A second topological concept that is introduced in analysis is compactness. It is a concept that is associated with the Bolzano-Weierstrass Theorem which is as follows

**THM. (Bolzano-Weierstrass). ** *Let be any infinite bounded set of . Then there is at least one such that every open ball centered on will contain at least one point in *.

The idea of the proof of this statement is to show that the intersection .

Insofar as compactness is concerned, there are a few different ways to introduce the concept. I will present the various definitions and show that they are all equivalent.

**Method 1: Open Covers and Finite Subcovers.**

In order to define compactness in this way, we need to define a few things; the first of which is an open cover.

**Definition. [Open Cover.]** Let be a metric space with the defined metric . Let . Then an open cover for is a collection of open sets such that

.

**N.B. **The collection of open subsets may be of infinite cardinality.

Another definition that we will need is the following:

**Definition. [Limit Point/Cluster Point.]** Let be a metric space and let , and let . Then is a limit point or a cluster point of if any open ball of center contains an infinite number of points from .

We need one more definition before we define compactness:

**Definition. [Finite Subcover.]** Given an open cover , a finite subcover is a finite subcollection of open sets from such that

.

Therefore, we can now definite compactness as follows:

**Definition. [Compact Set.]** Let be a metric space with the defined metric , and let . Then we say that is compact if every open cover for has a finite subcover.

To make this more concrete, consider the following example: **Example: ** Let and let Then the open interval is not a compact set. To see why consider the set of open subsets for . Note that . However,

. In other words, (or rather in words) what this says is that if we consider all of the open sets of the form (e.g. (1,1), (1/2,1), (1/3,1)…). We see that for each n =1,2,3,…, the open set increases in size*. Thus, if we consider all the elements that are in at least one of these increasing intervals, then the union contains the interval . However, note that if we take only a finite number , for simplicity say , then we have that the union does not contain all of the points that are contained in . Therefore, what this says is that while we can form an open cover for we cannot find a finite subcover for that set. Therefore, is not a compact set.

*: *There is a concept related to the size of an interval which lends itself to a field of study in analysis called measure theory (may post on this topic at a later time). *

**Method 2: Sequences and Subsequences: **

This approach has the benefit that we can just state the definition outright:

**Definition. [Compact Set.] ** Let is compact if every sequence in has a subsequence that converges to a limit that is also in .

There is one other type of “definition” used to understand compactness. Some books call this the Characterization of Compactness on the Real Line. **Theorem. [Compact Set.]** Let . Then is compact if and only if is closed and bounded.

The following theorem states that each of these different ways that are used to define compactness are in fact equivalent:

**Theorem. **Let . Then each of the following statements are equivalent:

(1.) is compact;

(2.) is closed and bounded;

(3.) Every open cover of has a finite subcover.

The implication of (2.) implies (1.) is what is referred to as the Heine-Borel Theorem. Furthermore, to circle back to the Bolzano-Weierstrass Theorem we can rewrite this statement in terms of compactness:

**Theorem. [Bolzano-Weierstrass Theorem.]** Let be a compact metric space, and let be an infinite subset of . Then has at least one cluster point.

The next post will discuss the proofs of the theorems in this post. Further posts will most likely be on astrophysics and/or cosmology. Until then, clear skies!