## Thoughts about compactness.

**Point of post: **In this post we discuss the notion of compactness for topological spaces.

Anyone who has spent more than twenty minutes doing analysis has surely notice that all subsets of are not created equally. For example, a lot more can be said about a function than a function . We know for example that is bounded and uniformly continuous. In fact, we know that assumes a maximum value on in the sense that for some . None of these things can be necessarily said about . In fact, has none of these properties. It is not surprising that analysts wanted to find a quality that differentiates these “special” sets, and this quality is *compactness*.

Before we define compactness for subsets of we first need some terminology.

**Open cover-** A class of sets is said to be an open cover of a set if each is open

**Subcover-** If is an open cover of a set then is called a subcover of if and is an open cover of

With this in mind we say a set is compact if every open cover of admits a finite subcover.

**Example:** The set is compact.

**Proof:** Let be an open cover of . Clearly by definition there exists some open set such that . But, by the formulation of open sets in this means there exists some such that . The Archimedean principle then furnishes us with some such that which of course implies that . So taking some open set such that we see that is a finite subcover of . Since was arbitrary the conclusion follows.

**Non-example: **The set is not compact.

**Proof:** Let . This is an open cover for since given any we have that, by the Archimedean principle, there exists some such that and so and so and since it follows that and so and so is an open cover for for . But given any finite subclass of say we see that but where . And thus we have exhibited an open cover for which admits no finite subcover, thus by definition is not compact.

Of course, there was nothing in the above definition that limits us to considering . The exact same formulation works for any metric space.

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We now wish to generalize this a tad further and talk about when a topological space is compact.

We call such a space compact if given any collection of open sets such that (an open cover)there exists a finite subclass such that (a finite subcover).

Some things are immediately obvious

**Theorem:** A topological space such that is finite is compact.

**Proof:** By necessity any open cover must be finite (since and thus serves as a finite subcover.

**Corollary 1: Any finite space is compact.**

**Corollary 2:** Any indiscrete space is compact.

**Theorem: **A discrete space is compact if and only if it is finite.

**Proof:**

: Let . This is clearly an open cover for with no proper subcover (i.e. the removal of any element results in a non-cover). It follows from ‘s compactness that itself is a finite subcover, which obviously implies that is compact.

: Follows from an earlier problem.

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We now give some alternative, and very useful, conditions that are necessary and sufficient for a topological space is compact.

**Theorem: **A topological space is compact if and only if every closed class of closed subsets of with empty intersection has a finite subclass with finite intersection.

**Proof**:

: We see that and so is an open cover for so so by assumption there exists some which is a finite subcover. Thus, and the conclusion follows.

: Let be an open cover of , it then follows (using the same logic as the previous problem) that and so there exists some such that and the conclusion follows.

We now introduce the term *finite intersection property* (F.I.P) to mean a class of sets such that every finite intersection is non-empty.

**Theorem:** A topological space is compact if and only if every class of closed subsets with the finite intersection property has non-empty interseciton

**Proof:**

: Let be class of closed subsets of with the F.I.P. and suppose that then and so is an open cover for but given any finite subcover we have that and since we see that and so is an open cover of with no finite subcover, which of course contradicts ‘s compactness.

: Suppose that was an open cover for which admitted no finite subcover. Then clearly, following the logic of the last problem (DeMorgan’s law), that but and so is a class of closed subsets of with the F.I.P but that has empty intersection. This is a contradiction.

From this we can derive kind of a nice little theorem:

**Theorem:** Call a sequence of sets decreasing if . Then, if is a decreasing sequence of non-empty closed sets in a compact topological space , then .

**Proof:** Clearly given any finite subclass of we have (since the subclass forms a finite chain) that where and since it follows that . Thus, is a class of closed sets with the F.I.P. and by the assumption that is compact we see that .

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