Fundamentals of Topology & Metric Spaces

In this brief post, we investigate the topological foundations of analysis and give some of its first applications. We limit the scope to the topology of a metric space. In particular, we provide the knowledge that is needed for the application to probability and measure theory.

The objective of the next section is to prepare the very basic concept for the definition of a topological space. We start by defining an open and a closed ball, a neighborhood, interior point, an open and a closed set. All these definitions are interlinked to each other as we will see later on.

Parts of this post are based on [1] and [2].

Contents

1 Basic Definitions & Properties
2 Topological Spaces
3 Applications

Basic Definitions & Properties

Balls, Open & Closed Sets & Neighborhoods

Let $X:=(X, d)$ denote a metric space.

Definition 1.1
An (open) ball $B$ of radius $r$ and center $x_0$ is the set of all points of distance less than $r$ from $x_0$ , i.e. $B(x_0, r):=\{x\in \mathbb{R} : d(x, x_0)<r \}$ .

A closed ball $\overline{B}$ of radius $r$ is the set of all points of distance less than or equal to $r$ from $x_0$ , i.e. $\overline{B}(x_0, r):=\{x\in \mathbb{R} : d(x, x_0)\leq r \}$ .

A subset $U$ of $X$ is called a neighborhood of $x_0\in X$ if there is some $r>0$ such that $B(x_0, r)\subseteq U$ .

An element $x_0\in A\subseteq X$ is called an interior point of $A$ if there is a neighborhood $U$ of $x_0$ such that $U\subseteq A$ .

A set $A$ of $X$ is called open if every point of $A$ is an interior point.

$\square$

The introduced concepts “interior point” and “open set” (and hence also the concept of “closed sets”) depend on the surrounding metric space $(X, d)$ . It is sometimes useful to make this explicit by saying “ $x_0$ is an interior point of $A$ with respect to $(X, d)$ ” or “ $A$ is open in $(X, d)$ “.

By definition, every superset of a ball is considered to be a neighborhood. Neighborhoods can be defined without mentioning explicitly the corresponding metric space.

Example 1.1
An open ball in $\mathbb{R}$ is an open interval $I$ . However, if we consider $\mathbb{R}$ embedded in $\mathbb{R}^2$ , then $I$ is apparently not an open ball in $\mathbb{R}^2$ .

For the sake of illustration we sketch a 2-dimensional open ball (i.e. an open interval) as a rectangle even though it would actually be a line without end-points. These end-points are illustrated by the lines crossing the x-axis.

The heuristic reasoning for that is that a one-dimensional interval cannot contain an open two-dimensional ball not matter how small the radius of this ball might be.

$\square$

Apparently, $x_0$ is an interior point of $A$ if and only if there is some real $\epsilon >0$ such that $B(x_0, \epsilon) \subseteq A$ . In addition, $A$ is open if and only if $A$ is a neighborhood of each of its points.

Example 1.2
Let us consider open and closed balls in $X:=\mathbb{R}^n$ along with the standard Euclidean metric $d$ . Every open ball $B(x_0, r)$ , $r>0$ in $(X, d)$ is an open set. Let $b \in B$ be an arbitrary point. Set $s := d(x_0, b)$ and consider $B(b, \epsilon)$ with $\epsilon:= r - s>0$ .

For all $x\in B(b, \epsilon)$ we have

$\begin{align*} d(x,b) \leq d(x, x_0) + d(x_0, b) < \epsilon +s = r, \end{align*}$

and so $B(b, \epsilon)$ is contained in $B(x_0, r)$ . This shows that the arbitrary point $b$ is an interior point of $B(x_0,r)$ and thus an open ball $B(x_0, r)$ is an open set.

$\square$

That is, a set $A$ is open if and only if for every $x_0\in A$ an open ball $B(x_0, \epsilon)$ exists such that $B(x_0, \epsilon) \subseteq A$ .

Definition 1.2 (Boundary Point)
Let $A\subseteq X$ and $x\in X$ . Then, $x$ is called boundary point for $A$ if

(1) $\begin{align*} B(x, \epsilon) &\cap A \neq \emptyset \quad \text{ and } \\ B(x, \epsilon) &\cap A^C \neq \emptyset \\ \end{align*}$

for all $\epsilon>0$ . The set of all boundary points is denoted by $\partial A$ .

$\square$

Boundary points are just points on the boundary between the set $A$ and the surrounding basis set $X$ (i.e. $A^C)$ of the metric space $(X, d)$ .

$\partial A$ can be contained in $A$ or in $A^C$ . If all boundary points $\partial A$ are outside of $A$ , i.e. if $A \cap \partial A = \emptyset$ , then it is an open set.

Definition 1.3 (Closed Set)
A set $A$ of $X$ is called closed in $X$ if its complement $A^C$ is open in $X$ .

$\square$

Equivalently, if all boundary points of $\partial A$ are inside the set $A$ , i.e. if $A \cup \partial A = A$ , then it is a closed set. This also means that the set $\overline{A}:= A \cup \partial A$ is always a closed set no matter what $A$ is. The set $\overline{A}$ is then also called the closure of $A$ .

Example 1.3
(a) The interval $A:= [a, \infty)$ is closed in $X:=(\mathbb{R}, d(x,y)=|x-y|)$ .
Let $y$ denote a real number that is not an element of $A$ . Then, $y<a$ and the ball $B(y, \epsilon)$ with $\epsilon:=(a-y)/2$ has empty intersection with $A$ .

Hence, the complement $A^C$ is open and so the set $A$ is closed.

For the sake of illustration we sketch a 2-dimensional open ball as a rectangle even though it would actually be a line without end-points.

