# Proving that if X is compact, X is closed

• math771
In summary: So in this context, the open cover is a subset of the set X that has all of the points between b_{0} and p. right?A similar argument holds if there is an infinite amount of points of X less than p.Again, what's the point of this?
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!

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.

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.

math771 said:
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,

Why?

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$_{0}$.

What's the point of continuing the process?

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.

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

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).

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

Thank you!

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.

Thanks, mathwonk. That one's neat.

## 1. What does it mean for a set to be "compact"?

In mathematics, a set is considered compact if it is both closed and bounded. This means that the set contains all of its limit points and can be contained within a finite region or interval.

## 2. How is compactness related to closedness?

If a set is compact, it automatically implies that the set is closed. This is because a compact set contains all of its limit points, which is one of the criteria for a set to be considered closed.

## 3. Can a set be compact but not closed?

No, a set cannot be compact but not closed. As mentioned earlier, compactness implies closedness. Therefore, if a set is compact, it must also be closed.

## 4. What is the significance of proving that if X is compact, X is closed?

Proving this statement is important because it helps us understand the relationship between compactness and closedness, which are fundamental concepts in mathematics. It also allows us to use the properties of compact sets in proving other theorems and solving mathematical problems.

## 5. Can this statement be applied to all types of sets?

Yes, this statement can be applied to all types of sets. It is a general property of compact sets and holds true for any set that satisfies the criteria of being compact (i.e. closed and bounded).

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