Fundamentals of Set-Theoretic Topology

Why would one want to generalize notions such as convergence and continuity to a setting even more abstract than metric spaces?

Topology deals with the relative position of objects to each other and their features. It is not about their concrete length, volume, and so on. Hence, topological features will not change if continuous transformations are applied to these objects. That is, topological features are preserved under stretching, squeezing, bending, and so on but they are not preserved under non-continuous transformations such as tearing apart, cutting and so on. Objects such as a circle, a rectangle and a triangle are from a topological point of view “equal” / homeomorphic even though the shapes are geometrically rather different.

What features are therefore of interest such that it is worth studying topology?

Assume that $A$ is a closed curve, i.e. a circle, a rectangle, a triangle or something like we can see in the graph above. As long as we transform a shape $A$ in a continuous fashion, the relative positions of the points $x\in A$ and $y\in A^C$ will be similar: for instance, points that have been inside of $A$ will still be inside after the continuous transformation. Points that have been on the boundary will still be on the boundary.

Hence, the generalization of continuity and the concept of convergence (i.e. points being ‘close to each other’) are the two most characterizing features in topological spaces.

However, convergence and continuity in a metric space $(X, d)$ were based on a notion of a distance function $d(x,y)$ for points $x,y \in X$ . Set-theoretic topology generalizes the features of topological metric space and ought to be based on an axiomatized notion of “closeness“.

This post is based on the literature [1] to [5]. For English-speaking beginners, [5] is recommended. The German lecture notes [3] is also a good introduction to topology.

Contents

1 Topology on a Set
2 Separation Properties
- 2.1 Hausdorff Space
- 2.2 Spaces with Weaker Separation Property
3 Convergent Sequences
4 Compactness
5 Continuous Functions
6 Appendix
7 Literature

Topology on a Set

The term “topology on a set” is based on an axiomatic description of so-called “open sets” with respect to some set-theoretic operators. It will turn out, that a topology is a set that has just enough structure to meaningful speak of convergence and continuous functions on it.

Open Sets

Definition 1.1 (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 $I$ .

$\square$

Let $(X, \mathcal{T})$ — short $X$ — be a topological space in this post.

The following video provides a rather unorthodox way of thinking about a topology. However, it might help to get a heuristic understanding. The connection between metrics and topologies is also mentioned.

Topology vs. “a” Topology by PBS Infinite Series

Some examples will further support the understanding.

Example 1.1 (Topologies)

(a) Let $X$ be a set and $\mathcal{T}_0=\{\emptyset, X\}$ then $\mathcal{T}_0$ is the so-called trivial, chaotic or indiscrete topology. The only open sets of the trivial topology are $\emptyset$ and $X$ .

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

(c) There are four topologies on the set $X:=\{0,1\}$ , i.e. $\mathcal{T}_0:=\{\emptyset, X\}$ , $\mathcal{T}_1:=\{\emptyset, X, \{0\}\}$ , $\mathcal{T}_2:=\{\emptyset, X, \{1\}\}$ , $\mathcal{T}_3:=\{\emptyset, X, \{0\}, \{1\} \} = 2^X$ .

It is still an open problem, which is related to combinatorics and lattice theory, to find a simple formula for the number of topologies on a finite set.

(d) Let $(X, \mathcal{T})$ be a topological space, and let $Y\subseteq X$ . The relative topology on $Y$ (or the topology inherited from $X$ ) is the collection

$\begin{align*} \mathcal{T}|_{Y}:= \{ Y \cap U: \ U\in \mathcal{T} \} \end{align*}$

of subsets of $Y$ . It is clearly a topology on $Y$ . The space $(Y,\mathcal{T}|_{Y})$ is then called a subspace of $X$ .

(e) Let $X$ be a non-empty infinite set and $\mathcal{T}:=\{A \subseteq X | A=\emptyset \text{ or } X \setminus A \text{ finite} \ \}$ be the family of open sets. Then $(X, \mathcal{T})$ is the so-called finite complement topology. First note, that an element $A\in \mathcal{T}$ , $A\neq \emptyset$ is of infinite cardinality. Think about what happens if we remove finitely many elements from an infinite set.

Apparently, $X\in \mathcal{T}$ since $X \setminus X = \emptyset$ can be considered as finite. $\emptyset \in \mathcal{T}$ due to the definition of the set $\mathcal{T}$ even though $X \setminus \emptyset = X$ is not finite.

The union of arbitrary open as well as the intersection of a finitely many open sets are open again. The corresponding proof employs De Morgan’s Laws.

(f) Let $X$ be a non-empty countable set and $\mathcal{T}:=\{A \subseteq X | A=\emptyset \text{ or } X \setminus A \text{ countable} \ \}$ be the family of open sets. Then $(X, \mathcal{T})$ is the so-called countable complement topology.

Apparently, $X\in \mathcal{T}$ since $X \setminus X = \emptyset$ is countable and $\emptyset \in \mathcal{T}$ due to the definition. The corresponding proofs of the union and intersection of open sets also employs De Morgan’s Laws.

$\square$

According to Proposition 1.1 – Fundamentals of Topology & Metric Spaces, the set $\mathcal{O}$ of all open sets in a metric space $(X, d)$ complies with the definition of a topology.

Definition 1.2 (Induced & Equivalent Topology)
A topology $(X,\mathcal{T}_d)$ induced by a metric space $(X, d)$ is defined as the set $\mathcal{O}$ of all open sets in $X$ .

Two metrics $d_1$ and $d_2$ on the same basic set $X$ are called topologically equivalent if $\mathcal{T}_{d_1}= \mathcal{T}_{d_2}$ .

$\square$

Let us define a very basic but important term.

Definition 1.3 (Neighborhood)
A neighborhood $A$ of a point $x_0$ in a topological space $(X, \mathcal{T})$ is any set containing $x_0$ as well as an open set $B\in \mathcal{T}$ of $x_0$ , i.e. $\{x_0\} \subseteq B \subseteq A$ .

$\square$

Some examples will improve the understanding.

Example 1.2 (Metrizable and Equivalent Topologies)

(a) If $(X, d)$ is a metric space and $\mathcal{T}_d:= \mathcal{O}$ is the set of all open sets, then $\mathcal{T}_d$ is a topology. The topology $\mathcal{T}_d$ does not depend on the particular metric $d$ since the proof of the referred Proposition 1.1 can also be done using the term neighborhood instead of the distance function. Hence, any metric on $X$ equivalent to $d$ yields the same topology. Topological spaces of this kind are called metrizable.

(b) Let us consider $X\subseteq \mathbb{R}^n$ equipped with the natural topology by taking the topology $\mathcal{T}_{d_E}$ induced by the Euclidean metric

$\begin{align*} d_E(x,y) = \sqrt{\sum_{i=1}^{n}{(x_i - y_i)^2} } \end{align*}$

with $x=(x_1, \ldots, x_n)$ and $y=(y_1, \ldots, y_n)$ . Instead of using the Euclidean metric, we could also employ the following distance functions:

$\begin{align*} d_M(x,y) &= \max\{|x_i - y_i| \ : i=1, \ldots, n \}, \\ d_S(x,y) &= \sum_{i=1}^{n}{(|x_i - y_i|}. \end{align*}$

All three induced topologies would be equivalent, i.e. $\mathcal{T}_E = \mathcal{T}_M = \mathcal{T}_S$ since

$\begin{align*} \frac{1}{n} d_E(x,y) \leq d_M(x,y) \leq d_E(x,y) \leq d_S(x,y) \end{align*}$

for all $x,y\in X$ . The corresponding unit open balls centered at $(0,0)\in \mathbb{R}^2$ are illustrated as follows.