Note that all closed intervals in $X=(\mathbb{R}, |.|)$ are of the form $[a,b]$ , $[a, \infty)$ , $(-\infty, b]$ , $\emptyset$ and $\mathbb{R}$ .

Let us now consider the set $A':= (a, \infty)$ . For all $x\in A'$ and $\epsilon':=(x-a)/2$ we have $B(x, \epsilon') \subseteq A$ . Hence, the set $A'$ is open in $X$ .

Note that all open intervals in $X=(\mathbb{R}, |.|)$ are of the form $(a,b)$ , $(a, \infty)$ , $(-\infty, b)$ , $\emptyset$ and $\mathbb{R}$ .

(b) Now consider the metric space $(X, d)$ with $X:= (0,1] \cup (5,\infty)$ with the same distance function $d(x,y)=|x-y|$ , $x,y\in X$ as in (a).

The set $A:=(0,1]$ is an open set since for all $x\in A$ we consider the open ball $B(x, \epsilon)$ with $\epsilon:=0.5 \cdot \min(d(x, 0), d(x, 1))$ . That is, we determine the shortest distance between $x$ and the two points $0$ and $1$ and divide it by 2. Then, the corresponding open ball $B(x, \epsilon)$ is contained within the set $A$ . For the point $x=1$ , we can consider the open ball $B(x, \epsilon)$ with $\epsilon=1$ , for instance. This open ball equals $B(x=1, \epsilon=1)=(0,1]$ since all the figures right from $x=1$ until 5 are not contained in $X$ . Hence, $A$ is open.

The set $A$ is also closed since $\partial A=\emptyset$ is included in $A$ . Note that $0$ is not a boundary point since it is not an element of the metric space $X$ . The point $x=1$ is not a boundary point since for $\epsilon=0.5$ , for instance, $B(x=1, \epsilon=0.5) \cap A^C= (0.5, 1] \cap (5, \infty) = \emptyset$ holds true.

(c) Let us consider the metric space $X:=\mathbb{Z}$ of all integers together with the discrete metric $d(x,y)=\begin{cases} 0 & x=y \\ 1 & x\neq y \end{cases}$ for all $x,y\in X$ . It follows that the open ball $B(x_0, r)=$ $\{x\in X| \ d(x_0, x)<r\}$ of radius $r>0, r\in \mathbb{R}$ and center at $x_0$ is either $\{x_0\}$ if $1\geq r >0$ or $X$ if $r>1$ .

(d) Let us consider open and closed balls in $X:=\mathbb{R}^n$ along with the standard Euclidean metric $d$ . Every closed ball $\overline{B(x_0, r)}$ , $r\geq 0$ in $(X, d)$ is a closed set. That is, we need to show that the complement set $B(x_0, r)^C =\mathbb{R}^n \setminus B(x_0, r)$ is open. Let $y_0\in B(x_0, r)^C$ . Set $\epsilon:= \frac{d(x_0, y_0) - r}{2}$ and consider the open ball $B(y_0, \epsilon)$ , that does not intersect with the closed ball $\overline{B(x_0, r)}$ due to the choice of $\epsilon$ : Let $x\in B(y_0, \epsilon)$ , then the following holds due to the triangular property.

$\begin{align*} d(x_0, y_0) \leq d(x,x_0)+d(x, y_0) \leq d(x, x_0)+\epsilon \end{align*}$

and $d(x, x_0) \geq \epsilon +r > r \geq 0$ because of $d(x, y_0)=2\epsilon+r$ .

Note that this also means that every isolated point in $\mathbb{R}^n$ along with the Euclidean metric is a closed set.

$\square$

In the following section we are going to outline basic properties of open & closed sets, which will provide the heuristic for the definition of a topological space.

Properties of Open & Closed Sets

The following two propositions state basic properties of open and closed sets. It turns out that these properties characterize a topological space.

Proposition 1.1 (Open Sets)
Let $\mathcal{T}:= \{ O \subseteq X: \ O \text{ is open}\}$ be a family of open sets.
(i) $\emptyset, X \in \mathcal{T}$ ;
(ii) $\bigcup_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for an arbitrary index set $i\in I$ ;
(iii) $\bigcap_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for a finite index set $I$ .

Proof:
(i) The empty set possesses every property since it does not have an element. Hence, the empty set is closed and open. Correspondingly, the set $X$ is open.

(ii) If $x\in \bigcup_{i\in I}{O_i}$ then there must be an index $i$ such that $x\in O_i$ with $B(x, \epsilon_i) \subseteq O_i$ . All the more $B(x, \epsilon_i) \subseteq \bigcup_{i\in I}{O_i}$ holds true.

(iii) Let $O_1, \ldots, O_n$ with $n\in \mathbb{N}$ be open sets. Since $x \in (O_1 \cap \ldots \cap O_n) \neq \emptyset$ there exists a $B(x, \epsilon_i)$ for each $i\in I:=\{1, \ldots, n\}$ with $B(x, \epsilon_i) \subseteq O_i$ . If we set $\epsilon:= \min\{\epsilon_1, \ldots, \epsilon_n\}$ then $B(x, \epsilon)\subseteq \bigcap_{i\in I}{O_i}$ and the assertion follows.

$\square$

Why (iii) holds only true for finite index sets can also be seen if we try to prove that the intersection of two open balls at $a\in X$ is open again. Just chose $\epsilon:= \min\{\epsilon_1, \epsilon_2\}$ for the open ball of the intersection $B(a, \epsilon_1)\cap B(a, \epsilon_2)$ . This can be iterated provided that the index set is finite.

