Topological Spaces

Star Issue Watch Follow

In the previous chapter we saw that in a metric space, continuity of functions is only indirectly determined by the metric itself. Instead, the structure that determines continuity is the collection open sets under the metric. This motivates the definition of a topological space, which abstracts the notion of open sets from metric spaces.

Topologies

First, we define topological spaces. These are sets equipped with a topology, a collection of subsets which we define to be open. Unlike in metric spaces, where we first defined open balls and then used them to define open sets, here we define open sets directly, and require they satisfy certain properties.

Definition 80 (Topological space)

A topological space is a set \(X,\) called the space, together with a collection \(\mathcal{U} \subseteq \mathcal{P}(X)\) of subsets of \(X,\) called the topology on \(X,\) such that

1. \(\emptyset, X \in \mathcal{U},\)

2. If \({U_i}_{i \in I} \subseteq \mathcal{U},\) then \(\bigcup_{i \in I} U_i \in \mathcal{U},\)

3. If \(U_1, \dots, U_n \in \mathcal{U},\) then \(\bigcap_{i=1}^n U_i \in \mathcal{U}.\)

The elements of \(X\) are called points, and the elements of \(\mathcal{U}\) are called open sets.

When working with specific spaces, they will often be already be equipped with a metric. We refer to the topology associated with a given metric as the induced topology.

Definition 81 (Induced topology)

Let \((X, d)\) be a metric space. Then, the topology induced by \(d\) is the set of all open sets in \(X\) with respect to the metric \(d.\)

We now also re-define continuity in terms of open sets.

Definition 82 (Continuous function)

Let \(f: X \to Y\) be a function between topological spaces. Then, \(f\) is continuous if for every open set \(U \subseteq Y,\) the pre-image \(f^{-1}(U)\) is an open set in \(X.\)

Lemma 15 (Composition preserves continuity)

If \(f: X \to Y\) and \(g: Y \to Z\) are continuous functions between topological spaces, then the composition \(g \circ f: X \to Z\) is continuous.

In topology, we are interested in studying the properties of spaces that are preserved under continuous deformations. Therefore, from a topology perspective, two spaces are considered essentially the same up to a continuous bijection. This is captured by the notion of homeomorphism.

Definition 83 (Homeomorphism)

A function \(f: X \to Y\) between topological spaces is a homeomorphism if it is bijective, continuous, and its inverse \(f^{-1}\) is also continuous. Equivalently, \(f\) is a homeomorphism if \(f\) is a bijection and \(U \subseteq X\) is open if and only if \(f(U) \subseteq Y\) is open. We say two spaces are homeomorphic if there exists a homeomorphism between them.

Lemma 16 (Homeomorphism is an equivalence relation)

Homeomorphism is an equivalence relation between topological spaces.

Proof: Homeomorphism is an equivalence relation

Reflexivity: The identity map \(I_X: X \to X\) is a homeomorphism, because it is bijective, continuous, and its inverse is itself. Therefore \(X \equiv X.\)

Symmetry: If \(f: X \to Y\) is a homeomorphism, then \(f^{-1}: Y \to X\) is also a homeomorphism. Therefore \(X \equiv Y\) implies \(Y \equiv X.\)

Transitivity: If \(f: X \to Y\) and \(g: Y \to Z\) are homeomorphisms, then \(g \circ f: X \to Z\) is a homeomorphism. Therefore \(X \equiv Y\) and \(Y \equiv Z\) implies \(X \equiv Z.\)

In general, the approach for showing that two spaces are homeomorphic is to find a homeomorphism between them. However, showing that two spaces are not homeomorphic is more difficult. In particular, there is no simple recipe for showing that two spaces are not homeomorphic. Instead, we resort to certain topological properties that are preserved under homeomorphisms. Whenever two spaces have different such properties, we can conclude that they are not homeomorphic. Two such properties are connectedness and compactness. In the remainder of this chapter we give definitions and results building up to these properties.

Sequences

We now turn to re-defining concepts from metric spaces in terms of topological spaces, starting with sequences. First we re-define the following shorthand for open sets.

Definition 84 (Open neighbourhood)

An open neighbourhood of a point \(x \in X\) in a topological space \((X, \mathcal{U})\) is an open set \(U \in \mathcal{U}\) such that \(x \in U.\)

In topological spaces, convergent sequences are defined directly in terms of open neighbourhoods, rather than using open balls.