Fig. 1: Unit balls centered at (0,0) by three equivalent metrics

An open ball $d(x,0)< 1$ with $x\in \mathbb{R}^2$ as shown in Fig. 1 is actually a set of points where each point has a distance of max. 1 to the origin $(0,0)$ . This, however, is nothing but the corresponding norm $||x||$ . For instance, the point $x=(0.5,0.6)$ is not an element of the unit ball induced by $d_S$ since $0.5+0.6>1$ . However, the same point $x$ is element of the unit balls induced by $d_E$ and $d_M$ .

From a topological point of view the shapes in Fig. 1 are all equivalent.

$\square$

Closed Sets

Directly linked via the definition to open sets and equivalent in their explanatory power are closed sets.

A set $A$ is open in an Euclidean metric space if and only if for every $x_0\in A$ an open ball $B(x_0, r_0)$ exists such that $B(x_0, r_0) \subseteq A$ . Let us therefore consider the situation in the real plane $\mathbb{R}^2$ employing the topology induced by the Euclidean metric. The open balls $B(x_1, r_1)$ and $B(x_2, r_2)$ are both contained in $A$ .

However, the open ball $B(x_0, r_0)$ cannot be fully contained in $A$ no matter how small we pick $r_0>0$ . Thus, $x_0$ is not an element of the open set $A$ .

Let us now study sets $A^C$ whenever $A$ is open.

Definition 1.4 (Closed Sets)
A set $A$ is called closed in $(X, \mathcal{T})$ if $A^C = X\setminus A$ is open in $X$ .

$\square$

The definition actually tells us that one just needs to consider the complement set $A^C$ to figure whether the set $A$ is closed.

Example 1.3 (Closed Sets)
(a) Let $(X, d)$ b a metric space. Then $A$ is closed in $(X, \mathcal{T}_d)$ if and only if $A$ is closed in $(X, d)$ . For instance, an arbitrary open ball $B(x_0, \epsilon)$ with $x\in \mathbb{R}^d$ is open. Thus, the set $\mathbb{R}^d \setminus B(x_0, \epsilon)$ is closed. The situation for $X=\mathbb{R}^2$ and the Euclidean topology is sketched in the next figure.

(b) The sets $\{X, \emptyset\}$ are not only open but also closed for any topological space $(X,\mathcal{T})$ since $\emptyset = X \setminus X$ and $X = X \setminus \emptyset$ .

The topological space is trivial / indiscrete if and only if these two sets are the only closed sets in $(X,\mathcal{T})$ . The closed sets of the indiscrete / trivial topology are the complements of the open sets. Hence, the closed sets are also $emptyset$ and $X$ .

(c) The topology $(X, \mathcal{T})$ is discrete if and only if every subset $A \subseteq X$ is closed. This can be seen by $A=X\setminus (X\setminus A) = (A^C)^C$ .

(d) The subset $[a,b]$ of $\mathbb{R}$ is closed because its complement $\mathbb{R} \setminus [a, b] = (-\infty, a) \cup (b, \infty)$ is open. Similarly, $[a, +\infty)$ is closed, because its complement $(-\infty, a)$ is open.
The subsets $[a, b)$ of $\mathbb{R}$ are neither open nor closed.

(e) In the finite complement topology on an infinite set $X$ , the closed sets consists of $X$ itself and all finite subsets of $X$ . This follows directly from the definition of the set $\mathcal{T}$ as set out in Example 1.1 (e). According to this definition, the sets $A^C = X\setminus A$ with $A\in \mathcal{T}$ need to be finite with the exception of $A^C=X$ .

$\square$

Let us characterize closed sets.

Proposition 1.1: (Characterization of Closed Sets)
(1) The set $\mathcal{S}$ of all closed sets of $(X, \mathcal{T})$ complies with the following conditions:
(i) $\emptyset \in \mathcal{S}$ and $X\in \mathcal{S}$ .
(ii) $A, B \in \mathcal{S}$ implies $A \cup B\in \mathcal{S}$ .
(iii) $\mathcal{A} \subseteq \mathcal{S}$ implies $\bigcap_{A\in \mathcal{A}}{A}\in \mathcal{S}$ .

(2) Let $\mathcal{A}$ be a family of sets that complies with (i), (ii) and (iii) then there exists a topology $(X, \mathcal{T})$ , such that $\mathcal{S}$ is the set of all closed sets in $(X, \mathcal{T})$ .

Proof.
(1) This follows directly from the definitions and applying the rules $X \setminus X=\emptyset$ , $X\setminus \emptyset=X$ , $X \setminus (A \cup B) = (X\setminus A) \cap (X \setminus B)$ as well as $X \setminus \bigcap_{A\in \mathcal{A}}{A} = \bigcup_{A\in \mathcal{A}}{X\setminus A}$ .

(2) The family of closed sets $\mathcal{S}$ fully determines the topology $\mathcal{T}$ on the same basic set $X$ since $\mathcal{S}=\{ A | \ (X\setminus A)\in \mathcal{T} \}$ . Its existence follows from the fact that $\mathcal{S}$ actually is a topology but this is clear given (1).

$\square$

The family of closed sets of a topology could also be used to define a topological space, i.e. the set of all closed sets contains exactly the same information as the set of all open sets that actually define the topology.

Points, that lie at the boundary between $A$ and $A^C$ , are crucial for the understanding and distinction of open and closed sets. The intuitive idea of a boundary point $x_0$ is depicted in the next figure.

Interior, Closure & Boundary

The points that lie close to both the “inside” and the “outside” of the set $A$ play an important role. However, before we can define a boundary point, we need to clarify what is meant by “inside” and “outside”.

Definition 1.5 (Interior & Closure)
Let $(X, \mathcal{T})$ be a topology. The interior of $A\subset X$ is defined as the union of all open sets contained in $A$ , i.e.

$\begin{align*} \text{Int} A = \bigcup_{T \subseteq A}{T}, \quad T\in \mathcal{T}. \end{align*}$

The closure of $A$ is defined as the intersection of all closed sets containing $A$ , i.e.

$\begin{align*} \text{Cl} A = \bigcap_{A \subseteq S}{S}, \quad S\in \mathcal{S}. \end{align*}$

$\square$

Apparently, the interior of $A$ is open and a subset of $A$ while the closure of $A$ is closed and contains $A$ . Thus, the following set relation is valid for any set $A\subseteq X$ in a topological space:

$\begin{align*} \text{Int}A \subseteq A \subseteq \text{Cl}A. \end{align*}$

The following theorem provides some useful relationships. However, we will not prove all of the statements and refer to section 2.1 in [5], for instance.

Theorem 1.1 (Properties of Closure and Interior)
For sets $A, B \subseteq X$ in a topological space $(X, \mathcal{T})$ , the following statements hold:
(i) If $O$ is an open set in $X$ and $O\subseteq A$ , then $O\subseteq$ Int $A$ ;
(ii) If $C$ is a closed set in $X$ and $A\subseteq C$ , then Cl $A\subseteq C$ ;
(iii) If $A\subseteq B$ then Int $A\subseteq$ Int $B$ and Cl $A\subseteq$ Cl $B$ ;
(iv) $A$ is open if and only if Int $A=A$ ;
(v) $A$ is closed if and only if Cl $A=A$ .

