K. Jänich, Topology. Chapter I: Fundamental Concepts. 1: The Concept of a Topological Space

Off we go with some actual mathematics, and the reason why I wanted to make these notes!

1. The Concept of a Topological Space

Definition. A topology on a set \(X\) is a set \(\mathcal{O} \subseteq \mathcal{P}X\) whose members we call open sets, such that:

1. For all \(\mathcal{U} \subseteq \mathcal{O}\) we have \(\bigcup \mathcal{U} \in \mathcal{O}\).
2. For all \(U,V \in \mathcal{O}\) we have \(U \cap V \in \mathcal{O}\).
3. \(X \in \mathcal{O}\)

A topological space is a pair \((X,\mathcal{O})\) such that \(\mathcal{O}\) is a topology on \(X\). We usually say just 'the topological space \(X\)' for 'the topological space \((X,\mathcal{O})\)'.

Definition. In a topological space \(X\):

• a set \(C \subseteq X\) is closed iff \(X - C\) is open;
• a set \(N \subseteq X\) is a neighbourhood of a point \(x \in X\), and \(x\) is an interior point of \(N\), iff there exists an open set \(U\) such that \(x \in U \subseteq N\);
• a point \(x \in X\) is an exterior point of a set \(B \subseteq X\) iff it is an interior point of \(X - B\);
• a point \(x \in X\) is a boundary point of a set \(B \subseteq X\) iff it is neither an interior point nor an exterior point of \(B\);
• the interior \(B^\circ\) of a set \(B \subseteq X\) is the set of all interior points of \(B\);
• the closure \(\overline{B}\) of a set \(B \subseteq X\) is the set of all points that are not exterior points of \(B\). 

These definitions are intended to capture our intuitive notion of closeness. Informally:

• a set \(U\) is open iff, for every point \(x \in U\), all the points close to \(x\) are also in \(U\).
• a set \(C\) is closed iff, for any point \(x\), if all the points close to \(x\) are in \(C\), then \(x \in C\).
• a set \(N\) is a neighbourhood of \(x\), and \(x\) is an interior point of \(N\), iff \(x \in N\) and all the points close to \(x\) are also in \(N\). (Thus a set is open iff it is a neighbourhood of all its points; see Proposition 6 below.)
• a point \(x\) is an exterior point of \(B\) iff \(x \notin B\) and all the points close to \(x\) are also outside \(B\).
• a point \(x\) is a boundary point of \(B\) iff some of the points close to \(x\) are in \(B\) and some of the points close to \(x\) are outside \(B\)
  

Some basic consequences of these definitions: 

Proposition 1. \(\emptyset\) is open.

Proof. Since \( \emptyset = \bigcup \emptyset \) is the union of a set of open sets. \(\Box\)

Proposition 2. The intersection of any nonempty set of closed sets is closed.

Proof. Let \(\mathcal{C}\) be a nonempty set of closed sets. Then \(X - \bigcap \mathcal{C} = \bigcup \{X - C : C \in \mathcal{C} \}\) is the union of a set of open sets, hence open. Therefore \(\bigcap \mathcal{C}\) is closed. \(\Box\)

Proposition 3. The union of two closed sets is closed.

Proof. Let \(C\) and \(D\) be closed sets. Then \(X - (C \cup D) = (X - C) \cap (X - D)\) is open, hence \(C \cup D\) is closed. \(\Box\)

Proposition 4. The interior of \(B\) is the union of all the open sets that are subsets of \(B\).

Proof. For any point \(x\), we have \(x\) is in the interior of \(B\) iff \(x\) is an interior point of \(B\), iff \(x\) is a member of some open set that is a subset of \(B\). \(\Box\)

Proposition 5. The closure of \(B\) is the intersection of all the closed sets that include \(B\).

Proof. For any point \(x\), we have:

\(x\) is in the closure of \(B\)

iff \(x\) is not an exterior point of \(B\)

iff \(x\) is not an interior point of \(X-B\)

iff, for every open set \(U\), if \(U \subseteq X - B\) then \(x \notin U\)

iff, for every open set \(U\), if \(B \subseteq X - U\) then \(x \in X - U\)

iff, for every closed set \(C\), if \(B \subseteq C\) then \(x \in C\). \(\Box\)

Proposition 6. For any set \(U \subseteq X\), the following are equivalent:2

1. \(U\) is open.
2. \(X-U\) is closed.
3. \(U\) is a neighbourhood of each of its points.
4. \(X-U = \overline{X - U} \)

Proof

\(1 \Leftrightarrow 2\): From the definition of closed set, since \( X - (X -U) = U \).