Definition 85 (Convergent sequence)

A sequence \(x_n \to x\) if for every open neighbourhood \(U\) of \(x,\) there exists \(N \in \mathbb{N}\) such that \(x_n \in U\) for all \(n > N.\)

We now turn to uniqueness of limits. In general, in a topological space limits need not be unique. For example, given a set \(X\) with the coarse topology \(\mathcal{U} = \{\emptyset, X\},\) every sequence converges to every point. However, further assumptions on the topology can result into unique limits.

Definition 86 (Hausdorff space)

A topological space \((X, \mathcal{U})\) is Hausdorff if for every pair of distinct points \(x_1, x_2 \in X,\) there exist open neighbourhoods \(U_1, U_2\) of \(x_1, x_2\) respectively such that \(U_1 \cap U_2 = \emptyset.\)

Earlier, we proved that limits in metric spaces are unique. The property we used in that proof was that, in a metric space, open balls centered around distinct points are disjoint if their radii are small enough. This was the Hausdorff property in disguise. Metric spaces are always Hausdorff, and therefore have unique limits.

Lemma 17 (Limits are unique in Hausdorff spaces)

If \(X\) is Hausdorff and \((x_n)\) is a sequence in \(X\) such that \(x_n \to x\) and \(x_n \to x',\) then \(x = x'.\)

Proof: Limits are unique in Hausdorff spaces

Let \((x_n)\) be a sequence in \(X\) such that \(x_n \to x\) and \(x_n \to x'.\) Suppose \(x \neq x'.\) Since \(X\) is Hausdorff, there exist open neighbourhoods \(U, U'\) of \(x, x'\) respectively such that \(U \cap U' = \emptyset.\) Since \(x_n \to x,\) there exists \(N \in \mathbb{N}\) such that \(x_n \in U\) for all \(n > N.\) Similarly, since \(x_n \to x',\) there exists \(N' \in \mathbb{N}\) such that \(x_n \in U'\) for all \(n > N'.\) Then, for all \(n > \max(N, N'),\) we have \(x_n \in U \cap U' = \emptyset,\) which is a contradiction. Therefore, \(x = x'.\)

We now revisit closed sets and limit points, this time in topological spaces.

Definition 87 (Closed set)

A set \(C \subseteq X\) in a topological space \((X, \mathcal{U})\) is closed if its complement \(X \setminus C\) is open.

We now prove some properties of closed sets.

Lemma 18 (Properties of closed sets)

Let \((X, \mathcal{U})\) be a topological space. Then,

1. \(\emptyset\) and \(X\) are closed,

2. If \(C_1, \dots, C_n\) are closed, then \(\bigcup_{i=1}^n C_i\) is closed,

3. If \(\{C_i\}_{i \in I}\) are closed, then \(\bigcap_{i \in I} C_i\) is closed.

Proof: Properties of closed sets

Property 1: Since \(\emptyset\) is open, its complement \(X\) is closed. Similarly, since \(X\) is open, its complement \(\emptyset\) is closed.

Property 2: If \(C_1, \dots, C_n\) are closed, then their complements \(X \setminus C_1, \dots, X \setminus C_n\) are open. Therefore, \(\bigcup_{i = 1}^n (X \setminus C_i) = X \setminus \bigcap_{i = 1}^n C_i\) is open, so \(\bigcap_{i = 1}^n C_i\) is closed.

Property 3: If \({C_i}_{i \in I}\) are closed, then their complements \({X \setminus C_i}_{i \in I}\) are open. Therefore, \(\bigcap_{i \in I} (X \setminus C_i) = X \setminus \bigcup_{i \in I} C_i\) is open, so \(\bigcup_{i \in I} C_i\) is closed.

Note the similarity of the properties above to the defining properties of open sets. In fact, we could have defined topologies in terms of closed sets rather than open sets.

Lemma 19 (Singleton set in Hausdorff space is closed)

If \(X\) is Hausdorff and \(x \in X,\) then \(\{x\}\) is closed.