Proof. (i) Since Int $A$ is the union of all of the open sets that are contained in $A$ , it follows that $O$ is one of the sets making up this union and therefore is a subset of the union. That is, $O\subseteq$ Int $A$ .
(ii) Since Cl $A$ is the intersection of all of the closed sets that contain $A$ , it follows that $C$ is one of the sets making up this intersection and therefore is Cl $A$ is contained in $C$ . That is, Cl $A \subseteq C$ .
(iii) Since $A \subseteq B$ , Int $A$ is an open set contained in $B$ . According to (i) every open set contained in $B$ is contained in Int $B$ . Therefore, Int $A\subseteq$ Int $B$ . One can use (ii) to show the second statement of (iii).
(v) If $A=$ Int $A$ , then $A$ is an open set since by definition Int $A$ is an open set. Now assume that $A$ is open. We show that $A=$ Int $A$ . First, Int $A\subseteq A$ by definition of Int $A$ . Furthermore, since $A$ is an open set contained in $A$ , it follows by (i) that $A\subseteq$ Int $A$ . Thus, $A=$ Int $A$ as we wished to show.

$\square$

Example 1.4 (Closure and Interior)
(a) Consider $A:=[-1,1)$ in the standard topology on $\mathbb{R}$ . Then $\text{Cl}A \setminus \text{Int} A = [-1,1]$ and Int $[-1,1)=(-1,1)$ .

(b) Consider $A:=[-1,1)$ as a subset of $X:=\mathbb{R}$ with the discrete topology $\mathcal{T}=2^{X}$ . Since all subsets are open and closed we have Int $A=$ Cl $A=[-1,1)$ .

(c) Consider $A:=[-1,1)$ in the finite complement topology on $\mathbb{R}$ . Since a closd set in this topology is either $X=\mathbb{R}$ or finite, it is clear that only $X$ is a closed set containing the infinite set $A$ . Hence, Cl $A=X=\mathbb{R}$ . The open sets are precisely the empty set and the cofinite subsets, i.e., the subsets whose complements are finite subsets of $X=\mathbb{R}$ . Hence, Int $A=\emptyset$ since there are no open sets in $X$ contained in $A$ .

$\square$

The last example highlights that not only the actual sets matter but also the sourounding topology. The next theorem provides a simple means for determining when a particular point $x$ is in the interior or in the closure of a given set $A$ .

Theorem 1.2 (Closure, Interior and Open Sets)
Let $(X, \mathcal{T})$ be a topological space, $A$ be a subset of $X$ , and $x\in X$ .
(i) Then $x\in$ Int $A$ if and only if there exists an open set $O$ such that $x\in O\subseteq A$ ;
(ii) Then $x\in$ Cl $A$ if and only if every open set containing $x$ intersects $A$ .

Proof. (i) First, suppose that there exists an open set $O$ such that $x\in O\subseteq A$ . Then, since $O$ is open and contained in $A$ , it follows that $O\subseteq$ Int $A$ . Thus, $x\in O\subseteq$ Int $A$ .

Next, if $x\in$ Int $A$ , and we set $O=$ Int $A$ , it follows that $O$ is an open set such that $x\in O \subseteq A$ .

(ii) Considering the contrapositive: $x\notin$ Cl $A$ if and only if there is one neighborhood that does not intersect $A$ . If $x\notin$ Cl $A$ then the set $O=X \setminus$ Cl $A$ is open and does contain $x$ , which is as claimed in the contrapositive.

Conversely, if there is a neighborhood $O$ of $x$ which does not intersects $A$ , then its complement $X\setminus O$ is a closed set that contains $A$ . By definition of the closure Cl $A$ , the set $X\setminus O$ must contain Cl $A$ . Since $x\in O$ it follows that $x\notin$ Cl $A$ . Contraction.

$\square$

Theorem 1.3 (Relations of Closure and Interior)
For sets $A, B \subseteq X$ in a topological space $(X, \mathcal{T})$ , the following statements hold:
(i) Int $(X\setminus A)$ = $X \setminus$ Cl $A$ ;
(ii) Cl $(X\setminus A)$ = $X \setminus$ Int $A$ ;
(iii) Int $A \cup$ Int $B \subseteq$ Int $(A \cup B)$ , and in general equality does not hold;
(iv) Int $A \cap$ Int $B$ = Int $(A\cap B)$ .

Proof. Refer to Theorem 2.6 in [5] and to Folgerung 1.2.25 in [3].

$\square$

Now, we have all ingredients to define the so-called boundary.

Definition 1.6 (Boundary Set & Points)
Let $(X, \mathcal{T})$ , $A\subseteq X$ . The boundary of $A$ , denoted by $\partial A := \text{Cl}A \setminus \text{Int}A$ . A point of $\partial A$ is called boundary point.

$\square$

There are situations that challenge or defy our intuitive understanding of boundary sets. For example, what is the boundary of $\mathbb{Q}$ as a subset of $\mathbb{R}$ in the standard topology?

Example 1.4 (Boundary Sets & Points)
(a) Consider $A:=[-1,1]$ in the standard topology on $\mathbb{R}$ . The boundary set $\partial A$ equals $\text{Cl}A \setminus \text{Int} A = [-1,1] \setminus (-1,1)=\{-1,+1\}$ .

(b) Consider $\mathbb{Q}$ in the standard topology on $\mathbb{R}$ . Since $\text{Cl}\mathbb{Q}= \mathbb{R}$ , and $\text{Int}\mathbb{Q}=\emptyset$ , it follows that $\partial \mathbb{Q}=\mathbb{R}$ . The entire real line is therefore the boundary of the rational numbers, which makes sense. Every real number is arbitrarily close to the set of rational numbers and to its complement, the set of irrational numbers.

(c) Let $A=[-1,1]$ in $\mathbb{R}$ with the discrete topology $2^\mathbb{R}$ . That is, every subset is open and closed at the same time. Hence, $\text{Int}[-1,1]=\text{Cl}[-1,1]=[-1,1]$ and $\partial [-1,1]=\emptyset$ .

$\square$

In a metric space, a point $x$ is a boundary point of $A$ 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 $\epsilon>0$ .

Let us now prove the generalized statement about boundary sets.

Proposition 1.2 (Characterization of Boundary Points)
Let $A$ be a subset of a topological space $(X, \mathcal{T})$ and let $x\in X$ . Then $x\in \partial A$ if and only if every neighborhood of $x$ intersects both $A$ and $A^C$ .

Proof. Suppose $x\in \partial A$ , then $x\in \text{Cl}A$ and $x\notin \text{Int}A$ due to the definition. Since $x\in \text{Cl}A$ , it follows that every neighborhood of $x$ intersects $A$ . Note that a neighborhood of $x$ contains $x$ . Furthermore, since $x\notin \text{Int}A$ , it follows that every neighborhood of $x$ intersects $A^C = X\setminus A$ . Thus, every neighborhood of $x$ intersects $A$ and $A^C$ .

Now suppose that every neighborhood of $x$ intersects $A$ and $X\setminus A$ . It follows that $x\in \text{Cl}A$ and $x\in$ Cl $(X\setminus A)$ . By

$\square$

Theorem 1.4 (Open Sets and Neighborhoods)
Let $X$ be a topological space and let $A$ be a subset of $X$ . Then $A$ is open in $X$ if and only if for each $x\in A$ , there is a neighborhood $U$ of $x$ such that $x\in U\subseteq A$ .

