Point-Set Topology — Chapter 1 Topological spaces

Chapter 1
Topological spaces

1.1. Topologies

We begin by defining topological spaces.

Definition 1.1.1. §

A topology on a set $X$ is a set $𝒯$ of subsets of $X$ such that

i.: the empty set $∅$ and the set $X$ are contained in $𝒯$ ,
ii.: if $𝒰$ is a subset of $𝒯$ , then $⋃_{U \in 𝒰} U \in 𝒯$ , and
iii.: if $U_{1}, \dots, U_{n} \in 𝒯$ for some $n \geq 1$ , then $U_{1} \cap \dots \cap U_{n} \in 𝒯$ .

In other words, a topology $𝒯$ is a collection of subsets of $X$ containing $\emptyset, X$ and which is closed under arbitrary unions and finite intersections.

Definition 1.1.2. §

A topological space is a pair $(X, 𝒯)$ consisting of a set $X$ and a topology $𝒯$ on $X$ .

Definition 1.1.3. §

Let $X$ be a set.

a.: The discrete topology on $X$ is the topology equal to the set of all subsets (i.e., the power set) of $X$ . We say that a topological space $X$ is discrete if its topology is the discrete topology on $X$ .
b.: The trivial topology on $X$ is the topology ${\emptyset, X}$ .

Definition 1.1.4. §

Let $(X, 𝒯)$ be a topological space. A subset $U$ of a topological space $X$ is called open if $U \in 𝒯$ .

Terminology 1.1.5. §

We often omit the notation of a topology $𝒯$ on a topological space $X$ and simply refer to $X$ as a topological space when its topology $𝒯$ is understood. At times, we say that a topological space $X$ is endowed with (or has) a topology $𝒯$ . We sometimes refer to a topological space more simply as a space.

Definition 1.1.6. §

An element of a topological space $X$ is called a point of $X$ .

Example 1.1.7. §

A topological space $X$ has the discrete topology if and only if every subset of $X$ is open.

Examples 1.1.8. §

a.: Let $X = {a, b}$ be a two-point set. Then there are $4$ distinct topologies on $X$ , all equal to ${\emptyset, X} \cup S$ , where $S$ is some subset of ${{a}, {b}}$ .
b.: Let $X = {a, b, c}$ be a three-point set. Then ${\emptyset, {a}, {b}, X}$ is not a topology on $X$ , as it is not closed under unions, while ${\emptyset, {a, b}, {a, c}, X}$ is not a topology on $X$ , as it is not closed under finite intersections.

Definition 1.1.9. §

A set $X$ has a topology under which a nonempty subset $U$ is open if and only if its complement $X - U = {x \in X ∣ x \notin U}$ is finite. This topology is known as the finite complement topology on $X$ .

Example 1.1.10. §

The Euclidean topology on $ℝ$ is the unique topology under which a set is open if it is a union of open intervals. This is a topology as both $∅$ and $ℝ$ are open intervals, any union of unions of open intervals is a union of open intervals, and any finite intersection of open intervals is an open interval (which, as the reader will check, implies that any finite intersection of unions of open intervals is a union of open intervals).

Definition 1.1.11. §

If $𝒯$ and $𝒯^{'}$ are topologies on a set $X$ with $𝒯 \subseteq 𝒯^{'}$ , we say that $𝒯^{'}$ is finer (or stronger) than $𝒯$ , and $𝒯$ is coarser (or weaker) than $𝒯^{'}$ . If, in addition, $𝒯 \neq 𝒯^{'}$ , we say that $𝒯^{'}$ is strictly finer (or strictly stronger) than $𝒯$ , and $𝒯$ is strictly coarser (or strictly weaker) than $𝒯^{'}$ .

Remark 1.1.12. §

We think of a topology with more open sets as being finer in that we think of open sets as separating points from each other, so a topology with more open sets is more “fine-grained”, in a sense. The discrete topology is the finest topology on any set, while the trivial topology is the coarsest.

Remark 1.1.13. §

The terminology of a “finer” topology including one that may be the same is in some sense unfortunate, but it is the most standard usage.

Example 1.1.14. §

Consider the three-point set $X = {a, b, c}$ with topologies $𝒯_{1} = {\emptyset, X}$ , $𝒯_{2} = {\emptyset, {a}, X}$ , $𝒯_{3} = {\emptyset, {a}, {b}, {a, b}, X}$ , and $𝒯_{4}$ the discrete topology. Then $𝒯_{i + 1}$ is strictly finer than $𝒯_{i}$ for each $1 \leq i \leq 3$ . If we set $𝒯_{3}^{'} = {\emptyset, {a}, {a, b}, {a, c}, X}$ , then $𝒯_{3}^{'}$ is strictly finer than $𝒯_{1}$ and $𝒯_{2}$ and strictly coarser than $𝒯_{4}$ but has no such relation with $𝒯_{3}$ .