\(1 \Rightarrow 3 \): If \(U\) is open and \(x \in U\), then we have \(U\) is an open set and \(x \in U \subseteq U \), hence \(U\) is a neighbourhood of \(x\).

\(3 \Rightarrow 1 \): If, for all \(x \in U\), there exists an open set \(V \subseteq U\) such that \(x \in V\), then \(U\) is the union of all the open sets \(V\) such that \(V \subseteq U\), and so \(U\) is open.

\(2 \Rightarrow 4\): If \(X - U\) is closed then \(X - U\) is the intersection of all the closed sets that include \(X - U\), so \(X - U = \overline{X - U}\) by Proposition 2.

\(4 \Rightarrow 2\): For any set \(B\) we have \(\overline{B}\) is closed by Proposition 1. Hence if \(X-U = \overline{X-U}\) then \(X-U\) is closed. \(\Box\)

Following Bourbaki, we defined a topology in terms of open sets. We could have chosen to define topologies in terms of closed sets, neighbourhoods, or closures instead:

Proposition 7. Let \(X\) be a set and \(\mathcal{C} \subseteq \mathcal{P} X\). Then there exists a topology on \(X\) with respect to which \(\mathcal{C}\) is the set of closed sets if and only if the following all hold:

1. For all nonempty \(\mathcal{D} \subseteq \mathcal{C}\) we have \(\bigcap \mathcal{D} \in \mathcal{C}\).
2. \(X \in \mathcal{C}\)
   
3. For any \(C,D \in \mathcal{C}\) we have \(C \cup D \in \mathcal{C}\).
4. \( \emptyset \in \mathcal{C} \)

In this case, the topology is unique.

Proof.  Given any topology on \(X\), we have:

1. The intersection of any nonempty set of closed sets is closed by Proposition 1.
2. \(X-X = \emptyset\) is open so \(X\) is closed.
3. The union of two closed sets is closed by Proposition 2.
4. \(X - \emptyset = X\) is open so \(\emptyset\) is closed.

Conversely, assume \(\mathcal{C}\) is a set that satisfies 1-4. Define \(\mathcal{O} = \{ X - C : C \in \mathcal{C} \}\). We show that \(\mathcal{O}\) is a topology:

1. Let \(\mathcal{U} \subseteq \mathcal{O}\). If \(\mathcal{U}\) is empty then we have \(X - \bigcup \mathcal{U} = X \in \mathcal{C}\) by 2. Otherwise we have \(X - \bigcup \mathcal{U} = \bigcap \{ X - U : U \in \mathcal{U} \} \in \mathcal{C} \) by 1. Hence \( \bigcup \mathcal{U} \in \mathcal{O} \) in either case.
   
2. Let \(U,V \in \mathcal{O}\). Then \(X - (U \cap V) = (X - U) \cup (X - V) \in \mathcal{C} \) by 3. Hence \(U \cap V \in \mathcal{O}\).
3. \(X - X = \emptyset \in \mathcal{C}\) by 4 and so \(X \in \mathcal{O}\).

With respect to this topology, a set \(C\) is closed iff \(X - C \in \mathcal{O}\) iff \(X - (X - C) \in \mathcal{C}\) iff \(C \in \mathcal{C}\). Thus, \(\mathcal{C}\) is the set of closed sets.

If \(\mathcal{O}'\) is any other topology with respect to which \(\mathcal{C}\) is the set of closed sets, then for any \(U\) we have \(U \in \mathcal{O}'\) iff \(X-U \in \mathcal{C}\) iff \(U \in \mathcal{O}\), and so \(\mathcal{O}' = \mathcal{O}\). This proves \(\mathcal{O}\) is unique. \(\Box\)

Proposition 8. Let \(X\) be a set and let \(\mathcal{N} : X \rightarrow \mathcal{P} X\). Then there exists a topology on \(X\) with respect to which \(\mathcal{N}(x)\) is the set of neighbourhoods of \(x\) for any point \(x\) if and only if the following all hold:

1. For all \(x \in X\) and \(N \in \mathcal{N}(x)\) we have \(x \in N\).
2. For all \(x \in X\) we have \(X \in \mathcal{N}(x)\).
3. For all \(x \in X\), \(M \in \mathcal{N}(x)\) and \(N \supseteq M\) we have \(N \in \mathcal{N}(x)\).
4. For all \(x \in X\) and \(M,N \in \mathcal{N}(x)\) we have \(M \cap N \in \mathcal{N}(x)\).
5. For all \(x \in X\) and \(N \in \mathcal{N}(x)\), there exists \(U \in \mathcal{N}(x)\) such that \(U \subseteq N\) and \( \forall y \in U. U \in \mathcal{N}(y) \).