Proof. First, suppose that $A$ is open in $X$ and $x\in A$ . If we let $U=A$ then $U$ is a neighborhood of $x$ for which $x\in U\subseteq A$ .

Now suppose that for every $x\in A$ there exists a neighborhood $U_x$ of $x$ such that $x\in U_x\subseteq A$ . Since we know that the union $A=\bigcup_{x\in A}{U_x}$ of open sets is open, the assertion follows.

$\square$

The set $\partial A$ 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.
If all boundary points are contained within the set, then it is a closed set.

Proposition 1.2 (Closure and Open Sets)
Let $X$ be a topological space, $A$ be a subset of $X$ , and $x$ be an element of $X$ . Then $x\in \text{Cl}A$ if and only if every open set containing $x$ intersects $A$ .

Proof. The closure $\text{Cl}A$ is the smallest closed set containing $A$ as a subset. Let $x\in \text{Cl}A$ and let $U$ be an open neighborhood of $x$ . If $U\cap A = \emptyset$ then $A\subseteq X \setminus U$ , and the latter set is closed (by definition), which leads to a contradiction. So $A \cap U\neq \emptyset$ .

$\square$

The property of the closure hitting an open set is so important that usually a new term is defined. Note, however, that we will not further use it in this post.

Definition 1.7 (Adherent Point)
Let $(X, \mathcal{T})$ , $A\subseteq X$ and $x\in A$ . The point $x$ is called adherent point if every open $U$ of $x$ has a non-empty intersection with $A$ , i.e. $A \cap U \neq \emptyset$ for all $U\in \mathcal{T}$ with $x\in U$ .

$\square$

Let us come back to Example 1.2 (a) – considering this example we could ask whether every topological space is metrizable?
The answer is no, and the root-cause is that topological spaces have different types of separation properties.

Separation Properties

A metric enables us to separate points in a metric space since any two distinct points have a strictly positive distance. In general topological spaces, separating points from each other is more subtle.

Hausdorff Space

Hausdorff spaces and the Hausdorff condition are named after Felix Hausdorff, one of the founders of topology. Let us first check out the formal definition.

Definition 2.1: (Hausdorff Space, $T_2$ Spaces)
A topological space $X$ is a Hausdorff or $T_2$ -space if, for any pair of distinct points $x, y\in X$ , $x\neq y$ there are disjoint open sets $U, V\in \mathcal{T}$ with $x\in U$ , $y\in V$ and $U\cap V = \emptyset$ .

$\square$

Every Euclidean space is Hausdorff since we can use the Euclidean metric to separate two distinct points. The following video outlines the Hausdorff condition and it provides a simple example of a Hausdorff space.

Hausdorff Condition incl. an Example and Counterexample by DanielChanMaths

Example 2.1 (Metric Space is Hausdorff)
(a) Let $(X, d)$ be a metric space, and let $x,y\in X$ be such that $x\neq y$ . It follows that $\epsilon:=0.5 \cdot d(x,y)>0$ . Let $U:=B(x, \epsilon)$ and $V:=B(y, \epsilon)$ . Then, $X$ is Hausdorff since $U\cap V=\emptyset$ .

(b) Consider the indiscrete / trivial topology $(\mathcal{T}_0=\{\emptyset, X\}, X)$ with $|X|>1$ . By definition, the only neighborhood of any two points $x,y\in X$ is the entire set $X$ . Thus, every neighborhood of $x$ will contain $y$ and vice verca. It follows that the indiscrete / trivial topology is neither Hausdorff nor metrizable.

$\square$

In a Hausdorff space, distinct points can be separated by open sets.

The situation in $X:=\mathbb{R}^2$ along with the topology implied by the Euclidean metric is illustrated in the graph above. For two distinct points $x_1, x_2\in X$ , we take half (or less) the distance to define $\epsilon$ to come up with two distinct open balls, that can also be seen as disjoint neighborhoods.

Proposition 2.1: (Subset of $T_2$ -spaces)
Let $(X, \mathcal{T})$ be a topological Hausdorff space. Then, each subset $A\subseteq X$ of a Hausdorff space is Hausdorff.

Proof. Let $x\neq y$ be in $X$ . The space being Hausdorff, let $U_x$ and $V_y$ be the two open separating sets as required in Definition 1.1. Then $U:=U_x \cap A$ as well as $V:=V_y \cap A$ are open since the difference of two open sets is open. In addition, $x\in U$ and $y\in V$ .

$\square$

Spaces with Weaker Separation Property

The following separation properties are weaker than the Hausdorff ( $T_2$ -) condition. This is also indicated by the index of the corresponding names of the separation axioms (from $T_0$ to $T_2$ ).

Definition 2.2. ( $T_0$ Space)
A topological space $(X, \mathcal{T})$ is called a $T_0$ – or Kolmogorov space if, for any $x,y\in X$ with $x\neq y$ , there is an open set $U\subseteq X$ with $x\in U$ and $y\notin U$ or $y\in U$ and $x\notin U$ .

$\square$

The most striking difference between a Hausdorff / $T_2$ – and $T_0$ -space is that only one open set $U$ , that contains only one of two distinct points, is required to fulfill the definition of a $T_0$ -space. Apparently, every $T_2$ -space is also a $T_0$ -space.

Example 2.2:
(a) Let $X=\{x,y\}$ be any set with at least two elements equipped with the so-called chaotic topology $\mathcal{T}:=\{\emptyset, X\}$ . Then, there is no open $U\in \mathcal{T}$ that separates the two distinct elements. Hence, this topology is not a $T_0$ -space and definitely also not a Hausdorff space. Hence, it is also not metrizable.

(b) Let $X=\{x,y\}$ be any set with the discrete topology $\mathcal{T}:=2^X$ . Then, $\{x\} \in \mathcal{T}$ separates the two elements $x$ and $y$ .

$\square$

In a $T_1$ -space two open sets are required to separate two distinct points, however, the two sets don’t need to be disjoint.

Definition 2.3. ( $T_1$ Space)
A topological space $(X, \mathcal{T})$ is called a $T_1$ -space if, for any $x,y\in X$ with $x\neq y$ , there are open sets $U, V\subseteq X$ with $x\in U$ and $y\notin U$ and $y\in U$ and $x\notin U$ .

$\square$

The main difference between a $T_2$ – and a $T_1$ -space is that the two required open sets do not have to be disjoint. However, one open set only contains one of the two distinct points.

Hence, every $T_1$ -space is also a $T_0$ -space. Just take one of the two open sets of the $T_1$ -space and it fulfills all requirements of a $T_0$ -space.

Proposition 2.2: (Characterization of $T_1$ -spaces)
(a) Let $(X, \mathcal{T})$ be a topological space. Then, $X$ is a $T_1$ -space if and only if $\{x\}$ is a closed set for each $x\in X$ ;
(b) Each Hausdorff space is a $T_1$ -space.

Proof. (a) Suppose $X$ is a $T_1$ -space, and let $x\in X$ . For any $y\in X$ with $y\neq x$ , there is an open subset $U_y$ of $X$ with $y\in U_y$ , but $x\notin U_y$ . It follows that $X\setminus \{x\}=\bigcup{ \{U_y: y\in X, x\neq y\} }$ .

Conversely, suppose that all singleton subsets of $X$ are closed, and let $x,y\in X$ be such that $x\neq y$ . Then, $V:= X\setminus \{x\}$ and $U:= X\setminus \{y\}$ fulfill the requirements of a $T_1$ -space.