Proof: Singleton set in Hausdorff space is closed

Proof 1: Let \(X\) be Hausdorff and \(x \in X.\) We want to show that \(\{x\}\) is closed, which we can do by showing that \(\{x\}\) is an intersection of closed sets. Let \(x' \in X \setminus \{x\}.\) Since \(X\) is Hausdorff, there exist open neighbourhoods \(U_x, U_{x'}\) of \(x, x'\) respectively such that \(U_x \cap U_{x'} = \emptyset.\) Then, \(X \setminus U_{x'}\) is closed, and \(\{x\} = X \setminus U_{x'}.\) Finally, we have \(\{x\} = \bigcap_{x' \in X \setminus \{x\}} (X \setminus U_{x'}),\) so \(\{x\}\) is closed.

Proof 2: Let \(X\) be Hausdorff and \(x \in X.\) We want to show that \(X \setminus \{x\}\) is open. Let \(x' \in X \setminus \{x\}.\) Then \(x \neq x'.\) Since \(X\) is Hausdorff and \(x \neq x',\) there exist open neighbourhoods \(U, U'\) of \(x, x'\) respectively such that \(U \cap U' = \emptyset.\) Then, \(U' \subseteq X \setminus \{x\},\) so \(X \setminus \{x\}\) is open.

Closure, interior and limit points

Definition 88 (Closure)

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U})\). Define

\[\mathcal{C}_A = \{C \subseteq X \mid A \subseteq C \text{ and } C \text{ is closed}\}.\]

Then, the closure of \(A,\) written \(\overline{A},\) is defned as

\[\overline{A} = \bigcap_{C \in \mathcal{C}_A} C.\]

Lemma 20 (Closure of a set is the smallest closed set containing it)

Given a set \(A \subseteq X\) in a topological space \((X, \mathcal{U}),\) its closure \(\overline{A}\) is the smallest closed set containing \(A.\)

Proof: Closure of a set is the smallest closed set containing it

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) We want to show that \(\overline{A}\) is the smallest closed set containing \(A.\) First, \(\overline{A}\) is an intersection of closed sets containing \(A,\) so it contains \(A\) and it is closed. Let \(C\) be a closed set containing \(A.\) Then \(C \in \mathcal{C}_A,\) so \(\overline{A} \subseteq C.\) We conclude, \(\overline{A}\) is the smallest closed set containing \(A.\)

The standard definition of limit point given here requires that the open neighbourhood includes a point different from the limit point itself. This is the standard definition of limit points, see e.g. [Munkres, 2014]. Dexter Chua’s notes for the Cambridge course on Metric and Topological Spaces, which I have been following, use a different definition of limit points, namely: a point in \(X\) is a limit point of \(A,\) if there exists a sequence \((x_n)\) in \(A\) such that \(x_n \to x.\) Under this definition, all elements of \(A\) are limit points of \(A,\) which does not match the standard defintion. I have chosen to follow the standard definition, ammending the results provided in the notes as needed.

Definition 89 (Limit point)

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) Then \(x \in X\) is a limit point of \(A\) if every open neighbourhood of \(x\) contains a point of \(A\) different from \(x.\)

Lemma 21 (Set is closed if and only if it contains its limit points)

Given a set \(A \subseteq X\) in a topological space \((X, \mathcal{U}),\) the set \(A\) is closed if and only if \(A\) contains all its limit points.

Proof: Set is closed if and only if it contains its limit points

First, suppose \(A\) is closed. We will show it contains all its limit points. Let \(x\) be a limit point of \(A.\) Suppose \(x \notin A.\) Since \(A\) is closed, \(X \setminus A\) is open, so there exists an open neighbourhood \(U\) of \(x\) such that \(U \cap A = \emptyset.\) However, since \(x\) is a limit point of \(A,\) every open neighbourhood of \(x\) contains a point of \(A\) different from \(x,\) which contradicts \(U \cap A = \emptyset.\)

Next, suppose \(A\) contains all its limit points. We will show it is closed. Suppose \(x \in X \setminus A.\) This means \(x\) is not a limit point of \(A,\) so there exists an open neighbourhood \(U\) of \(x\) such that \(U \cap A\) contains no points other than, possibly, \(x.\) Therefore, \(U \subseteq X \setminus A,\) so \(X \setminus A\) is open, and \(A\) is closed.

