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!

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