(b) Let $x$ be a given point. By assumption, each $y\neq x$ belongs to an open set $U_y$ such that $x\notin G_y$ . Consequently, $X \setminus \{x\} = \bigcup_{y\neq x}{U_y}$ . Thus, $X\setminus \{x\}$ is open, and $\{x\}$ is closed.

$\square$

Convergent Sequences

One of the key features of topological spaces is the generalization of the convergence concept.

A sequence in a (metric) space $X$ is a function $x_n:\mathbb{N} \rightarrow X$ that we also denote by $(x_n)$ , in particular, if we want to refer to the elements of the sequence. Given a sequence $x_n$ in a metric space, a sub-sequence is the restriction of $x_n$ to an infinite subset $S \subseteq \mathbb{N}$ . If we exhibit $S$ as $n_1<n_2<\ldots <n_k<\ldots$ , then we write the subsequence $(x_{n_k})$ .

We say that a sequence $(x_n)$ converges to $x\in X$ if given $\epsilon>0$ , there exist $N\in \mathbb{N}$ such that for all $n\in \mathbb{N}$ , we have $x_n\in B(x, \epsilon)$ .

In other words, for all $n\geq N$ , we have $x_n\in B(x, \epsilon)$ as illustrated in Figure above. The finitely many elements $x_1, \ldots, x_{n-1}$ of the sequence $(x_n)$ are, however, not contained in $B(x, \epsilon)$ . We take this property to define what a convergent sequence in a topological space is.

Definition 3.1. (Convergent Sequence)
Let $(X, \mathcal{T}_d)$ be a topological space. A sequence $(x_n)$ converges to $x\in X$ if $\mathcal{T}_d$ , the set $\{k \ | \ x_k\notin U\}$ is finite for any open set $U\ni x$ .

The point $x\in X$ is then called the limit of the sequence $(x_n)$ and we denote it by $x = \lim_{n\rightarrow \infty}{x_n}$ or by $x_n \rightarrow x$ .

$\square$

Note that the set of $\{k \ | \ x_k\in U\} = \{k \ | \ x_k\notin U\}^C$ is infinite for all open sets $U\ni x$ while (at the same time) the set $\{k \ | \ x_k\notin U\}$ is finite. Both sets/conditions matter in this situation as we will see further below!

Lemma 3.1. (Limit of a sequence is unique)
The limit $x$ of a convergent sequence $(x_n)\rightarrow x$ in a Hausdorff space $(X, d)$ is unique.

Proof. Assume that this is not the case and $x_n \rightarrow x$ as well as $x_n \rightarrow y$ with $x\neq y$ holds true. A metric space is Hausdorff, that is, we find two disjoint open balls $B(x, r)$ and $B(y, r)$ . Given that $x$ and $y$ are the limit points almost all elements must lie in the disjoint balls, which contradicts the initial assumption of $x\neq y$ .

$\square$

Lemma 1.1 is false in arbitrary topological spaces.

Every $x\in X$ of a topological space $(X, \mathcal{T})$ is the limit of a certain sequence $(x_n)$ . Apparently, we could simply use the constant sequence $x_1=x, x_2=x, \ldots$ or we could define $x_k:=x$ for all $k\geq N$ , $N\in \mathbb{N}$ . This fact should be also considered in the following examples.

Example 3.1:
(a) Let $(X, \mathcal{T})$ be the discrete topology with $\mathcal{T}=2^X$ . Further, let $x_n \rightarrow x$ . Recall that in this topology every set is open by definition. Hence, also $\{x\}$ is an open set that must be contained in all other (open) supersets. Hence, the set $\{k \ | \ x_k\neq x\}$ has to be finite and $\{k \ | \ x_k= x\}$ has to be infinite.

(b) Let $(X, \mathcal{T})$ be the indiscrete topology with $\mathcal{T}=\{\emptyset, X\}$ . Further, let $x_n \rightarrow x$ . Since $U=X$ is the only set that contains $x\in X$ , the set $\{k \ | \ x_k \notin U\}$ has to be finite for every sequence $(x_n)$ . Hence, every sequence converges to every point in $x\in X$ .

$\square$

Closely related to converging sequences and their limits are accumulation points.

Definition 3.2. (Accumulation Point)
An element $x$ of a sequence $(x_n)$ is called accumulation point (sometimes also cluster or limit point) if $\{k \ | \ x_k\in U\}$ is infinite for every open set $U$ of $x$ .

$\square$

The subtle but important difference between an accumulation point and a limit is that the complement set of $\{k \ | \ x_k\in U\}$ can also be infinite. Let us consider a simple example.

Example 3.2:
Let us consider the sequence $(x_n):= \{(-1)^n \ | \ n\in \mathbb{N}\}$ in the topology induced by the Euclidean space $(X=\mathbb{R}, d_E)$ on the real line. There are two accumulation points $\pm 1$ but no limit of the sequence. Note that the sequence is alternating between $+1$ and $-1$ , such that $\{k \ | \ (-1)^{2k} = 1\}$ and $\{k \ | \ (-1)^{2k} = -1\}$ are both infinite but disjoint to each other. In addition, the set $\{k \ | \ x_k\in U\}$ for an open set $U$ of $x=\pm 1$ are both infinite.

$\square$

A sub-sequence $(x_{n_k})$ of a convergent sequence $(x_n)$ converges to the same limit $x$ . This is evident since if the condition of a convergent sequence is fulfilled for all elements $x_n$ , $n\in \mathbb{N}$ of the series $(x_n)$ . Hence, the condition is also fulfilled for a subset $(x_{n_k}) \subseteq (x_n)$ that represents the sub-sequence.

Due to the fact that the finiteness of $\{k \ | \ x_k\notin U\}$ implies the infiniteness of $\{k \ | \ x_k\in U\}$ in $\mathbb{N}$ every limit is an accumulation point. The converse is not true as we can see in Example 3.2.

Theorem 3.1 (Convergence in Topological Spaces)
Let $(X, \mathcal{T})$ be a topological space.
(i) Every limit of a convergent sequence is also the limit of any sub-sequence.
(ii) Every accumulation point of any sub-sequence $(x_{n_k})$ is also an accumulation point of $(x_n)$ .
(iii) Every accumulation point of a sequence $(x_n)$ in $A$ is an adherence point of the set $A\subseteq X$ .

Proof. (i) If $x_n \rightarrow x$ then $\{k | x_k\notin U\}$ is finite for all open $U$ of $x$ . In particular, this holds true for any sub-sequence $(x_{n_k})\subseteq (x_n)$ and thus $x_{n_k} \rightarrow x$ .

(ii) Let $p\in X$ be an accumulation point of the sub-sequence, i.e. the set $\{k \ | \ x_{k}\in U\}$ is infinite for every open set $U$ of $x$ . Since the sub-sequence is only a subset of the element of the sequence, the assertion follows directly.

(iii) Let $x$ be an accumulation point of $(x_n)$ and $U$ be an open set of $x$ . Apparently, $U$ intersects $(x_n)$ since the set $\{k|\ x_k\in U\}$ is infinite. Thus, $U$ will also intersect $A$ , which implies that $x\in$ Cl $A$ .

$\square$

Let us now assume that $A=\text{Cl}A$ and $(x_n) \rightarrow x$ is a converging sequence with $x_n\in A$ for all $n\in \mathbb{N}$ . Then, the limit point $x$ is also contained in $A$ . A closed set contains all its limit points.

Compactness

The concept of compactness is not as intuitive as others topics such as continuity. In $\mathbb{R}^d$ , the compact sets are the closed and bounded sets, but in a general topology compact sets are not as simple to describe.