In this case, the topology \(\mathcal{O}\) is unique.

Proof. Take any topology on \(X\) and let \(\mathcal{N}(x)\) be the set of neighbourhoods of \(x\) for all \(x \in X\). We prove 1-5 hold.

1. By definition, if \(N\) is a neighbourhood of \(x\) then \(x \in N\).
2. For any \(x\), we have that \(X\) is an open set and \(x \in X \subseteq X\). Thus, \(X\) is a neighbourhood of \(x\).
3. Suppose \(M\) is a neighbourhood of \(x\) and \(M \subseteq N\). Pick an open set \(U\) such that \(x \in U \subseteq M\). Then \(x \in U \subseteq N\), and so \(N\) is a neighbourhood of \(x\).
4. Let \(M\) and \(N\) be neighbourhoods of \(x\). Pick open sets \(U\) and \(V\) such that \(x \in U \subeseteq M\) and \(x \in V \subseteq N\). Then \(U \cap V\) is open and \(x \in U \cap V \subseteq M \cap N\). Thus, \(M \cap N\) is a neighbourhood of \(x\).
5. Let \(N\) be a neighbourhood of \(x\). Pick an open set \(U\) such that \(x \in U \subseteq N \). Then \(U\) is a neighbourhood of each of its points, including \(x\), by Proposition 6.

Conversely, let \(\mathcal{N} : X \rightarrow \mathcal{P} X\) and assume 1-5 hold. Let \(\mathcal{O} = \{U \in \mathcal{P} X : \forall x \in U. U \in \mathcal{N}(x) \}\). We prove \(\mathcal{O}\) is a topology.

1. Let \(\mathcal{U} \subseteq \mathcal{O}\). Let \(x \in \bigcup \mathcal{U}\); we prove \(\bigcup \mathcal{U} \in \mathcal{N}(x)\). Pick \(U \in \mathcal{U}\) such that \(x \in U\). Then \(U \in \mathcal{N}(x)\), and so \(\bigcup \mathcal{U} \in \mathcal{N}(x)\) by 3.
2. Let \(U,V \in \mathcal{O}\). Let \(x \in U \cap V\); we prove \(U \cap V \in \mathcal{N}(x)\). We have \(U,V \in \mathcal{N}(x)\) and so \(U \cap V \in \mathcal{N}(x)\) by 4.
3. For all \(x \in X\) we have \(X \in \mathcal{N}(x)\) by 2.

Now, let \(x \in X\) and \(N \subseteq X\). We prove that \(N\) is a neighbourhood of \(x\) (with respect to \(\mathcal{O}\)) if and only if \(N \in \mathcal{N}(x)\).

Assume \(N\) is a neighbourhood of \(x\). Pick \(U \in \mathcal{O}\) such that \(x \in U \subseteq N\). Then we have \(U \in \mathcal{N}(x)\) and so \(N \in \mathcal{N}(x)\) by 3.

Conversely, assume \(N \in \mathcal{N}(x)\). By 5, pick \(U \in \mathcal{N}(x)\) such that \( \forall y \in U. U \in \mathcal{N}(y) \). Then \(U \in \mathcal{O}\) and \(x \in U \subseteq N\) (using 1), so \(N\) is a neighbourhood of \(x\). 

Finally, let \(\mathcal{O}'\) be any topology with respect to which \(\mathcal{N}(x)\) is the set of neighbourhoods of \(x\) for all \(x\). Then \(U \in \mathcal{O}'\) iff \(\forall x \in U. U \in \mathcal{N}(x)\) by Proposition 6, and so \( \mathcal{O}' = \mathcal{O} \). \(\Box\)

Proposition 9. Let \(X\) be a set and \( \overline{\ } : \mathcal{P} X \rightarrow \mathcal{P} X \). Then there exists a topology on \(X\) with respect to which \(\overline{A}\) is the closure of \(A\) for all \(A \subseteq \mathcal{P} X\) if and only if the following all hold, called the Kuratowski closure axioms:

1. \( \overline{\emptyset} = \emptyset \)
2. For all \(A\) we have \(A \subseteq \overline{A}\)
3. For all \(A\) we have \(\overline{\overline{A}} = \overline{A}\)
4. For all \(A\), \(B\) we have \( \overline{A \cup B} = \overline{A} \cup \overline{B} \)

In this case, the topology is unique.