Please note that the proof above would also be valid if we use neighborhoods instead of balls. This would mean that the proof is not dependent on the used metric space and suggests that we could derive a new mathematical structure from it, where –in general– no algebraic operations, distance functions or orders need to be defined. However, given that we restrict our-self to topology for metric spaces, we will not further outline this idea.

The proof of the corresponding proposition for closed sets is apparent given the definition of a closed set.

Proposition 1.2 (Closed Sets)
Let $\mathcal{T}:= \{ C \subseteq X: \ C \text{ is closed}\}$ be a family of closed sets.
(i) $\emptyset, X \in \mathcal{T}$ ;
(ii) $\bigcap_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for an arbitrary index set $i\in I$ ;
(iii) $\bigcup_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for a finite index set $I$ .

Proof:
Apply the definition and Proposition 1.1 (Open Sets).

$\square$

Example 1.4
Let us continue Example 1.3. The intersection of a finite number of intervals $[a_i, \infty)$ is closed in $X:=(\mathbb{R}, |.|)$ . Without loss of generality, let $a_1 \leq a_2 \leq \ldots \leq a_n$ . All the sets $[a_i, \infty)$ are closed and so is $\bigcap_{i=1}^{n}{[a_i, \infty)} = [\max\{a_1, \ldots, a_n\}, \infty)$ in $X$ .

Let us now consider the family of open sets $\{ (a_i, \infty) | \ i\in \mathbb{N}\}$ . For all elements $x_i$ of all open sets $(a_i, \infty)$ , we can find an open ball $B(x_i, \epsilon_i) \subseteq (a_i, \infty)$ . Hence, the union of these sets also contain open balls for all elements and is thus an open set itself.

$\square$

Example 1.5
An infinite intersection of open sets need not to be open. Consider $\bigcap_{i=1}^{\infty}{B(0, \frac{1}{n})} = \{0\}$ in $(\mathbb{R}, |.|)$ . Apparently, there cannot be an $\epsilon>0$ such that $B(0, \epsilon)\subseteq \{0\}$ .

Correspondingly, infinite unions of closed sets need not to be closed. To this end, consider $\bigcup_{i=1}^{\infty}{B(0, \frac{1}{n})^C} = \mathbb{R} \setminus \{0\}$ in $(\mathbb{R}, |.|)$ .

$\square$

Accumulation and Limit Points

Closely related to closed sets are so-called accumulation and limit points. In addition, this section will pave the way to one of the below application of a topological space — limits.

Definition 1.2 (Accumulation & Limit Point)
Let $A \subseteq X$ and $x\in X$ . We call $x$ an accumulation point of $A$ if every neighborhood of $x$ in $X$ has non-empty intersection with $A$ .

The element $x\in X$ is called limit point of $A$ if every neighborhood of $x$ in $X$ contains a point of $A$ other than $x$ . Finally we set

$\begin{align*} \overline{A}:= \{ x\in X | x \text{ is an accumulation point of } A \}. \end{align*}$

$\square$

Every point within an open set but also every limit point of this set is an accumulation point as illustrated in the following example. Here, $A=(0,1)$ is the open interval in the standard metric space $(X=\mathbb{R}, |\cdot|)$ .

Every point in $A=(-1,1)$ is an interior, an accumulation and a limit point of $A$ . Note that an interior point of $A$ requires to have a neighborhood $U$ that is contained in $A$ . Given that $A$ is an open set this holds true for all points within the set $A$ by definition.

The ‘boundary’ points $x=-1$ and $x=1$ are no interior points but limit and accumulation points of $A$ . Why?
Because we can define any open ball around it and the intersection with $A$ will always contain a point different from $x=\pm 1$ .

Let us now consider a set $A=\{0\}$ with a single isolated point in the metric space $(\mathbb{R}, |\cdot|)$ . Is $0$ , that is the only element of the set, an accumulation and/or a limit point of $A$ ?

Apparently, every neighborhood $U$ of $x$ in $X$ has non-empty intersection with $A$ , i.e. $A \cap U = \{0\}$ . However, $0$ is not a limit point since the intersection $A \cap U = {0}$ does not contain a point other than $0$ .

According to our thoughts above, elements of $A$ are accumulation points of $A$ . This fact is also reflected in the next proposition.

Proposition 1.3: Let $A$ be a subset of a metric space $X=(X, d)$ .
(i) $A \subseteq \overline{A}$ ;
(ii) $A = \overline{A}$ if and only if $A$ is closed.

Proof: (i) See argumentation above.
(ii) ‘ $\Rightarrow'$ Let $x\in A^c=(\overline{A}^c)$ . Since $x$ is not an accumulation point of $A$ , there is some $U\in \mathcal{U}(x)$ of all neighborhoods of $x$ such that $U \cap A = \emptyset$ . Thus, $U \subseteq A^c$ , that is, $x$ is an interior point of $A^c$ . Consequently $A^c$ is open and $A$ is closed in $X$ .

‘ $\Leftarrow'$ Let $A$ be closed in $X$ . Then $A^c$ is open in $X$ according to its definition. For any $x\in A^c$ , there is some $U \in \mathcal{U}(x)$ such that $U\subseteq A^c$ . This means that $U$ and $A$ are disjoint, and so $x$ is not an accumulation point of $A$ , i.e., $x\in (\overline{A})^c$ . Hence, we have proved the inclusion $A^c \subseteq (\overline{A})^c$ , which is equivalent to $(\overline{A})^c \subseteq A$ . With (i), this implies $\overline{A} = A$ .

$\square$

The limit points of a set $A$ are the limits of certain sequences in $A$ .

Proposition 1.4: An element $x$ of $X$ is a limit point of $A$ if and only if there is a sequence $(x_k)$ in $A \setminus \{x\}$ which converges to $x$ .