Definition 1.1.15. §

An open neighborhood of a point $x$ in a topological space $X$ is an open set containing $x$ . We say that an open neighborhood $V$ of $x$ is an open neighborhood of $V$ in a subset $A$ if $V$ is contained in $A$ .

Lemma 1.1.16. §

A subset $U$ of a topological space $X$ is open if and only if every point of $U$ has an open neighborhood in $U$ .

Proof.

If $U$ is open and $x \in U$ , then $U$ is an open neighborhood of $x$ in $U$ . Conversely, if every point $x$ of a subset $U$ of $X$ has an open neighborhood $V_{x}$ in $U$ , then $U = ⋃_{x \in U} V_{x}$ , so $U$ is open as a union of open sets in $X$ . □

At times, we may wish to speak of closed neighborhoods, in which case the following definition is useful.

Definition 1.1.17. §

A neighborhood of a point $x$ in a topological space $X$ is any subset of $X$ containing an open neighborhood of $x$ .

Proposition 1.1.18. §

A set of subsets $𝒯$ of $X$ is a topology on $X$ if and only if $\emptyset, X \in 𝒯$ , the set $𝒯$ is closed under arbitrary unions, and for all $U_{1}, \dots, U_{n} \in 𝒯$ with $n \geq 1$ and $x \in ⋂_{i = 1}^{n} U_{i}$ , there exists an open neighborhood of $x$ contained in $⋂_{i = 1}^{n} U_{i}$ .

Proof.

Suppose that $U_{1}, \dots, U_{n} \in 𝒯$ for some $n \geq 1$ , and set $W = ⋂_{i = 1}^{n} U_{i}$ . If $𝒯$ is a topology, then $W$ is open, so we may take $W$ to be the open neighborhood of the proposition. If on the other hand we have that for each $x \in W$ , there exists an open neighborhood $V_{x}$ of $x$ in $W$ , then $W = ⋃_{x \in W} V_{x}$ is an element of $𝒯$ if $𝒯$ is closed under unions. Thus, $𝒯$ is a topology under the conditions of the proposition. □

Notation 1.1.19. §

Given a set $X$ and a subset $A$ , we write $A^{c}$ for the complement $X - A$ of $A$ when $X$ is understood.

Definition 1.1.20. §

A subset $A$ of a topological space $X$ is closed, or a closed subset of $X$ , if its complement $X - A$ is a open.

Examples 1.1.21. §

a.: Every subset of a discrete space is closed.
b.: The only closed subsets of a space $X$ with the trivial topology are $∅$ and $X$ .
c.: In $ℝ$ with its Euclidean topology, closed intervals are closed subsets, as are their finite unions. Some but not all infinite unions of closed intervals are also closed: e.g., the set $ℤ$ of integers inside $ℝ$ is an infinite union of closed intervals of length zero that is closed.

Proposition 1.1.22. §

Let $X$ be a set. A set $𝒯$ of subsets of $X$ forms topology on $X$ if and only if the set $𝒞 = {U^{c} ∣ U \in 𝒯}$ has the properties that it contains $∅$ and $X$ , intersections of elements of $𝒞$ are contained in $𝒞$ , and finite unions of elements of $𝒞$ are contained in $𝒞$ .

Proof.

Let $𝒯$ be a set of subsets of $X$ and $𝒞$ be the set of complements of elements of $𝒯$ . We have $\emptyset = X^{c}$ and $X = X^{c}$ , so $\emptyset, X \in 𝒞$ if and only if $\emptyset, X \in 𝒯$ . We have

{(⋂_{A \in 𝒞} A)}^{c} = ⋃_{A \in 𝒞} A^{c},

so $𝒞$ is closed under intersections if and only if $𝒯$ is closed under unions. If $A_{1}, \dots, A_{n} \in 𝒞$ for some $n \geq 1$ , then

{(⋃_{i = 1}^{n} A_{i})}^{c} = ⋂_{i = 1}^{n} A_{i}^{c},

so $𝒞$ is closed under finite unions if and only if $𝒯$ is closed under finite intersections. Thus, $𝒯$ is a topology if and only if $𝒞$ satisfies the conditions of the proposition. □

1.2. Subspaces

Definition 1.2.1. §

Let $(X, 𝒯)$ be a topological space and $A$ be a subset of $X$ . The subspace topology on $A$ is the set $𝒯_{A} = {U \cap A ∣ U \in 𝒯}$ of subsets of $A$ .

