X and would thus contradict the definition of limit point.

Now that I reflect on it, I'm not sure that continuing the process is necessary. I guess I was just trying to visualize the situation.

The union of intervals is a cover because for any point of X, called x, such that p < x < b$_{0}$, there is an n such that p + (1/n) < x < b$_{0}$. To show this, we observe that x - p > 0, and that by the archimedean property, n(x - p) > 1 for some natural number n. Thus, x > p + (1/n).

5. Jun 23, 2011

### micromass

OK, sounds like you've got it!! Nice proof by the way.

6. Jun 23, 2011

### math771

Thank you!

7. Jun 23, 2011

### mathwonk

Here is a usual argument. Let K be compact in a hausdorff space and let p be outside K and q inside K. By hypothesis there is an open set Uq containing q whose closure does not contain P, (because some open set around p misses U). By compactness, there is a finite number of such Uq that cover K, none of whose closures contain p. Thus the union of their closures is a closed set containing K but not p. Hence for every p outside K, there is a closed set Cp containing K but not p. The intersection of these sets Cp is a closed set containing K but no point outside K, hence equals K.

but an argument you do yourself is always better.

8. Jun 23, 2011

### math771

Thanks, mathwonk. That one's neat.

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