Proof: Let $x$ be a limit point of $A$ . For each $k\in \{1,2, \ldots \}$ , choose some element $x_k \neq x$ in $B(x, \frac{1}{k})$ . Then $(x_k)$ is a sequence in $A\setminus \{x\}$ such that $x_k \rightarrow x$ .

Conversely, let $(x_k)$ be a sequence in $A \setminus \{x\}$ such that $x_k \rightarrow x$ . Then, for each neighborhood $U$ of $x$ , there is some $k\in \mathbb{N}$ such that $x_k\in U$ . This means that $x_k\in U\cap (A\setminus \{x\})$ . Hence, each neighborhood of $x$ contains an element of $A$ other than $x$ .

$\square$

Corollary 1.1: An element $x\in X$ is an accumulation point of $A$ if and only if there is a sequence $(x_k)$ in $A$ such that $x_k \rightarrow x$ .

Proof: If $x$ is a limit point, then the claim follows from Proposition 1.4. Otherwise, if $x$ is an accumulation point, but not a limit point of $A$ , then there is a neighborhood $U$ of $x$ such that $U \cap A = \{x\}$ . Thus, $x\in A$ , and the constant sequence $(x_k)$ with $x_k=x$ for all $k\in \mathbb{N}$ has the desired property.

$\square$

We can also characterize closed sets using convergent sequences by simply using the fact that a closed set is the complement of an open set.

Proposition 1.5: For $A \subseteq X$ , the following are equivalent:
(i) $A$ is closed;
(ii) $A$ contains all its limit points;
(iii) Every sequence in $A$ , which converges in $X$ , has its limit in $A$ .

Proof: ‘(i) $\Rightarrow$ (ii)’ Any limit point of $A$ is also an accumulation point and so is contained in $\overline{A}$ . Since $A$ is closed and Proposition 1.3, $A=\overline{A}$ , and so all limit points are in $A$ .

‘(ii) $\Rightarrow$ (iii)’ Let $(x_k)$ be a sequence in $A$ such that $x_k \rightarrow x$ in $X$ . Then, by Corollary 1.1, $x$ is an accumulation point of $A$ . This means that, either $x$ is in $A$ , or $x$ is a limit point of $A$ , so, by assumption, $x\in A$ .

‘(iii) $\Rightarrow$ (i)’ This implication follows from Proposition 1.3 and Corollary 1.1.

$\square$

Compact Sets

This section is about a specific class of topological sets, that will turn out to have important properties, in particular, in connection with continuity.

Definition 1.3 (Open Cover)
By an open cover $\mathcal{O}$ of a set $D$ in a metric space $X$ , we mean a collection $\mathcal{O}=\{O_i | \ i\in I\}$ of open subsets of $X$ such that $D \subseteq \bigcup_{i\in I}{O_i}$ .

A sub-collection $\mathcal{O'}:=\{O_j| \ j\in J\}$ with $J \subseteq I$ of an open cover is called an open subcover of $\mathcal{O}$ .

$\square$

An open cover of $D$ is a collection of possibly overlapping open sets in $X$ which, after considering their union globally, contains the set $D$ inside.

Example 1.6 (Real Line)
Let us consider the set $D:=\mathbb{R}$ of the metric space $(\mathbb{R}, |\cdot|)$ . Then, $\mathcal{O}:=\{(-n, n) | n\in \mathbb{N}\}$ is an open cover of $D$ because $D \subseteq \bigcup_{n\in \mathbb{N}}{(-n, n)}$ . This cover apparently contains infinitely many open sets $(-n, n)$ .

$\mathcal{O'}=\{ (-2n, 2n) | \ n\in \mathbb{N}\}$ is an open subcover of $D$ since $D\subseteq \bigcup_{n\in \mathbb{N}}{(-2n, 2n)}$ but $\mathcal{O'} \subset \mathcal{O}$ . This subcover, however, still contains infinitely many open sets.

$\square$

Now we have the ingredients for the central definition of this section.

Definition 1.3 (Compact Set)
A subset $K$ of a metric space $X$ is said to be compact if every open cover $\mathcal{O}$ of $K$ contains a finite subcover.

$\square$

It is clear that every finite set is compact.

Example 1.7 (Finite Set & Compactness)
Let $D:=\{x_0, \ldots, x_n\}$ be a finite set in the metric space $(\mathbb{R}, |\cdot|)$ and $\mathcal{O}:=\{B(x_i, \epsilon=0.1) | \ i\in \{0, 1, \ldots, n\} \}$ . Note that we could chose any $\epsilon>0$ for this purpose. Apparently, $\mathcal{O}$ is a finite and open cover of $D \subseteq \bigcup_{i=0}^{n}{B(x_i, 0.1)}$ .

Each point of the finite set $D$ is contained in some open set by the cover depending upon that point. This directly implies that there is a finite sub-cover most the size of the set $D$ .

$\square$

Let us generalize the approach taken in Example 1.7 for an arbitrary non-empty set $D \subseteq \mathbb{R}$ . By setting

$\begin{align*} \delta:D \rightarrow (0,\infty) \quad \text{ defined by } x \mapsto \delta(x):= B(x, \delta) \end{align*}$

with $\delta>0$ , an open set $B(x, \delta)$ is assigned to each point $x\in D$ . Hence, the set $\mathcal{O}:=\{\delta(x) | \ x\in D\}$ is an open cover for any set $D$ .

Let us consider another simple example.

Example 1.8 (Compact Unit Interval)
Consider the set $D:=[0, 1] \subseteq \mathbb{R}$ . Apparently, this set is compact since $\mathcal{O}:=\{(-0.1,0.6), (0.5,1.1)\}$ is a finite open cover of $D$ .