Compact sets are so important since they possess important properties, that are known from finite sets:

Set is bounded;
Set contains a maximal and minimal element;
An infinite sequence contains a constant subsequence.

The famous Heine-Borel Theorem shows that compact sets in metric spaces do indeed have these properties. This analogy is also outlined in this really nice video (in German only) by Prof. Dr. Edmund Weitz.

Let $(X, \mathcal{T})$ be a topological space.

Definition 4.1 (Cover)
The collection $\mathcal{A}$ is said to cover a set $A\subseteq X$ or to be a cover of $A$ if the union of the elements of $\mathcal{A}$ contains $A$ .

If $\mathcal{A}$ covers $A$ , and each set in $\mathcal{A}$ is open, then we call $\mathcal{A}$ an open cover of $A$ .

A sub-collection $\mathcal{A}'$ of a cover is called a subcover of $\mathcal{A}$ .

$\square$

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

Example 4.1 (Real Line)
Let us consider the set $A:=\mathbb{R}$ of the topological space $(X=\mathbb{R}, \mathcal{O})$ , where $\mathcal{O}$ are the open sets of $\mathbb{R}$ . Note that any nonempty open subset of $\mathbb{R}$ can be written as a finite or countable union of open mutually disjoint intervals.

Then, $\mathcal{A}:=\{(-n, n) | n\in \mathbb{N}\}$ is an open cover of $A$ because $A \subseteq \bigcup_{n\in \mathbb{N}}{(-n, n)}$ . This cover apparently contains infinitely many open sets $(-n, n)$ .

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

$\square$

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

Definition 4.2 (Compact Set)
A subset $K$ of a topological space $(X, \mathcal{T})$ is said to be compact if every open cover $\mathcal{O}$ of $K$ has a finite (open) subcover.

A topological space $(X, \mathcal{T})$ is called compact if $X$ is compact.

$\square$

It is clear that every finite set is compact and the following example is going to illustrate that.

Example 4.2 (Finite Set & Compactness)
Let $A:=\{a_0, \ldots, a_n\}$ be a finite set in the standard Euclidean topological space $(\mathbb{R}, \mathcal{T})$ . By setting

$\begin{align*} a &\mapsto B(a, \epsilon) \end{align*}$

with $\epsilon>0$ , an open set $B(a, \epsilon)\in \mathcal{T}$ is assigned bijectively to each point $a\in A$ . Hence, the set $\mathcal{A}':=\{B(a, \epsilon) | \ a\in A\}$ is an open and finite cover for any set $A\subseteq \mathbb{R}$ . Each point of the finite set $A$ is contained in one of the elements of $\mathcal{A}'$ . Given any open cover $\mathcal{A}$ , we can always use $\mathcal{A}'$ with $\epsilon>0$ small enough simply because there are only finitely many elements in $A$ .

$\square$

Let us consider another simple example.

Example 4.3 (Real Line)
The real line $\mathbb{R}$ in the standard Euclidean topology is not compact since the set $\mathcal{A}$ of Example 5.1 is an open cover, but no finite sub-collection of $\mathcal{A}$ covers $\mathbb{R}$ . If we picked finitely many open sets of $\mathcal{A}$ the points before the minimum / after the maximum of this sub-collection would not be covered.

$\square$

The last example directly used the definition of a compact set to show that it is not compact since every open cover needs to have a finite sub-cover such that the set can be compact.

Even though $\{ (-\infty, 3), (0, \infty) \}$ is a finite open cover of $A=\mathbb{R}$ , this does not mean that $\mathbb{R}$ is compact: if it were compact, all open covers (including the one defined in Example 5.2/5.3) would have to have a finite sub-cover.

Example 4.4 (Converging Sequence & Compact Set)
(a) The subset $A:=\{\frac{1}{n}| \ n\in \mathbb{N}\}$ $\cup \{0\}$ is compact in the standard topology on $\mathbb{R}$ .

Given any open cover $\mathcal{A}$ of $A$ , there is an element set $U_0\in \mathcal{A}$ containing $0$ . The set $U_0$ contains either all points or all but finitely many of $A$ . If $U_0$ contains all of the points in $A$ , then $U_0$ , by itself, is a finite subcover of $\mathcal{A}$ . Otherwise, let $\frac{1}{k}$ be the smallest of the points in $A$ that are not in $U_0$ . Then we can pick open sets $U_1$ for $\frac{1}{1}\in A$ , $U_2$ for $\frac{1}{2}\in A$ , $\ldots$ , and $U_{k}$ for $\frac{1}{k}\in A$ such that $U_0, U_1, \ldots, U_k$ is a finite sub-cover of $\mathcal{A}$ . Hence, $A$ is compact.

(b) The compact sets of the discrete topology $(X, \mathcal{T})$ are finite. To see this realize that all subsets of $X$ are open and closed. Thus, all possible families of subsets can be open covers. For instance, the family $\mathcal{O}:=\{ \{x\} \ | \ x\in X \}$ is an open cover and if $X$ is of infinite cardinality, so is $\mathcal{O}$ . If we leave out one of the elements of $\mathcal{O}$ it would not be a cover. Hence, there can be no open subcover, which implies that the compact sets are finite. Refer also to this formal proof.

$\square$

Let us now extend the definition of compactness to subsets of topological spaces.

Definition 4.1 (Subspace Topology)
Let $(X, \mathcal{T})$ be a topological space. If $Y \subseteq X$ , the collection

$\begin{align*} \mathcal{T}_Y := \{ Y \cap O | \ O\in \mathcal{T} \} \end{align*}$

is a topology on $Y$ , called the subspace topology. With this topology, $(Y,\mathcal{T}_Y)$ is called topological subspace of $(X,\mathcal{T})$ .

$\square$

Let us check that $(Y,\mathcal{T}_Y)$ is indeed a topology. It contains $\emptyset$ and $Y$ because $\emptyset = Y \cap \emptyset$ and $Y = Y \cap X$ , where $\emptyset$ and $X$ on the right-hand side of the $\cap$ -symbol are elements of $\mathcal{T}$ . The fact that it is closed under finite intersections and arbitrary unions follows from the equations

$\begin{align*} (Y \cap O_1) \cap \ldots \cap (Y \cap O_n) &=Y \cap (O_1 \cap \ldots \cap O_n), \\ \bigcup_{i\in I}{Y \cap O_i} &= Y \cap \bigcup_{i\in I}{O_i}. \end{align*}$

Lemma 4.1 (Compactness & Subspaces)
Let $Y$ be a subspace of $X$ . Then $Y$ is compact in $X$ if and only if every open cover of $Y$ by open sets in $X$ contains a finite subcover of $Y$ .

Proof. Suppose that $Y$ is compact and $\{O_i\}_{i\in I}$ is a cover of $Y$ by sets open in $X$ . Then the collection $\{Y \cap O_i | i\in J\}$ is a covering of $Y$ by sets open in $Y$ . Due to the assumption that $Y$ is compact in $X$ , there exists a finite subcover

$\begin{align*} O_{i_1}', \ldots, O_{i_n}' \end{align*}$

in $Y$ . For each $i\in I$ chose a set $O_i':=O_i \cap Y$ . Then

$\begin{align*} Y \cap (\bigcap_{k=1}{n}{O_{i_k}} = (Y \cap O_{i_1}), \ldots, (Y \cap O_{i_n}) \end{align*}$

covers $Y$ .