Definition 1.2.2. §

A topological space $A$ with underlying set a subset of a topological space $X$ is called a subspace if its topology is the subspace topology from $X$ .

We verify that the subspace topology on a topological space is in fact a topology.

Proposition 1.2.3. §

Let $X$ be a topological space and $A$ be a subset of $X$ . Then the subspace topology on $A$ is a topology on $A$ .

Proof.

Let $𝒯$ denote the topology on $X$ and $𝒯_{A}$ the subspace topology on $A$ . We have $\emptyset = \emptyset \cap A \in 𝒯_{A}$ and $A = X \cap A \in 𝒯_{A}$ , so $𝒯_{A}$ satisfies property (i) of a topology.

If $𝒱 \subseteq 𝒯_{A}$ , then for each $V \in 𝒱$ , there exists $U_{V} \in 𝒯$ with $V = U_{V} \cap A$ . We then have

⋃_{V \in 𝒱} V = ⋃_{V \in 𝒱} (U_{V} \cap A) = (⋃_{V \in 𝒱} U_{V}) \cap A \in 𝒯_{A}

since $⋃_{V \in 𝒱} U_{V} \in 𝒯$ as $𝒯$ is a topology. Thus, $𝒯_{A}$ satisfies property (ii) of a topology.

If $V_{1}, \dots, V_{n} \in 𝒯_{A}$ for some $n \geq 1$ , then $V_{i} = U_{i} \cap A$ for some $U_{i} \in 𝒯$ for $1 \leq i \leq n$ . We then have

⋂_{i = 1}^{n} V_{i} = ⋂_{i = 1}^{n} (U_{i} \cap A) = (⋂_{i = 1}^{n} U_{i}) \cap A \in 𝒯_{A},

since $⋂_{i = 1}^{n} U_{i} \in 𝒯$ as $𝒯$ is a topology. Thus, $𝒯_{A}$ satisfies property (iii) of a topology. □

The subspace topology on an open subset of a topological space is a subset of the topology on the space.

Proposition 1.2.4. §

If $X$ is a topological space and $U$ is an open subset of $X$ , then the subspace topology on $U$ is the set of open subsets of $X$ that are contained in $U$ .

Proof.

If $V$ is an open subset of $U$ , then $V = W \cap U$ for some open subset $W$ of $X$ , so $V$ is open in $X$ as an intersection of two of its open subsets. Conversely, if $V$ is an open subset of $X$ contained in $U$ , then $V = V \cap U$ , so $V$ is open in $U$ as well. □

Examples 1.2.5. §

a.: Every subspace of a discrete space $X$ is discrete.
b.: The subspace topology on an open interval $(a, b) \in ℝ$ with $a < b$ consists of $∅$ and all unions of open intervals $(a^{'}, b^{'})$ with $a \leq a^{'} \leq b^{'} \leq b$ .
c.: Consider the four-point set $X = {a, b, c, d}$ with the topology ${\emptyset, {a}, {b}, {a, b}, X}$ . Then the subspace topology on ${a, b}$ is the discrete topology, while the subspace topology on ${c, d}$ is the trivial topology.

The reader may check the following lemma.

Lemma 1.2.6. §

Let $A$ be a closed subset of a topological space $X$ . Then the closed sets of $A$ under the subspace topology are exactly the intersections of closed subsets of $X$ with $A$ .

Terminology 1.2.7. §

An open (resp., closed) subset of a topological space $X$ , when endowed with the subspace topology, is called an open (resp., closed) subspace of $X$ .

1.3. Bases

Definition 1.3.1. §

A subset $B$ of a topology $𝒯$ on a topological space $X$ is said to be a base, or basis, for the topology $𝒯$ on $X$ if every open set $U \in 𝒯$ is a (possibly empty) union of elements of $B$ .

Example 1.3.2. §

The set of open intervals in $ℝ$ is a base for the Euclidean topology on $ℝ$ .

Example 1.3.3. §

If $X$ is a set with the discrete topology, then the set ${{x} ∣ x \in X}$ is a base of the topology on $X$ .

Definition 1.3.4. §

We say that a set $𝒮$ of subsets of $X$ covers a subset $A$ of $X$ if $A \subseteq ⋃_{S \in 𝒮} S$ , and $𝒮$ is a cover of $A$ if $𝒮$ covers $A$ .

Theorem 1.3.5. §

Let $X$ be a set, and let $B$ be a set of subsets of $X$ such that

i.: $B$ covers $X$ and
ii.: for every $U, V \in B$ and $x \in U \cap V$ , there exists $W \in B$ with $x \in W$ and $W \subseteq U \cap V$ .