$\square$

The Heine-Borel theorem states that every closed and bounded set in $\mathbb{R}^n$ , $n\in \mathbb{N}$ is compact. This theorem is very important for analysis, measure theory and beyond.

Theorem 1.1 (Heine-Borel)
Let $D$ be a subset of the standard Euclidean metric space $(\mathbb{R}^n, L_n)$ with $n\in \mathbb{N}$ . Then the following statements are eqivalent:
(i) $D$ is compact;
(ii) $D$ is closed and bounded;
(iii) Every infinite subset of $D$ has an accumulation point in $D$ .

Proof: Refer to Theorem 3.31 in [1] or see there. Also note that a visualization by the University of Hannover of one proof can be found on YouTube. There is also a proof analysis of one direction of the Heine-Borel Theorem.

$\square$

Topological Spaces

Topological spaces can be defined in many ways, however, given our focus on metric spaces and open sets the following definition is the most natural one. Please note that we could also use the properties of closed sets to define a topological space.

Definition 2.1 (Topological Space)
A topological space is a pair $(X, \mathcal{T})$ , where $X$ is a set and $\mathcal{T}$ is a family of subsets that satisfies

(i) $\emptyset, X \in \mathcal{T}$ ;
(ii) $\bigcup_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for an arbitrary index set $i\in I$ ;
(iii) $\bigcap_{i\in I}{O_i}\in \mathcal{T}$ if $O_i \in \mathcal{T}$ for a finite index set $I$ .

$\square$

The following video provides a rather unorthodox way of thinking about a topology. However, it might help to get a heuristic for topological spaces. It also mentions the connection between metrics and a topology.

Topology vs. “a” Topology by PBS Infinite Series

Example 2.1 (Topologies):

a) If $X$ is a metric space and $\mathcal{T}_X$ is the set of all open sets, then $\mathcal{T}_X$ is a topology according to Proposition 1.1.

b) Let $X$ be a set and $\mathcal{T}_0=\{\emptyset, X\}$ then $\mathcal{T}_0$ is the so-called trivial or indiscrete topology.

c) The power set $2^X$ of a set $X$ is the so-called discrete topology. In this topology every subset it open.

$\square$

Applications

General Topology has to do with, among other things, notions of convergence. Convergence ca be defined in many different ways. In the following section, we study the most popular way to define convergence by a metric. In order to define other types of convergence (e.g. point-wise convergence of functions) one needs to extend the following approach based on open sets. This, however, is not in scope of this post.

Limits of a Sequence

In this section, we apply our knowledge about metrics, open and closed sets to limits. We thereby restrict ourselves to the basics of limits. Refer to [1] for further details.

Let $X:=(X, d)$ denote a metric space. If the metric is not specified, we assume that the standard Euclidean metric is assigned.

Definition 3.1 (Sequence):
A sequence in $X$ is a function from $\mathbb{N}$ to $X$ by assigning a value $f(n)\in X$ to each natural number $n\in \mathbb{N}$ . The set $(x_n)_{n\in \mathbb{N}}$ $= (x_n) =:$ $\{x_n | \ n\in\mathbb{N} \}=$ $\{ x_1, x_2, \ldots, x_n, \ldots\}$ is called sequence of $X$ . The elements of $(x_n)_{n\in \mathbb{N}}$ are called terms of the sequence.

$\square$

Sequences in $X \subseteq \mathbb{R}^1$ are called real number sequences. Sequences in $X \subseteq \mathbb{R}^d$ , $d \in \mathbb{N}$ are called real tuple sequences. Note that latter definition is simply a generalization since number sequences are, of course, $d$ -tuple sequences with $d=1$ .

Let $(x_n)$ be a $d$ -tuple sequence in $X$ equipped with property $E$ . Property $E$ holds for almost all terms of $(x_n)$ if there is some $m\in \mathbb{N}$ such that $E$ is true for infinitely many of the terms $x_k$ with $k\geq m$ .

Note that a sequence can be considered as a function with domain $\mathbb{N}$ . We need to distinguish this from functions that map sequences to corresponding function values. Latter concept is very closely related to continuity at a point.

Sequences are, basically, countably many ( $1$ – or higher-dimensional) vectors arranged in an ordered set that may or may not exhibit certain patterns.

Example 3.1:
a) The sequence $\{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots \}$ can be written as $( \frac{1}{n} )_{n\in \mathbb{N}}$ and is nothing but a function $h:\mathbb{N} \rightarrow \mathbb{R}$ defined by $h(n)=\frac{1}{n}$ .

Plot of harmonic series — Plot of $h(n)$ for $n\in \{1, 2, \ldots, 49\}$

b) Let us now consider the sequence $\{ +1, -1, +1, -1, \ldots \}$ that can be denoted by $\{(-1)^n\}_{ n\in \mathbb{N} \cup \{ 0 \} }$ . The range of the function only comprises two real figures $\{+1,-1\}$ .

c) Now, let us consider the sequence $\{1, 2, 3, \ldots \} = \{n\}_{n\in \mathbb{N}}$ . Here, each natural $n$ is mapped on itself.

d) Consider $(\frac{n-1}{n})_{n\in \mathbb{N}} := \{0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots \}$ , that can also be written as $\{\frac{n-1}{n}\}_{n\in \mathbb{N}} = \{1-\frac{1}{n}\}_{n\in \mathbb{N}}$ .

Plot of the sequence $(\frac{n-1}{n})_{n\in \mathbb{N}}$

e) Consider the 2-tuple sequence $(x_n)_{n\in \mathbb{N}} := ((1+\frac{1}{n})^n, 1-\frac{1}{n})$ in $\mathbb{R}^2$ .

