# Topology/Compactness

The notion of **Compactness** appears in a wide variety of contexts. In particular, compactness is a "tameness property" that tells you that the objects you are dealing with are in some sense well-behaved.

## Definition[edit]

Let be a topological space and let

A collection of open sets is said to be an **Open Cover** of if

is said to be **Compact** if and only if every open cover of has a finite subcover. More formally, is compact iff for every open cover of , there exists a finite subset of that is also an open cover of .

If the set itself is compact, we say that is a *Compact Topological Space*.

Compactness of topological spaces can also be expressed by one of the following equivalent characterisations:

- Every filter on containing a filter basis of closed sets has a nonempty intersection.
- Every ultrafilter on converges.

## Important Properties[edit]

- Every closed subset of a compact set is compact

**Proof**:

Let be a compact set, and let be a closed subset of . Consider any open cover of . Observe that being open, the collection of open sets is an open cover of . As is compact, this open cover has a finite subcover .

Now, consider the collection . This collection is obviously finite and is also a subcover of . Hence, it is a finite subcover of

- Every compact subset of a Hausdorff space is closed.

**Proof**:

Let be compact. If the complement is empty, then is the same as the space; thus closed. Suppose not; that is, there is a point . Then for each , by the Hausdorff separation axiom we can find and disjoint, open and such that and . Since is compact and the collection covers , we can find a finite number of points in such that:

It then follows that:

. Hence, every has an open neighbourhood .

As can be represented as the union of open sets , is open and is closed.

- Every compact set in a metric space is bounded.

**Proof**:

Let be a metric space and let be compact.

Consider the collection of open balls for some (fixed) . We see that is an open cover of . As is compact, it has a finite subcover, say . Let . We see that , and hence, is bounded.

*Heine-Borel Theorem*: For any interval , and for any open cover of that interval, there exists a finite subcover of .

**Proof**:

Let be the set of all such that has a finite subcover of . is non-empty because is within the set. Define .

Assume if possible, . Then there is a finite cover of sets within for . is within a set within the cover . Thus, there exists a such that . Then is also within , contradicting the definition of . Thus, . Therefore, has a finite subcover.

Sources differ as to what exactly should be called the 'Heine-Borel Theorem'. It seems that Emile Borel proved the most relevant result, dealing with compact subsets of a Euclidean Space. However, we provide the simpler case, for reals.

- Let be topological spaces. If is continuous, and is compact, then the image of , , is compact.

**Proof:**

Let be any open cover of . Consider the inverses where . These inverses are open because is continuous. This covers , and thus there is a finite subcover of , . Then the images**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \{ f(B_i)\}}**is a finite subcover of**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle f(A)}**.

- If a set is compact and Hausdorff, then it is normal.

**Proof:**

Let be compact and Hausdorff. Consider two closed subsets and**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): B**which are themselves compact by theorem 1 above. For every**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): a\in A**and , there exist two disjoint sets and such that and . The union of all such for a fixed is a cover for , and thus it has a finite subcover, say, and let be the union of its members.

Let , and let . Observe that being finite, is open. The union covers , and therefore it has a finite subcover . Let be the union of all members of this subcover.

Let denote the set of all elements such that . Take the intersection , which is open.

Then is an open superset of , is an open superset of , and they are disjoint. Thus, is normal.

- In a compact metric space X, a function from X to Y is uniformly continuous if and only if it is continuous.

**Proof:**

- If two topological spaces are compact, then their product space is also compact.

**Proof:**

Let X_{1}and X_{2}be two compact spaces. Let S be a cover of X_{1}×X_{2}. Let x be an element of X_{1}. Consider the sets A_{x,y}within S that contain (x,y) for each y in X_{2}. forms a cover for X_{2}, with a finite subcover {A_{x,yi}}. Let B_{x}be the intersection of within {A_{yi}}, which is open. Thus, {B_{x}} forms an open cover, which has a finite subcover, {B_{xi}}. The corresponding sets {A_{xi,yi}} is finite, and forms an open subcover of the set.

- All closed and bounded sets in the Euclidean Space are compact.

**Proof:**

Let S by any bounded closed set in . Then since S is bounded, it is contained in some "box" of the products of closed intervals of R. Since those closed intervals are compact, their product is also compact. Therefore, S is a closed set in a compact set, and is therefore also compact.

### Tychonoff's Theorem[edit]

The more general result on the compactness of product spaces is called Tychonoff's Theorem. Unlike the compactness of the product of two spaces, however, Tychonoff's Theorem requires Zorn's Lemma. (In fact, it is equivalent to Axiom of Choice.)

Theorem: Let , and let each be compact. Then the X is also compact.

Proof: The proof is in terms of nets. Recall the following facts:

*Lemma 1* - A net in converges to if and only if each coordinate converges to .

*Lemma 2* - A topological space is compact if and only if every net in has a convergent subnet.

*Lemma 3* - Every net has a universal subnet.

*Lemma 4* - A universal net in a compact space is convergent.

We now prove Tychonoff's theorem.

Let be a net in .

Using Lemma 3 we can find a universal subnet of .

It is easily seen that each coordinate net is a universal net in .

Using Lemma 4 we see that each coordinate net converges, because is compact.

Using Lemma 1 we see that the whole net converges in .

We conclude that every net in has a convergent subnet, so, by Lemma 2, must be compact.

## Relative Compactness[edit]

Relative compactness is another property of interest.

Definition: A subset S of a topological space X is relative compact when the closure Cl(x) is compact.

Note that relative compactness does not carry over to topological subspaces. For example, the open interval (0,1) is relatively compact in R with the usual topology, but is not relatively compact in itself.

## Local Compactness[edit]

The idea of local compactness is based on the idea of relative compactness.

If, in a topological space X, every element has a neighborhood that is relatively compact, then X is **locally compact.**

It can be shown that all compact sets are locally compact, but not conversely.

## Exercises[edit]

- It is not true in general for a metric space that a closed and bounded set is compact. Take the following metric on a set X:

1 if x is not equal to y d(x,y) = 0 if x=y

a) Show that this is a metric

b) Which subspaces of X are compact

c) Show that if Y is a subspace of X and Y is compact, then Y is closed and bounded

d) Show that for any metric space, compact sets are always closed and bounded

e) Show that with this particular metric, closed and bounded sets need not be compact