Proof. Given any topology on \(X\), let \( \overline{A} \) be the closure of \(A\) for all \(A\). We prove 1-4 hold.

1. Since \(\emptyset\) is closed (Proposition 7), we have \(\overline{\emptyset} = \emptyset\).
2. The closure \(\overline{A}\) is the intersection of all the closed sets that include \(A\), and so includes \(A\).
3. The closure \(\overline{A}\) is the intersection of a set of closed sets, hence closed (Proposition 2), and so \(\overline{\overline{A}} = \overline{A}\) by Proposition 6.
4. We have \(A \subseteq \overline{A}\) and \(B \subseteq \overline{B}\), hence \(A \cup B \subseteq \overline{A} \cup \overline{B}\). Thus \( \overline{A} \cup \overline{B}\) is a closed set that includes \(A \cup B\), and so \( \overline{A \cup B} \subseteq \overline{A} \cup \overline{B} \).
   Conversely, we have \(A \subseteq A \cup B\) and so \(\overline{A} \subseteq \overline{A \cup B}\) (because every closed set that includes \(A \cup B\) includes \(A\)). Similarly \(\overline{B} \subseteq \overline{A \cup B}\), and so \(\overline{A} \cup \overline{B} \subseteq \overline{A \cup B} \).

Conversely, let \( \overline{\ } : \mathcal{P} X \rightarrow \mathcal{P} X\) be a function that satisfies 1-4. We first show that, if \(A \subseteq B\), then \(\overline{A} \subseteq \overline{B}\). This holds because \(B = A \cup B\) and so \(\overline{B} = \overline{A} \cup \overline{B}\) by 4, hence \(\overline{A} \subseteq \overline{B} \).

Now, let \( \mathcal{C} = \{ A \in \mathcal{P} X : \overline{A} = A \} \). We prove that \(\mathcal{C}\) satisfies the conditions in Proposition 7.

1. Let \( \mathcal{D} \subseteq \mathcal{C} \) be nonempty. We have \( \bigcap \mathcal{D} \subseteq \overline{\bigcap \mathcal{D}} \) by 2. Conversely, for any \(D \in \mathcal{D}\) we have \(\bigcap \mathcal{D} \subseteq D\), and so \(\overline{\bigcap \mathcal{D}} \subseteq \overline{D} = D \). Thus, \( \overline{\bigcap \mathcal{D}} \subseteq \bigcap \mathcal{D} \), and so \( \overline{\bigcap \mathcal{D}} = \bigcap \mathcal{D} \) as required.
2. \( \overline{X} = X \) by 2.
3. Let \(C, D \in \mathcal{C} \). Then \(\overline{C \cup D} = \overline{C} \cup \overline{D} = C \cup D \) and so \( C \cup D \in \mathcal{C} \).
4. \( \overline{\emptyset} = \emptyset \) by 2.

Hence there is a unique topology for which \( \mathcal{C} \) is the set of closed sets. With respect to this topology, we prove \( \overline{A} \) is the closure of \(A\).

If \(C\) is a closed set and \(A \subseteq C\) then we have \( \overline{A} \subseteq \overline{C} = C \). Conversely, \( \overline{A} \) is a closed set (by 3) that includes \(A\) (by 2). Thus \( \overline{A} \) is the intersection of all the closed sets that include \(A\).

Finally, if there is any other topology for which \( \overline{A} \) is the closure of \(A\) for every \(A\), then we have that \(A\) is closed if and only if \( \overline{A} = A \) and so \(\mathcal{C}\) is the set of closed sets. Thus, the topology is unique. \(\Box\)

Definition 10 Let \(\mathcal{O}\) and \(\mathcal{O}'\) be two topologies on the same set \(X\). We say that \(\mathcal{O}\) is coarser than \(\mathcal{O}'\), and \(\mathcal{O}'\) is finer than \(\mathcal{O}\), iff \( \mathcal{O} \subseteq \mathcal{O}' \).

Example 11 On any set \(X\), we have that \(\{\emptyset, X \}\) is a topology on \(X\), called the trivial or indiscrete topology. It is the coarsest topology on \(X\).

Example 12 On any set \(X\), we have that \(\mathcal{P} X \) is a topology on \(X\), called the discrete topology. It is the finest topology on \(X\).