Plot of 2-tuple sequence for the first 1000 points that seems to head towards a specific point in $\mathbb{R}^2$ .

$\square$

We might think that some of the above examples contain patterns of vectors (e.g. number) that are “getting close” to some other vector (e.g. number). Other sequences may not give us that impression. We are interested in what the long-term behavior of the sequence is:

What happens for large values of $n$ ?
Does the sequence approach a (real) vector/number?

Convergence of a Sequence

Now, let us try to formalize our heuristic thoughts about a sequence approaching a number $x\in \mathbb{R}^n$ arbitrarily close by employing mathematical terms.

By writing

(2) $\begin{align*}\lim_{n \rightarrow \infty}{x_n} = x,\end{align*}$

we mean that for every real number $\epsilon>0$ there is an integer $N\in \mathbb{N}$ , such that

(3) $\begin{align*}d(x_n, x) < \epsilon\end{align*}$

whenever $n\geq N$ . A sequence $(x_n)$ that fulfills this requirement is called convergent. We can illustrate that on the real line using balls (i.e. open intervals) as follows.

Note that $d(x_n, x) < \epsilon$ represents an open ball $B(x, \epsilon)$ centered at the convergence point or limit x. For instance, for $\epsilon=0.5$ we have the following situation, that all points (i.e. an infinite number) smaller than $\frac{1}{2}$ lie within the open ball $B(0, \epsilon=0.5)$ . Those points are sketched smaller than the ones outside of the open ball $B(0, \epsilon=0.5)$ .

Accordingly, a real number sequence is convergent if the absolute amount is getting arbitrarily close to some (potentially unknown) number $x$ , i.e. if there is an integer $N\in \mathbb{N}$ such that $|x_n - x|<\epsilon$ whenever $n\geq N$ .

Convergence actually means that the corresponding sequence gets as close as it is desired without actually reaching its limit. Hence, it might be that the limit $\lim_{n \rightarrow \infty}{x_n}=x_0$ of the sequence $x_n$ is not defined at $x_0$ but it has to be defined in a neighborhood of $x_0$ .

This limit process conveys the intuitive idea that $x_n$ can be made arbitrarily close to $x\in \mathbb{R}$ provided that $n$ is sufficiently large. “Arbitrarily close to the limit $x$ ” can also be reflected by corresponding open balls $B(x, \epsilon)$ , where the radius $\epsilon>0$ needs to be adjusted accordingly.

The sequence can also be considered as a function $f:\mathbb{N} \rightarrow \mathbb{R}$ defined by $f(n)=x_n$ with

(4) $\begin{align*}\lim_{n \rightarrow \infty}{x_n} = x\end{align*}$

whenever $n\geq N$ .

If there is no such $x\in \mathbb{R}$ , the sequence $\{x_n\}$ is said to diverge. Please note that it also important in what space the process is considered. It might be that a sequence is heading to a number that is not in the range of the sequence (i.e. not part of the considered space). For instance, the sequence Example 3.1 a) converges in $\mathbb{R}$ to 0, however, fails to converge in the set of all positive real numbers (excluding zero).

The definition of convergence implies that $x_n \rightarrow x$ if and only if $d(x_n, x) \rightarrow 0$ . The convergence of the sequence $\{d(x_n, x) \}$ to 0 takes place in the standard Euclidean metric space $(\mathbb{R}^1, |\cdot|)$ .

While a sequence $\{x_n\}$ in a metric space $(V, d)$ does not need to converge, if $\{x_n\} \rightarrow x \in \mathbb{R}$ its limit is unique. Notice, that a ‘detour’ via another convergence point $y\in V$ (triangle property) would turn out to be the direct path with respect to the metric as $n \rightarrow \infty$ .

A convergent sequence $x_n \rightarrow x$ is also bounded. We can prove this intuitive statement by setting $\epsilon:=1$ . Hence, it exists a $N\in \mathbb{N}$ , such that $|x_n - x| < 1$ for all $n>N$ . This implies

(5) $\begin{align*}|x_n - x| &<1 \ \text{for all } n>N \\\Rightarrow |x_n| &< |x_n - x| +|x| < 1 +|x|\end{align*}$

for all $n > N$ . Let $m := \max(|x_k| , 1 +|x|)$ for $k\in\{1, \ldots, N\}$ then the assertion $|x_n|<m$ follows immediately.
$\square$

Let us re-consider Example 3.1, where the sequence a) apparently converges towards $zero$ . Sequence b) instead is alternating between $+1$ and $-1$ and, hence, does not converge. Note that example b) is a bounded sequence that is not convergent. Sequence c) does not have a limit in $\mathbb{R}$ as it is growing towards $\infty$ and is therefore not bounded. However, sequence d) converges towards 1. Finally, 2-tuple sequence e) converges to the vector $(e, 1)\in \mathbb{R}^2$ .

If we consider one of the converging examples carefully, we will notice that we can chose an arbitrarily small $\epsilon >0$ and we will find a correspondingly large $N\in \mathbb{N}$ , such that $|x_n - x| < \epsilon$ with $n>N$ .

For instance, let us define $\epsilon$ to be $0,0000001$ in Example 3.1 a). Then, we can set $N:=1,000,000$ and $\frac{1}{n} - 0| < 0,0000001$ provided that $n>1,000,000$ .

$\square$

A sequence $\{x_n\}$ is called increasing if $x_n \leq x_{n+1}$ for all $n\in \mathbb{N}$ . If an increasing sequence is bounded above, then $\{x_n\}$ converges to the supremum $\sup\{x_1, x_2, \ldots, x_n\}$ of its range.

