Compactness in Metric Spaces

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Compact Sets in a Metric Space

If is a metric space and then a cover or covering of is a collection of subsets in such that:

Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.

We also said that a subset is a subcover/subcovering (or open subcover/subcovering if is an open covering) if is also a cover of , that is:

We can now define the concept of a compact set using the definitions above.

Definition: Let be a metric space. The subset is said to be Compact if every open covering of has a finite subcovering of .

In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.

Consider the metric space where is the Euclidean metric and consider the set . We claim that this set is not compact. To show that is not compact, we need to find an open covering of that does not have a finite subcovering. Consider the following open covering:

Clearly is an infinite subcovering of and furthermore:

Let be a finite subset of containing elements. Then:

Let . Then due to the nesting of the open covering , we see that:

But for we need . But , so and , so . Therefore any finite subset of cannot cover . Hence, is not compact.

Boundedness of Compact Sets in a Metric Space

Recall that if is a metric space then a subset is said to be compact in if for every open covering of there exists a finite subcovering of .

We will now look at a rather important theorem which will tell us that if is a compact subset of then we can further deduce that is also a bounded subset.

Theorem 1: If be a metric space and is a compact subset of then is bounded.

  • Proof: For a fixed and for , consider the ball centered at with radius , i.e., . Let denote the collection of balls centered at with varying radii :
  • It should not be hard to see that is an open covering of , since for all we have that , so .
  • Now since is compact and since is an open covering of , there exists a finite open subcovering subset that covers . Since is finite, we have that:
  • And by definition covers so:
  • Each of the open balls in the open subcovering is centered at with . Since the set is a finite set, there exists a maximum value. Let:
  • Then for all we have that and therefore:
  • Hence is bounded.

Basic Theorems Regarding Compact Sets in a Metric Space

Recall that if is a metric space then a set is said to be compact in if for every open covering of there exists a finite subcovering of .

We will now look at some theorems regarding compact sets in a metric space.

Theorem 1: Let be a metric space and let . Then if is closed and is compact in then is compact in .

  • Proof: Let be closed and let be compact in . Notice that:
  • Furthermore, is closed. This is because is given as closed, and since is compact we know that is closed (and bounded). So the finite intersection is closed. But any closed subset of a compact set is also compact, so is compact in .

Theorem 2: Let be a metric space and let be a finite collection of compact sets in . Then is also compact in .

  • Proof: Let be a finite collection of compact sets in . Consider the union and let be any open covering of , that is:
  • Now since for all we see that is also an open covering of and so there exists a finite subcollection that also covers , i.e.:
  • Let . Then is finite since it is equal to a finite union of finite sets. Furthermore:
  • So is a finite open subcovering of . So for all open coverings of there exists a finite open subcovering of , so is compact in .

Theorem 3: Let be a metric space and let be an arbitrary collection of compact sets in . Then is also compact in .

  • Proof: Let be an arbitrary collection of compact sets in . Notice that for all that:
  • Furthermore, since each is compact, then each is closed (and bounded). An arbitrary intersection of closed sets is closed, and so is a closed subset of the compact set . Therefore by the theorem referenced earlier, is compact in .

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