Definition 90 (Dense subset)

A subset \(A \subseteq X\) of a topological space \((X, \mathcal{U})\) is dense in \(X\) if \(\overline{A} = X.\)

Definition 91 (Interior)

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) Define

\[\mathcal{O}_A = \{U \subseteq X \mid U \subseteq A \text{ and } U \text{ is open}\}.\]

Then, the interior of \(A,\) written \(\text{Int}(A),\) is defined as

\[\text{Int}(A) = \bigcup_{U \in \mathcal{O}_A} U.\]

Lemma 22 (Interior of a set is the largest open set contained in it)

Given a set \(A \subseteq X\) in a topological space \((X, \mathcal{U}),\) its interior \(\text{Int}(A)\) is the largest open set contained in \(A.\)

Proof: Interior of a set is the largest open set contained in it

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) Then, since \(\text{Int}(A)\) is a union of open sets contained in \(A,\) it is open and contained in \(A.\) Suppose \(U\) is an open set contained in \(A.\) Then \(U \in \mathcal{O}_A,\) so \(U \subseteq \text{Int}(A).\) Therefore, \(\text{Int}(A)\) is the largest open set contained in \(A.\)

Lemma 23 (Complement of interior is closure of complement)

Given a set \(A \subseteq X\) in a topological space \((X, \mathcal{U}),\) we have

\[X \setminus \text{Int}(A) = \overline{X \setminus A}.\]

Proof: Complement of interior is closure of complement

Let \(A \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) We have that

\[X \setminus \text{Int}(A) = X \setminus \bigcup_{U \in \mathcal{O}_A} U = \bigcap_{U \in \mathcal{O}_A} (X \setminus U) = \bigcap_{C \in \mathcal{C}_{X \setminus A}} C = \overline{X \setminus A}.\]

New topologies from old

Now we look into how existing topological spaces can be used to define new ones.

Subspace topology

First, we define the subspace topology.

Definition 92 (Subspace topology)

Let \((X, \mathcal{U})\) be a topological space, and let \(Y \subseteq X\) be a subset. Then, the subspace topology on \(Y\) is the collection of subsets of \(Y\) given by: \(V\) is an open set in \(Y\) if there exists an open set \(U \in \mathcal{U}\) such that \(V = U \cap Y.\)

Lemma 24 (Subspace topology is a topology)

The subspace topology on a subset \(Y \subseteq X\) of a topological space \((X, \mathcal{U})\) is a topology.

Proof: Subspace topology is a topology

Let \((X, \mathcal{U})\) be a topological space, and let \(Y \subseteq X\) be a subset. We want to show that the subspace topology on \(Y\) is a topology. Let \(\mathcal{U}_Y\) be the subspace topology on \(Y.\) We show each of the properties of a topology are satisfied by the subspace topology.

First, \(\emptyset = \emptyset \cap Y\) and \(Y = X \cap Y,\) so \(\emptyset, Y \in \mathcal{U}_Y.\) Second, suppose \(V_\alpha \in \mathcal{U}_Y\) for all \(\alpha \in A.\) Then, there exist open sets \(U_\alpha \in \mathcal{U}\) such that \(V_\alpha = U_\alpha \cap Y.\) Then

\[\bigcup_{\alpha \in I} V_\alpha = \bigcup_{\alpha \in I} (U_\alpha \cap Y) = (\bigcup_{\alpha \in I} U_\alpha) \cap Y \in \mathcal{U}_Y.\]

Finally, suppose \(V_1, \dots, V_n \in \mathcal{U}_Y.\) Then, there exist open sets \(U_1, \dots, U_n \in \mathcal{U}\) such that \(V_i = U_i \cap Y.\) Then

\[\bigcap_{i=1}^n V_i = \bigcap_{i=1}^n (U_i \cap Y) = (\bigcap_{i=1}^n U_i) \cap Y \in \mathcal{U}_Y.\]

Putting these three parts together, we conclude that the subspace topology on \(Y\) is a topology.

Definition 93 (Inclusion function)

Let \(Y \subseteq X\) be a subset of a topological space \((X, \mathcal{U})\). Then, the inclusion function \(\iota: Y \to X\) is the function defined by \(\iota(y) = y\) for all \(y \in Y.\)

Lemma 25 (Condition for continuity in subspace topology)

Let \(Y \subseteq X\) be a subset of a topological space \((X, \mathcal{U})\). Then, a function \(f: Z \to Y\) is continuous if and only if the composition of \(f\) with the inclusion function \(\iota: Y \to X,\) is also continuous.

