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Proving that if X is compact, X is closed

  1. Jun 23, 2011 #1
    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?
  2. jcsd
  3. Jun 23, 2011 #2


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    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.
  4. Jun 23, 2011 #3
    Hi math771! :smile:

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


    What's the point of continuing the process?

    Why is this a cover?
    The rest of the proof sounds ok...
  5. Jun 23, 2011 #4
    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[itex]_{0}[/itex], there is an n such that p + (1/n) < x < b[itex]_{0}[/itex]. 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).
  6. Jun 23, 2011 #5
    OK, sounds like you've got it!! Nice proof by the way.
  7. Jun 23, 2011 #6
    Thank you! :smile:
  8. Jun 23, 2011 #7


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    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.
  9. Jun 23, 2011 #8
    Thanks, mathwonk. That one's neat. :smile:
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