# 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!

# CONVERGENT SEQUENCES, CAUCHY SEQUENCES, COMPLETENESS

If one takes quantum mechanics, when they first encounter the wavefunction which is a complex-valued function, they learn that the arena in which quantum mechanics is a Hilbert space. If one goes further in order to understand what a Hilbert space is they find that it is a complete inner product space. While many physicists take advantage of this fact, they do not really interest themselves with what this means in a rigorous mathematical sense. When I first encountered this, I was unsatisfied with the so-called “definition” of a Hilbert space. So I found that I had to learn more advanced mathematics; more specifically, real analysis. To that end, the purpose of this post is to understand what the term “complete” means. To remedy any confusion of what an inner product space is, an inner product space is a vector space $V$ that equipped with an inner product $\langle u, v \rangle$.

In order to understand what completeness is, we require a couple of definitions:

Definition. Let $\{p_{n}\}_{n=1}^{\infty}$ where $n\in \mathbb{N}$ be a sequence of points in the metric space $(E,d)$. A point $p\in \mathbb{E}$ is called a limit of the sequence of points if for any $\epsilon>0$, there exists $N\in \mathbb{N}$ such that if $n>N$,$d(p,p_{n})< \epsilon$. If such a limit exists, then we say that the sequence of points $\{p_{n}\}_{n=1}^{\infty}$ converges to the point $p\in E$.

What this says intuitively, is that in the sequence of points above there exists a term for which $n=N$ which corresponds to the point $p_{N}$ in the metric space $E$, beyond which any later terms in the sequence will be contained in what we call an open ball which is defined to be the set given by $B_{p}(\epsilon)= \{q\in E|d(q,p)<\epsilon\}$. We can regard the term $p_{N}$ as a “boundary point”.

Definition. A sequence of points $\{p_{n}\}_{n=1}^{\infty}$ in a metric space $(E,d)$ is said to be a Cauchy sequence, if for any $\epsilon>0$, there exists $N\in \mathbb{N}$ such that whenever $n,m>N, d(p_{n},p_{m})< \epsilon$.

The intuitive idea behind this concept is that suppose we take two terms in the sequence of points $\{p_{n}\}_{n=1}^{\infty}$, we say that it is a Cauchy sequence if whenever these two chosen terms are “beyond the boundary” the distance between these two terms are within $\epsilon$ of each other in the metric space $(E,d)$.

One important result that I am not going to prove is the following:

Theorem. If $\{p_{n}\}_{n=1}^{\infty}$ is a convergent sequence of points in the metric space $(E,d)$, then such a sequence is Cauchy.

An important note: the converse of this theorem is not necessarily true. If the converse is indeed true, we get the following definition:

Definition. A metric space $(E,d)$ is said to be complete if every Cauchy sequence of points in the metric space $(E,d)$ converges to a point $p\in E$.

An example of this is that $\mathbb{R}$ with the metric $d(p,q)=|p-q|$ is a complete metric space. Intuitively, what this means is that given a Cauchy sequence that converges in $\mathbb{R}$ to a real number. In other words, any possible Cauchy sequence will converge to some real number $p$.

The next post will discuss compactness in the context of metric spaces, covers, and open covers.

Clear Skies!

# Introduction to Metric Spaces

Metric Spaces are one of those mathematical topics that everyone intuitively understands. The best example of this is that of three dimensional Euclidean space $E^{3}$. This serves as the basis for the intuitive concept of a “space”, and our ability to ascribe a distance between to points in three-dimensional space can be described by a distance function $d: E\times E \rightarrow E$, or a metric. The underlying set $E$ together with the metric $d$ form what is called a metric space $(E,d)$.

To be more mathematically precise, we make the following definition.

Definition. A metric space is a set $E$ together with a rule which associates with pair $p,q\in E$ a real number $d(p,q)$ such that
$\displaystyle d(p,q)\geq 0, \forall p,q\in E$
$\displaystyle d(p,q) = 0 \iff p=q$
$\displaystyle d(p,q)=d(q,p) \forall p,q\in E$
$\displaystyle d(p,r) \leq d(p,q)+ d(q,r)$

As an example, suppose that the underlying set $E = \mathbb{R}$ and the metric coupled with this set is defined by $d(p,q)= |p-q|$. To verify that this indeed a metric space we must show that the four axioms are satisfied.

Claim. The mathematical structure $(\mathbb{R},d)$ in which $d:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ is defined by $d(p,q)=|p-q|$ is a metric space.
Proof. Let $p,q\in \mathbb{R}$. Then by definition of $d(p,q)$ and by definition of the absolute value function, we have that $|p-q|\geq 0$, so that axiom 1 is satisfied. Suppose now that the points in $\mathbb{R}$ are equal, i.e. that $p=q\in \mathbb{R}$. Then by definition of $d(p,q)$, we have that $|p-q|=|0|=0.$ Conversely, suppose that $|p-q|= 0.$ By the triangle inequality we have that $|p-q|\geq |p|-|q|=0$ This implies that $|p|= |q|$. Thus, condition (2.) is satisfied and hence the distance between two points in $\mathbb{R}$ is zero if and only if the two points are the same. To prove condition (3.), let $p,q\in \mathbb{R}$, so that $d(p,q) = |p-q|$. By virtue of the definition of the absolute value, we can say that
$\displaystyle d(p,q) = |p-q| = |-(p+q)|= |-1||q-p|= |q-p|= d(q,p).$
Thus, we see that the arguments of the proposed distance function is symmetric with respect to its arguments, namely any real numbers $p,q\in \mathbb{R}$. To prove condition (4.), consider any three points $p,q,r\in \mathbb{R}$, then the distance function between the points $p,r \in \mathbb{R}$ becomes
$\displaystyle d(p,r) = |p-r| = |p-q+q-r|,$
wherein we add zero in the form of adding and subtracting the point q (a very common trick in analysis). Then by the properties of the absolute value it follows that
$\displaystyle d(p,r) = |p-q+q+r| \leq |p-q|+|q-r| = d(p,q)+d(q,r)$,
where by the last equality follows from the definition of the distance function. Therefore, all four conditions have been satisfied, and hence by our definition of a metric space, it follows that $(\mathbb{R},d)$ whose distance function $d:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ is defined by $d(p,q) = |p-q|$ is indeed a metric space. $\square$