Proof: Condition for continuity in subspace topology

Let \(Y \subseteq X\) be a subset of a topological space \((X, \mathcal{U}).\) Let \(f: Y \to Z\) be a function and \(\iota: Y \to X\) be the inclusion function.

To show the first part, suppose \(f\) is continuous. Note that the inclusion function \(\iota\) is continuous, and this follows directly from the definition of the subspace topology. In particular, if \(U\) is open in \(X,\) then \(U \cap Y\) is open in the subspace topology on \(Y,\) and since \(\iota^{-1}(U) = U \cap Y,\) it follows that \(\iota\) is continuous. Since \(f\) is continuous, the composition \(f \circ \iota: Y \to Z\) is continuous.

Conversely, suppose \(\iota \circ f\) is continuous and let \(U \subseteq X\) be open in \(X.\) Note that \(\iota^{-1}(U) = U \cap Y\) which is open in the subspace topology on \(Y.\) Because \(\iota \circ f\) is continuous, it follows that \((\iota \circ f)^{-1}(U) = f^{-1}(U \cap Y)\) is open in \(Z.\) Therefore, since any open set in the subspace topology on \(Y\) is of the form \(U \cap Y\) for some open set \(U\) in \(X,\) it follows that \(f\) is continuous.

Product topology

Now we turn to defining product topologies. Before doing so, we define products of sets and projection functions. These may be familiar from other contexts, and are included for completeness.

Definition 94 (Product of sets)

If \(X\) and \(Y\) are sets, then the product \(X \times Y\) is the set of all ordered pairs \((x, y)\) with \(x \in X\) and \(y \in Y,\) that is

\[X \times Y = \{(x, y) \mid x \in X, y \in Y\}.\]

Definition 95 (Projection function)

Given sets \(X\) and \(Y,\) the projection functions \(\pi_1 : X \times Y \to X\) and \(\pi_2 : X \times Y \to Y\) are defined by

\[\pi_1(x, y) = x \quad \text{and} \quad \pi_2(x, y) = y.\]

Now we are ready to define the product topology.

Definition 96 (Product topology)

Let \((X, \mathcal{U})\) and \((Y, \mathcal{V})\) be topological spaces. Then, the product topology on \(X \times Y\) is the collection of subsets of \(X \times Y\) given by: \(U\) is open in \(X \times Y\) if for every \((x, y) \in U,\) there exist open sets \(U_x \in \mathcal{U}\) and \(U_y \in \mathcal{V}\) such that \(x \in U_x,\) \(y \in U_y,\) and \(U_x \times U_y \subseteq U.\)

The product topology can therefore be thought as a collection in which each open set is a union of products of open sets that is contained in the set. In particular, \(U\) is open in \(X \times Y\) then

\[U = \bigcup_{(x, y) \in U} U_x \times U_y,\]

where \(U_x \times U_y \subseteq U\) for all \((x, y) \in U.\) Conversely, if \(U\) can be expressed in the form of the union above, it is open. In this sense, we can think of the sets over which we are taking the unions as a basis that generates the topology.

Quotient topology

Finally, we define the quotient topology. First, we define quotients and projections onto equivalence classes for completeness.

Definition 97 (Equivalence classes and quotient)

Let \(X\) be a set and \(\sim\) be an equivalence relation on \(X.\) Then, the quotient of \(X / \sim\) is the set of equivalence classes of \(\sim.\) The projection \(\pi: X \to X / \sim\) is defined by \(\pi(x) = [x],\) where \([x]\) is the equivalence class of \(x \in X.\)

Now we define the quotient topology. The quotient topology can be thought of as a topology on the shape that results when we identify, i.e. glue together, a subset of its points.

Definition 98 (Quotient topology)

Let \((X, \mathcal{U})\) be a topological space, and let \(\sim\) be an equivalence relation on \(X.\) The quotient topology on \(X / \sim\) is the set of all subsets \(U \subseteq X / \sim\) such that \(\pi^{-1}(U)\) is open in \(X.\)

[Mun14]

James Munkres. Topology. Pearson, 2nd edition, 2014.