Suppose that every open cover of $Y$ by sets open in $X$ contains a finite open subcover of $Y$ . We would like to show that $Y$ is compact in $X$ . Let $\{O_i'\}_{i\in I}$ be an open cover of $Y$ by sets open in $X$ (and thus in $Y$ ). By hypothesis, some finite subcollection $O_{i_1}', \ldots, O_{i_n}'$ exists that covers $Y$ . Then, by definition of the subspace topology, for each $O_{i}'$ , $i\in I$ we can chose an open set $O_i$ via $O_i'=Y \cap O_i$ . It follows that $O_{i_1}, \ldots, O_{i_n}$ is a finite subcover of $\{O_i\}_{i\in I}$ in $X$ . Hence, $Y$ is compact in $X$ .

$\square$

A closed subspace is a subspace $Y$ , that when treated as a subset of the original space $X$ is a closed set in the original topology $\mathcal{T}$ .

Theorem 4.1 (Compactness & Closed Spaces)
Every closed subspace $Y$ of a compact space $X$ is compact.

Proof. Let $Y$ be a closed subspace of the compact space $X$ . Given a covering $\mathcal{O}$ of $Y$ by sets open in $X$ , let us form an open cover $\mathcal{P}$ by adjoining to $\mathcal{O}$ the single open set $X\setminus Y$ . Note that $Y$ is closed such that its complement needs to be open.

Thus, $\mathcal{P}:= \mathcal{O} \cup (X\setminus Y)$ is an open cover of $X$ . Due to the fact that $X$ is compact, some finite subcollection of $\mathcal{P}$ covers $X$ . If this subcollection contains the set $X \setminus Y$ , discard $X\setminus Y$ . Otherwise, leave the subcollection alone. The resulting collection is a finite cover of $\mathcal{O}$ that covers $Y$ . Hence, $Y$ is compact.

$\square$

Theorem 4.2 (Compactness in Hausdorff Spaces)
Every compact subspace $Y$ of a Hausdorff space $X$ is closed.

Proof. Let $Y$ be a compact subspace of a Hausdorff space $X$ . We are going to prove that $X\setminus Y$ is open, so that $Y$ is closed.

Let $x_0\in X\setminus Y$ . We show there is a neighborhood of $x_0$ that is disjoint from $Y$ . For each point $y\in Y$ , let us choose disjoint neighborhoods $U_y$ and $V_y$ of the points $x_0$ and $y$ , respectively (using the Hausdorff condition). The collection $\{V_y | y\in Y\}$ is a covering of $Y$ by sets open in $X$ . Therefore, finitely many of them $V_{y_1}, \ldots, V_{y_n}$ cover $Y$ (since $Y$ is assumed to be compact). The open set $V=\bigcup_{i=1}^{n}{V_{y_i}}$ contains $Y$ . The open set $U=\bigcap_{i=1}^{n}{U_{y_i}}$ contains all points that are disjoint from any of the open sets $V_{y_i}$ and thus $U\cap V = \emptyset$ . Hence, $U$ is a neighborhood of some arbitrary $x_0\in Y^C$ and thus $Y$ is closed.

$\square$

Example 4.5 (Theorems 4.1 and 4.2)
(a) Once we prove that the interval $[a,b]$ in $\mathcal{R}$ is compact, it follows from Theorem 4.1 that any closed subspace of $[a,b]$ is compact. On the other hand, it follows from Theorem 4.2 that the intervals $(a, b]$ and $(a,b)$ in $\mathbb{R}$ cannot be compact because they are not closed in the Hausdorff space $\mathbb{R}$ .

(b) The Hausdorff condition in Theorem 4.2 is necessary. Consider the finite complement topology on the real line. The only proper subsets of $\mathbb{R}$ that are closed in this topology are the finite sets. But every subset of $\mathbb{R}$ is compact in this topology since you can always find an open finite subcover given any open cover.

(c) The interval $(0,1]$ is not compact in the Euclidean standard topology of $\mathbb{R}$ . The open cover $\mathcal{A}:=\{ \frac{1}{n} | n\in \mathbb{N} \}$ contains no finite sub-collection covering $(0,1]$ .

$\square$

Continuous Functions

Topological spaces have been introduced because they are the natural habitat for continuous functions. These spaces have been built such that the topological structure is respected. Continuous functions therefore take the same role on topological spaces as linear maps within vector spaces.

The notion of continuity is particularly easy to formulate in terms of open (and closed) sets and the following version is called the open set definition of continuity.

Definition 5.1: (Continuous Function on Topological Space)
Let $(X, \mathcal{T})$ and $(Y, \mathcal{S})$ be two topological spaces. A function $f:X \rightarrow Y$ is continuous if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$ .

$\square$

Before we will illustrate this definition let us recall the definitions of an image $f(A)$ and a preimage $f^{-1}(B)$ of a function $f:X \rightarrow Y$ with $A\subseteq X$ , $B\subseteq Y$ :

$\begin{align*} f(A) &= \{f(x) | \ x\in A\} \\ & = \{y\in B| \ (\exists x\in A) f(x)=y\} \\ f^{-1}(B) &= \{ x\in X | \ f(x) \in B\}. \end{align*}$

The condition that $f^{-1}$ be continuous says that for each open set $V$ of $Y$ , the inverse image of $V$ under the map $f^{-1}:f(X) \rightarrow X$ is open in $Y$

Example 5.1: (Simple Continuous Function)
Let $(X=\{a, b, c, d\}, \mathcal{T})$ and $(Y:=\{1,2,3\}, \mathcal{S})$ be two topological spaces defined by

$\begin{align*} \mathcal{T}:=&\{X, \emptyset, \{c\}, \{c,d\}, \{a,b\}, \{a,b,c\} \} \\ \mathcal{S}:=&\{Y, \emptyset, \{1\}, \{2\}, \{1,2\} \} \end{align*}$

Let $f, g, h: X \rightarrow Y$ be three functions defined by

$\begin{align*} f(a)= 1, \ f(b)=1, \ f(c)=2, \ f(d)&=2, \\ g(a)= 2, \ g(b)=2, \ g(c)=1, \ g(d)&=3, \\ h(a)= 1, \ h(b)=2, \ h(c)=2, \ h(d)&=3. \\ \end{align*}$

The functions $f$ is continuous since the pre-image of each open set in $\mathcal{S}$ is an element of $\mathcal{T}$ . Similarly, $g$ is continuous but note that $\{3\}$ is not an open set. The function $h$ , however, is not continuous since $\{2\}$ is open in $Y$ , but $h^{-1}(\{2\})={b, c}$ is not open in $X$ .

Let us now study how the functions map closed sets. By definitions these are the complements of $\mathcal{T}$ and $\mathcal{S}$ , i.e.

$\begin{align*} \mathcal{T}^C:=&\{X, \emptyset, \{a,b,d\}, \{a,b\}, \{c,d\}, \{d\} \} \\ \mathcal{S}^C:=&\{Y, \emptyset, \{1,2,3\}, \{1,3\}, \{3\} \} \end{align*}$

Recognize that the image of a closed set in $X$ under $f$ and $g$ is contained in a closed set in $Y$ . For instance, $\{a,b,d\}\in \mathcal{T}^C$ $\Rightarrow$ $f(a)=f(b)=1$ and $f(d)=2$ and $\{1,2\} \subset \text{Cl}A = \{1,2,3\}\in \mathcal{S}^C$ .

$\square$

The last example made the definition of a continuous function between simple topological spaces rather clear.

