# Open covers, Finite Subcovers, and COMPACTNESS

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 $A$ be any infinite bounded set of $\mathbb{R}$. Then there is at least one $x\in \mathbb{R}$ such that every open ball centered on $x$ will contain at least one point in $A$.

The idea of the proof of this statement is to show that the intersection $B_{x}(\epsilon)\cap A \neq \emptyset$.

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 $(X,d)$ be a metric space with the defined metric $d$. Let $A\subset (X,d)$. Then an open cover for $A$ is a collection of open sets $\{O_{\alpha}|\alpha \in \mathbb{N}\}$ such that
$\displaystyle A \subset \bigcup_{\alpha\in \mathbb{N}}O_{\alpha}$.

N.B. The collection of open subsets $O_{\alpha}$ may be of infinite cardinality.

Another definition that we will need is the following:

Definition. [Limit Point/Cluster Point.] Let $(X,d)$ be a metric space and let $B\subset (X,d)$, and let $x_{0}\in X$. Then $x_{0}$ is a limit point or a cluster point of $A$ if any open ball of center $x_{0}$ contains an infinite number of points from $A$.

We need one more definition before we define compactness:

Definition. [Finite Subcover.] Given an open cover $\{O_{\alpha}\}$, a finite subcover is a finite subcollection of open sets from $\{O_{\alpha}\}$ such that
$\displaystyle \bigcup_{\alpha = 1}^{n}O_{\alpha}$.

Therefore, we can now definite compactness as follows:

Definition. [Compact Set.] Let $(X,d)$ be a metric space with the defined metric $d$, and let $A\subset (X,d)$. Then we say that $A$ is compact if every open cover for $A$ has a finite subcover.

To make this more concrete, consider the following example:
Example: Let $X= \mathbb{R}$ and let $d:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R} \triangleq d(p,q)=|p-q|.$ Then the open interval $(0,1)$ is not a compact set. To see why consider the set of open subsets $(1/n,1)$ for $n\in \mathbb{N}$. Note that $(0,1)\subset \bigcup_{n \in \mathbb{N}}(1/n,1)$. However,
$\displaystyle (0,1) \not\subset \bigcup_{n=1}^{m}(1/n,1)$. In other words, (or rather in words) what this says is that if we consider all of the open sets of the form $(1/n,1)$ (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 $(1,1)\cup (1/2,1) \cup (1/3,1) \cup ... \cup (1/n,1)$ contains the interval $(0.1)$. However, note that if we take only a finite number $m$, for simplicity say $m=3$, then we have that the union $(1,1) \cup (1/2,1) \cup (1/3,1)$ does not contain all of the points that are contained in $(0,1)$. Therefore, what this says is that while we can form an open cover for $(0,1)$ we cannot find a finite subcover for that set. Therefore, $(0,1)$ 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 $A\subset \mathbb{R}$ is compact if every sequence in $A$ has a subsequence that converges to a limit that is also in $A$.

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 $A\subset \mathbb{R}$. Then $A$ is compact if and only if $A$ 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 $A\subset \mathbb{R}$. Then each of the following statements are equivalent:
(1.) $A$ is compact;
(2.) $A$ is closed and bounded;
(3.) Every open cover $\{O_{\alpha}\}$ of $A$ 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 $(X,d)$ be a compact metric space, and let $A$ be an infinite subset of $(X,d)$. Then $A$ 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!