Then the collection

𝒯 = {⋃_{U \in 𝒞} U ∣ 𝒞 \subseteq B}

of arbitrary unions of elements of $B$ is a topology on $X$ , and $B$ is a base for the topology $𝒯$ . Moreover, if $X$ is a topological space and $B$ is a base of open sets in $X$ , then $𝒯$ is the topology on $X$ .

Proof.

If $𝒯$ is a topology, then $B$ will by definition be a base, so we need only verify that $𝒯$ is a topology. Note that $∅$ is the empty union of elements of $B$ and $X \in 𝒯$ by property (i) of $B$ . We also have that $𝒯$ is closed under arbitrary unions, as its elements are just the arbitrary unions of elements of $B$ . To check that $𝒯$ is closed under finite intersections, it suffices by recursion to show that it is closed under intersections of two elements, so let $U_{1}, U_{2} \in 𝒯$ . Any $x \in U_{1} \cap U_{2}$ lies in the intersection of some $V_{1} \in B$ contained in $U_{1}$ and some $V_{2} \in B$ contained in $U_{2}$ . So, by property (ii) of $B$ , there exists $W \in B$ with

x \in W \subseteq V_{1} \cap V_{2} \subseteq U_{1} \cap U_{2} .

Since $B \subset 𝒯$ , we have $W \in 𝒯$ as well. By Proposition 1.1.18, the set $𝒯$ is a topology on $X$ .

If $X$ is a topological space with topology $𝒯^{'}$ and $B$ is a base of open sets in $X$ , then both $𝒯$ above and the topology $𝒯^{'}$ are the set of unions of elements of $B$ , the first by definition of $𝒯$ , and the second by definition of a base, so $𝒯 = 𝒯^{'}$ . □

Definition 1.3.6. §

The topology $𝒯$ on a set $X$ given by arbitrary unions of elements of a set of subsets $B$ of $X$ satisfying the two conditions of Theorem 1.3.5 is called the topology generated by the base $B$ . We say that $B$ generates the topology $𝒯$ .

We may now easily define the Euclidean topology on $ℝ^{n}$ .

Example 1.3.7. §

The Euclidean topology on $ℝ^{n}$ is the topology generated by the base consisting of open balls in $ℝ^{n}$ of finite radius. To see this is a topology, one need only note that around any point inside any nonempty intersection of two open balls, there exists another open ball.

Definition 1.3.8. §

Let $X$ be a topological space. A base of open neighborhoods of a point $x$ is a set $B_{x}$ of open neighborhoods of $x$ such that for every open neighborhood $U$ of $x$ in $X$ , there exists $V \in B_{x}$ with $V \subseteq U$ . An element of $B_{x}$ is said to be a basic open neighborhood of $x$ .

Lemma 1.3.9. §

Let $X$ be a topological space, and for each $x \in X$ , let $B_{x}$ be a base of open neighborhoods of $x$ . Then $B = ⋃_{x \in X} B_{x}$ is a base for the topology on $X$ .

Proof.

Since each $B_{x}$ is nonempty, we have $X = ⋃_{U \in B} U$ . If $U \in B_{x}$ and $V \in B_{y}$ for some $x, y \in X$ and $z \in U \cap V$ , then there exists $W \in B_{z}$ with $W \subseteq U \cap V$ . By Theorem 1.3.5, $B$ is a base for a topology $𝒯^{'}$ on $X$ . We have $𝒯^{'} \subseteq 𝒯$ since every set in $B$ is open and $𝒯$ is closed under unions. We claim that $𝒯^{'}$ is in fact the original topology $𝒯$ .

Let $W \in 𝒯$ . For $x \in W$ , since $B_{x}$ is a base of open neighborhoods of $x$ , there exists $U_{x} \in B_{x}$ with $U_{x} \subseteq W$ . Then $W \in 𝒯^{'}$ by Lemma 1.1.16. Thus $𝒯 = 𝒯^{'}$ . □

Example 1.3.10. §

In $ℝ^{n}$ , the set of open balls centered at a point $x$ forms a base of open neighborhoods of $x$ .

In fact, any set of subsets of $X$ with union $X$ gives rise to a topology on $X$ .

Definition 1.3.11. §

A set $𝒮$ of open subsets of a topological space $X$ is called a subbase, or subbasis, for the topology on $X$ if every proper open set in $X$ is a union of finite intersections of elements of $𝒮$ .

The following is an simple consequence of Theorem 1.3.5.

Proposition 1.3.12. §

Let $𝒮$ be a set of subsets of $X$ . Let $B$ be the set consisting of $X$ and all finite intersections of elements of $𝒮$ . Then $B$ is a base for a topology on $X$ for which $𝒮$ is a subbase.