Sometimes it is also helpful to study the properties that will not be preserved: a continuous function does not necessarily map open sets to open sets.
For example, the function $f:\mathbb{R} \rightarrow \mathbb{R}$ , given by $f(x)=x^2$ , is continuous, but the image of the open set $(-1,1)$ is $[0,1)$ , which is neither open nor closed. Let us also double-check that in Example 4.1. The function $g$ maps the open set $\{c,d\}$ to $\{1,3\}$ , which is not open.

Now, let us have a look at more general examples.

Example 5.2: (Identity and Constant Function)
(a) The identity function id: $X\rightarrow X$ , given by id $(x)=x$ , is continuous in all topological spaces. If a function $f:X \rightarrow X$ is continuous at $x$ , i.e. if $\{x\}$ is open so is its preimage $\{x\}$ . This argument can be generalized onto subsets $U\subseteq X$ .

(b) The constant function $\hat{c}:X \rightarrow Y$ defined by $\hat{c} (x)=c\in Y$ for every $x\in X$ . Suppose $V$ is open in $Y$ , then $\hat{c}^{-1}(V) = X$ if $c\in V$ , and $\hat{c}^{-1}(V) = \emptyset$ if $c\notin V$ . In either case, the preimage is open in $X$ , and therefore $\hat{c}$ is continuous.

$\square$

Continuous functions preserve proximity as we can see in the next theorem. Also refer to Example 4.1.

Theorem 5.1 (Continuous Functions & Closeness)
Let $f:X\rightarrow Y$ be continuous and assume that $A\subseteq X$ . If $x\in \text{Cl}A$ , then $f(x)\in \text{Cl}(f(A))$ .

Proof. Suppose that $f:X\rightarrow Y$ is continuous, $x\in X$ , and $A \subseteq X$ . We prove that if $f(x)\notin \text{Cl}(f(A))$ , then $x\notin \text{Cl}(A)$ .

Hence, suppose that $f(x)\notin \text{Cl}(A)$ . By Proposition 1.2 there exists an open set $U$ containing $f(x)$ , but not intersecting $f(A)$ . It follows that $f^{-1}(U)$ is an open set containing $x$ that does not intersect $A$ . Thus, $x\notin \text{Cl}A$ , and the result follows.

$\square$

The next theorem translates the well-known $\epsilon$ – $\delta$ definition of continuity with Definition 4.1.

Theorem 5.2 (Continuity & $\epsilon$ – $\delta$ Condition)
A function $f:X\rightarrow Y$ is continuous in the open set definition of continuity if and only if, for every $x\in X$ and every open set $U$ containing $f(x)$ , there exists a neighborhood $V$ of $x$ such that $f(V)\subseteq U$ .

Proof. First, suppose that the open set definition holds for functions $f:X\rightarrow Y$ . Let $x\in X$ and an open set $V\subseteq Y$ containing $f(x)$ be given. Set $U=f^{-1}(V)$ . It follows that $x\in U$ and that $U$ is open in $X$ since $f$ is continuous by the open set definition 4.1. Clearly, $f(U)\subseteq V$ , and therefore we have shown the desired result.

Now assume that for every $x\in X$ and every open set $V$ containing $f(x)$ , there exist a neighborhood $U$ of $x$ such that $f(U) \subseteq V$ . We show that $f^{-1}(W)$ is open in $X$ for every open set $W$ in $Y$ . Hence, let $W$ be an arbitrary open set in $Y$ . To show that $f^{-1}(W)$ is open in $X$ , choose an arbitrary $x\in f^{-1}(W)$ . It follows that $f(x)\in W$ , and therefore exists a neighborhood $U_x$ of $x$ in $X$ such that $f(U_x)\subseteq W$ , or, equivalently, such that $U_x\subseteq f^{-1}(W)$ . Thus, for an arbitrary $x\in f^{-1}(W)$ there exists an open set $U_x$ such that $x\in U_x\subseteq f^{-1}(W)$ . Then, the assertion follows applying Theorem 1.1.

$\square$

Theorem 4.2 generalizes this idea of continuity in metrizable topological spaces to general topological spaces. In a metric space, we can consider an open ball as an open set and therefore as a neighborhood. That is, for each $\epsilon$ -ball $B(f(x_0), \epsilon)$ there need to be a suitable $\delta$ -ball $B(x_0, \delta)$ , such that $f(B(x_0, \delta)) \subseteq B(f(x_0), \epsilon)$ . For further details please refer to this deep-mind.org post and this Wikipedia article.

The second important property that is preserved by continuous functions is the concept of convergence.

Theorem 5.3 (Continuity & Convergent Sequences)
Assume that $f: X \rightarrow Y$ is continuous. If a sequence $(x_1, x_2, \ldots)$ in $X$ converges to a point $x$ , then the sequence $(f(x_1), f(x_2), \ldots)$ in $Y$ converges to $f(x)$ .

Proof. Let $V$ be an arbitrary neighborhood of $f(x)$ in $Y$ . Since $f$ is continuous, $f^{-1}(V)$ is open in $X$ . Furthermore, $f(x)\in V$ implies that $x\in f^{-1}(V)$ . The sequence $(x_1, x_2, \ldots)$ converges to $x$ ; thus, there exists $N\in \mathbb{N}$ such that $x_n\in f^{-1}(V)$ for all $n\geq N$ . It follows that $f(x_n)\in V$ for all $n\geq N$ , and therefore the sequence $(f(x_1), f(x_2), \ldots)$ converges to $f(x)$ .

$\square$

Another important property of continuous functions is directly linked to the actual definition.

Theorem 5.4 (Continuity & Pre-Image of Closed Sets)
Let $X$ and $Y$ be topological spaces. A function $f:X\rightarrow Y$ is continuous if and only if $f^{-1}(B)$ is closed in $X$ for every closed set $B\subseteq Y$ .

Proof. Let $B$ be closed in $Y$ and let $A:=f^{-1}(B)$ . We wish to prove that $A$ is closed in $X$ . We show that Cl $A=A$ . By elementary set theory, we have $f(A)=f(f^{-1}(B))\subseteq B$ . Therefore, if $x\in \text{Cl}A$ then $f(x)\in f(\text{Cl}A)$ $\subseteq \text{Cl}f(A)\subseteq \text{Cl}B=B$ . Hence, $x\in f^{-1}(B)=A$ and Cl $A\subseteq A$ and thus Cl $A=A$ .

$\square$

Theorem 5.5 (Continuous Functions & Compact Sets)
Let $f:X \rightarrow Y$ be continuous, and let $A$ be compact in $X$ . Then $f(A)$ is compact in $Y$ .

Proof. Let $f:X\rightarrow Y$ be continuous, and assume that $A$ is compact in $X$ . To show that $f(A)$ is compact in $Y$ , let $\mathcal{B}$ be a cover of $f(A)$ by open sets in $Y$ . Then $f^{-1}(V)$ is open in $X$ for every set $V\in \mathcal{B}$ . Hence, $\mathcal{A}:=\{ f^{-1}(V) |\ V\in \mathcal{B} \}$ is a cover of $A$ by open sets in $X$ . Since $A$ is compact there is a finite sub-collection of $\mathcal{A}$ that covers $A$ . Thus, $\mathcal{B}$ has a finite sub-cover, implying that $f(A)$ is compact in $Y$ .

$\square$

Appendix

A really nice introduction to the abstract concept of a topology, however, in German language only.

Literature

[1]

[2]

[3]

[4]

[5]