Cauchy Sequences

If a sequence $\{x_n\}_{n\in \mathbb{N}}$ converges to a limit $x$ , its terms must ultimately become close to its limit $x$ and hence close to each other. That is, two arbitrary terms $x_l$ and $x_k$ of a convergent sequence $\{x_n\}$ become closer and closer to each other provided that the index of both are sufficiently large.

Theorem 3.1 (Convergent and Cauchy Sequences):

Assume that $\{x_n\}_{n\in \mathbb{N}}$ converges in a metric space $(X, d)$ to a limit $x$ . Then for every real $\epsilon^*>0$ there is an integer $N^*$ such that

$\begin{align*}d(x_l, x_k) < \epsilon^* \quad \text{whenever } l,k\geq N^*.\end{align*}$

Proof: $x$ is the limit of $\{x_n\}_{n\in \mathbb{N}}$ , i.e. $\lim_{n \rightarrow \infty} x_n = x$ . Assume $\epsilon^* > 0$ is given. Due to the fact that $\lim_{n \rightarrow \infty} x_n = x$ , we can choose an integer $N$ , such that $d(x_n, x)< \epsilon=\frac{\epsilon^*}{2}$ for all $n>N$ . If $l, k>N$ is valid, we can conclude

$\begin{align*}d(x_l, x_k) \leq d(x_l, x) + d(x_k, x) < \frac{\epsilon^*}{2} + \frac{\epsilon^*}{2} = \epsilon^*\end{align*}$

by employing the triangle inequality of the metric.

$\square$

The last proposition proved that two terms of a convergent sequence becomes arbitrarily close to each other. This property was used by Cauchy to construct the real number system by adding new points to a metric space until it is ‘completed‘. Sequences that fulfill this property are called Cauchy sequence.

Definition 3.2 (Cauchy Sequence):
A sequence $\{x_n\}_{n\in \mathbb{R}}$ is called Cauchy sequence, if the following condition holds true:

(6) $\begin{align*} \forall \epsilon \ \exists N^* \in \mathbb{N} \text{ such that } \forall l,k > N^*:\\ d(x_l, x_k) < \epsilon \end{align*}$

$\square$

Note that all pairs of terms with index greater than $N^*$ need to get close together. It is not sufficient to require that two consecutive terms get close together.

In the following example, we consider the function and sequences that are interpreted as attributes of this function. If we consider the points of the domain and the function values of the range, we get two sequences that correspond to each other via the function. This concept is closely related to continuity.

Example 3.2 (Non-Complete Space):
If we consider $\mathbb{Q}$ embedded in $\mathbb{R}$ such that symbols such as $\sqrt{2}$ and $e$ can be interpreted and used. The letter $x$ is assumed to represent a rational number in this example.

$\begin{align*} f(x):= \frac{1}{x^2-2} \quad \text{with } 0\leq x\leq 2.\end{align*}$

Example for incompletness of rational numbers — Function graph of $f$ with singularities at $\pm$ 2

Considering the sequence $\lim_{x\rightarrow 2}{f(x)}$ in $\mathbb{Q}$ shows that the actual limit $\sqrt{2}\in \mathbb{R} \setminus \mathbb{Q}$ is not contained in $\mathbb{Q}$ .

$\square$

Hence, a key question is:

What condition on a sequence of numbers is necessary and sufficient for the sequence to converge to a limit but does not explicitly involve the limit?

If we already knew the limit in advance, the answer would be trivial. In general, however, the limit is not known and thus the question not easy to answer. It turns out that the Cauchy-property of a sequence is not only necessary but also sufficient. That is, every convergent Cauchy sequence is convergent (sufficient) and every convergent sequence is a Cauchy sequence (necessary).

Note that every Cauchy sequence is bounded. To see this set $\epsilon:=1$ , then there is a $N^* \in \mathbb{N}$ : $(l, k>N^* \Rightarrow d(x_l, x_k)<1$ and thus $d(x_k, x_{N^*})<1$ for all $k>N^*$ . This means that all points $x_k$ with $k>N^*$ lies within a ball of radius 1 with $x_{N^*}$ as its center.

Theorem 3.2 (Cauchy Sequences & Convergence):

In an Euclidean space $\mathbb{R}^d$ every Cauchy sequence is convergent.

Proof: Let $\{x_n\}$ be a Cauchy sequence in $\mathbb{R}^d$ and let $T:=\{x_1, x_2, \ldots\}$ be the range of the sequence. If $T$ is finite, then all except a finite number of the terms $\{x_n\}$ are equal and hence $\{x_n\}$ converges to this common value.

Now suppose $T$ is infinite. We use the Balzano-Weierstrass Theorem to show that $T$ has an accumulation point $x_0$ , and then we show that $\{x_n\}$ converges to $x_0$ . First, recall that each Cauchy sequence is bounded. Hence, since $T$ is infinite there must be an accumulation point $x_0$ according to the Bolzano-Weierstrass Theorem.
Now let $\epsilon >0$ , then there exist a $N^*$ such that $d(x_k, x_l)< \frac{\epsilon}{2}$ whenever $k,l\geq N^*$ . Hence, if $k\geq N^*$ we have

$d(x_k, x_0) \leq d(x_k, x_l) + d(x_l, x_0) < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ ,

so $\lim_{k \rightarrow \infty}{x_k}=x_0$ .
$\square$

Let us furthermore connect the concepts of metric spaces and Cauchy sequences.

Definition 3.3: (Complete Metric Space & Banach Space)
A metric space $(X, d)$ is called complete (or a Cauchy space) if every Cauchy sequence of points in $X$ has a limit that is also in $X$ .
A Banach space is a complete normed vector space, i.e. a real or complex vector space on which a norm is defined.

