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Compactness and Metric Spaces

  1. Sep 29, 2010 #1
    In R we have the following result:

    A set C is compact if and only if it is closed and bounded. However, the converse of this statement isn't always true in a general metric space. What makes R so special? Or in other words, what conditions would I need on an arbitrary metric space for the converse to hold?

    This is something I was just thinking about, any ideas would be appreciated!
     
  2. jcsd
  3. Sep 29, 2010 #2
    It is complete.

    You'll want to consider sets that are not just bounded but totally bounded (or pre-compact).
     
  4. Sep 30, 2010 #3
    Hmm I'm going to have to take a closer look at this. Thanks
     
  5. Oct 4, 2010 #4
    Having a complete metric space isn't enough. For instance, consider the complete metric space ℓ of bounded sequences of real numbers, with the sup norm. Then the closed unit ball is closed and bounded, but not compact.

    Now, if you have a complete metric space where bounded sets are totally bounded, then closed and bounded subsets are compact.
     
  6. Oct 5, 2010 #5
    Hmm I'm having troubles understanding what totally bounded means...can someone help me out? It just seems that 'totally bounded' has the same definition of compactness :S So maybe I'm not understanding something?
     
  7. Oct 5, 2010 #6
    A subset A of a metric space X is totally bounded if for every r > 0, there is a finite open cover of A by open balls of radius r.
     
  8. Oct 5, 2010 #7

    lavinia

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    What about this metric space?

    take the unit disk in the Euclidean plane and define the distance between two points to be the sum of the lengths of the two line segments that connect the two points to the origin. The disk is bounded and closed under this metric.

    Consider an open cover by intervals that lie completely on the line segments through the origin. There is no finite subcover. In fact there is no countable subcover.
     
    Last edited: Oct 5, 2010
  9. Oct 5, 2010 #8
    That's not a metric: if a point x is not at the origin, then d(x,x) is not zero.
     
  10. Oct 6, 2010 #9

    lavinia

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    OK mia culpa for being a little inaccurate. You get a metric if you take the usual distance between two points if they are on the same segment to the origin and take the sum if they are not. This is a good example of a complete bounded metric space that is not compact I think.
     
    Last edited: Oct 6, 2010
  11. Oct 6, 2010 #10
    So wait I'm getting confused now lol Is complete and totally bounded enough then?
     
  12. Oct 6, 2010 #11

    lavinia

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    I don't think the example I posted is totally bounded.
     
  13. Oct 6, 2010 #12

    HallsofIvy

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    Here is a simpler example in which "closed and bounded" does not imply "compact. Let A be an infinite set and define d(x,y)= 1 if [itex]x\ne y[/itex], d(x, x)= 0. That is a metric and, since the distance between any two distinct points is never less than 1, the neighborhood of point x, with radius 1/2 (or any number less than 1) is just the singleton set {x} itself.

    That means that every singleton set {x} is open and, since any set can be written as a union of its singleton subsets, every set is open. Since a set is closed if an only if it is the complement of an open set, and every set is the complement of some set, it follows that every set is closed. That is, every set is both closed and bounded. This is the "discrete topology" on A.

    Since the distance between two points is never larger than one, every set is bounded. In this metric, every set is both closed and bounded.

    But it is easy to see that infinite sets are not compact. Given an infinite set, X, the collection of all singleton sub-sets is an open cover. Since every point in the set X is in one and only one of the subsets, we cannot remove any of them and so cannot have a finite subcover.

    But the really important example of a space in which there exist closed and bounded sets that are not compact is the Rational numbers, with the metric topology defined by d(x,y)= |x- y|.

    Let [itex]A= \{x \in Q| x\ge 0, x^2\le 2[/itex]. That is a closed and bounded set of rational numbers. Let [itex]\{x_n\}[/itex] be a sequence of rational numbers converging to the irrational number [itex]\sqrt{2}[/itex]. Then the collection of open sets [itex]\{0< x< x_n, x\in Q\} is an open cover for A which has no finite subcover.
     
  14. Oct 6, 2010 #13

    lavinia

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    A metric space must be complete and totally bounded to be compact. The example of the rationals is not complete. Closed is not enough since every topological space is closed. If the space is not complete then a Cauchy sequence that fails to converge will inherit the discrete topology.
     
    Last edited: Oct 6, 2010
  15. Oct 7, 2010 #14
    I was talking to a professor the other day and he said the space had to be complete and separable for this to be true. Is that correct? I have a feeling it isn't lol
     
  16. Oct 11, 2010 #15
    You should have asked him for a proof.
    Every compact metric space is separable.
     
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