$\square$

Complete and Banach space will become important in Functional Analysis, for instance.

One-Sided Limit of a Function

In this section it is about the limit of a sequence that is mapped via a function to a corresponding sequence of the range. As mentioned before, this concept is closely related to continuity. Let $X:=(X, d)$ denote the standard metric space on the real line with $X=(\mathbb{R}, |\cdot|)$ and $f:X \rightarrow \mathbb{R}$ .

Consider the Heaviside function as shown below. In the one-dimensional metric space $X=(\mathbb{R}, |\cdot|)$ there are only two ways to approach a certain point $x_0\in \mathbb{R}$ on the real line. For instance, the point $x_0=0$ can be either be approached from the negative (denoted by $\nearrow$ ) or from the positive (denoted by $\swarrow$ ) part of the real line. Sometimes this is stated as the limit $x_0$ is approached “from the left/righ” or “from below/above”.

The Heaviside function does not have a limit at $x_0=0$ , because if you approach 0 from positive numbers the value is 1 while if you approach from negative numbers the value is 0. As we know, the limit needs to be unique if it exists.

Hence, by writing the left-sided limit

(7) $\begin{align*}\lim_{x \nearrow a}{f(x)} = y_0,\end{align*}$

we mean that for every real number $\epsilon>0$ , there is a $\delta>0$ , such that

(8) $\begin{align*}0 < a-x < \delta \ \Rightarrow \ | f(x), y_0| < \epsilon.\end{align*}$

Left-sided means that the $x$ -value increases on the real axis and approaches from the left to the limit point $x_0$ . Hence, $x_0-x$ is positive.

Accordingly, by writing the right-sided limit

(9) $\begin{align*}\lim_{x \swarrow a}{f(x)} = y_0,\end{align*}$

we mean that for every real number $\epsilon>0$ , there is a $\delta>0$ , such that

(10) $\begin{align*}0 < x-a < \delta \ \Rightarrow \ | f(x), y_0| < \epsilon.\end{align*}$

Right-sided means that the $x$ -value decreases on the real axis and approaches from the right to the limit point $x_0$ . Hence, $x-x_0$ is positive.

$\square$

Consider that the left-sided and right-sided limits are just the restricted functions, where the domain is constrained to the “right-hand side” or “left-hand side” of the domain relative to its limit point $x_0\in X$ .

That is, for $X$ being the metric space the left-sided and the right-sided domains are $X_l :=\{x\in X| x<a\}$ and $X_r :={x\in X| x<a}$ , respectively. If we then consider the limit of the restricted functions $f_{|X_l}$ and $f_{|X_r}$ , we get an equivalent to the definitions above.

Having said that, it is clear that all the rules and principles also apply to this type of convergence. In particular, this type will be of interest in the context of continuity.

Continuous Functions, Open and Closed Sets

Now, we connect two important concepts via the following theorem.

Theorem 3.3 (Continuous Functions & Closed/Open Sets):
Let $f:X \rightarrow Y$ be a function between metric spaces $X$ and $Y$ . Then the following are equivalent:

(i) $f$ is continuous;
(ii) $f^{-1}(O)$ is open in $X$ for each open set $O\in Y$ ;
(iii) $f^{-1}(C)$ is closed in $X$ for each closed set $C\in Y$ .

Proof: ‘(i) $\Rightarrow$ (ii)’ Suppose $f$ is continuous on $X$ and let $O\subseteq Y$ be an open set. If $f^{-1}(O)=\emptyset$ , then the claim follows from Proposition 1.1 (i). Thus, we suppose that $f^{-1}(O)\neq \emptyset$ .

Let $x_0\in f^{-1}(O) \subseteq X$ . Due to the fact that $O$ is open, we can chose at least one $\epsilon >0$ such that $B(f(x_0), \epsilon)\subseteq O$ for every $f(x_0)\in O \subseteq Y$ . Given that $f$ is continuous on $X$ , a corresponding $\delta >0$ exists such that

$\begin{align*} f(B(x_0, \delta) \cap X) \subseteq B(f(x_0), \epsilon) \end{align*}$

and $x_0\in f^{-1}(O)$ . That is, $x \in B(x_0, \delta)$ implies $f(x)\in B(f(x_0), \epsilon)$ . Thus, $B(x_0, \delta) \subseteq f^{-1}(O)$ , which means that $f^{-1}(O)$ is an open set.

‘(ii) $\Rightarrow$ (i)’ Let us suppose that $f^{-1}(O)$ is open whenever $O\subseteq Y$ is open and we aim to prove that $f$ is continuous at each point of $X$ . Given $x_0\in X$ and $\epsilon>0$ , we know that the ball $B(f(x_0), \epsilon)$ is open in $Y$ . By assumption so is $f^{-1}(B(f(x_0), \epsilon)$ . Since $x_0\in f^{-1}(B(f(x_0), \epsilon)$ and $f^{-1}(B(f(x_0), \epsilon)$ is an open set, there must be a $\delta>0$ such that $B(x_0, \delta) \subseteq f^{-1}(B(f(x_0), \epsilon)$ . This proves the assertion.

$\square$

Please also refer to some counterexamples illustrating that the image of an open (closed) set under a continuous function does not need to be open (closed).

Literature:
[1]

[2]

Basic Definitions & Properties

Balls, Open & Closed Sets & Neighborhoods

Properties of Open & Closed Sets

Accumulation and Limit Points

Compact Sets

Topological Spaces

Applications

Convergence of a Sequence

Cauchy Sequences

One-Sided Limit of a Function

Continuous Functions, Open and Closed Sets

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