Example 1.3.13. §

The set of all intervals $(a, \infty)$ and $(- \infty, b)$ with $a, b \in ℝ$ is a subbase for the Euclidean topology on $ℝ$ .

1.4. Closure

There is a smallest closed set containing a given subset of a topological space, known as its closure.

Definition 1.4.1. §

The closure $\bar{A}$ of a subset $A$ of a topological space $X$ is the intersection of all closed subsets of $X$ containing $A$ .

Lemma 1.4.2. §

The closure $\bar{A}$ of a subset $A$ of $X$ is the smallest closed subset of $X$ containing $A$ in the sense that $\bar{A}$ is closed and contains $A$ and, if $B$ is a closed subset of $X$ with $A \subseteq B$ , then $\bar{A} \subseteq B$ .

Proof.

First, we remark that $\bar{A}$ is closed as an intersection of closed sets and contains $A$ , as all of these closed sets contain $A$ . Moreover, if $B$ is closed and contains $A$ , then $B$ contains the intersection $\bar{A}$ of all closed sets containing $A$ , since $B$ is one of the sets over which the intersection is taken. □

Example 1.4.3. §

The closure of an open interval $(a, b)$ in $ℝ$ is the closed interval $[a, b]$ , as $[a, b]$ is a closed set containing $(a, b)$ , and none of $(a, b)$ , $[a, b)$ , and $(a, b]$ is closed.

We have the following alternative characterization of the closure.

Proposition 1.4.4. §

Let $A$ be a subset of a topological space $X$ . Then $x \in \bar{A}$ if and only if every (open) neighborhood of $x$ in $X$ has nonempty intersection with $A$ .

Proof.

We have $x \in \bar{A}$ if and only if $x \in B$ for all closed sets $B$ containing $A$ . The latter holds if and only if $x \notin U$ for all open sets $U \subseteq A^{c}$ . And this holds if and only if every open set $U$ containing $x$ is not contained in $A^{c}$ , hence has nonempty intersection with $A$ . □

Example 1.4.5. §

If $x \in (a, b)$ , then clearly every open neighborhood of $x$ intersects $(a, b)$ , in particular in $x$ . Any open interval of the form $(a - 𝜖, a + 𝜖)$ intersects $(a, b)$ in ${a + δ ∣ 0 < δ < 𝜖}$ . Similarly, any interval $(b - 𝜖, b + 𝜖)$ intersects $(a, b)$ in ${b - δ ∣ 0 < δ < 𝜖}$ . On the other hand, if $x > b$ or $x < a$ , then there exists a sufficiently small interval centered at $x$ that does not intersect $(a, b)$ . Since every open neighborhood of a point in $ℝ$ contains an open interval (centered at the point), Proposition 1.4.4 again tells us that the closure of $(a, b)$ is $[a, b]$ .

Definition 1.4.6. §

A subset $A$ of a topological space $X$ is dense in $X$ if $\bar{A} = X$ .

Example 1.4.7. §

The rational numbers $ℚ$ are dense in $ℝ$ with its Euclidean topology. That is, every open interval containing a real number contains a rational number, being that the interval has finite nonzero length.

We also have the notion of an interior of a set.

Definition 1.4.8. §

The interior $A^{\circ}$ of a subset $A$ of a topological space $X$ is the union of all open sets of $X$ contained in $A$ .

Note that $A^{\circ} = {(\bar{A^{c}})}^{c}$ by its definition. We then have the following as a consequence of Lemma 1.4.2.

Lemma 1.4.9. §

The interior of a subset $A$ of a topological space $X$ is the largest open subset of $X$ contained in $A$ .

We also have a notion of boundary.

Definition 1.4.10. §

The boundary $∂𝐴$ of a subset $A$ of $X$ is the complement of the interior $A^{\circ}$ of $A$ in the closure $\bar{A}$ .

Example 1.4.11. §

In $ℝ$ , the closure of an open interval $(a, b)$ with $a, b \in ℝ$ and $a < b$ is the closed interval $[a, b]$ , and the interior of $[a, b]$ is $(a, b)$ . The boundary of $(a, b)$ , or $[a, b]$ , is ${a} \cup {b}$ .

Example 1.4.12. §

In $ℝ^{n}$ , the closure of the open ball of radius $𝜖$ about a point $x$ is the closed ball of radius $𝜖$ about the point $x$ , while the boundary is the sphere of radius $𝜖$ centered at $x$ .

Example 1.4.13. §

If $X$ has the trivial topology, then the closure of any nonempty subset $A$ is $X$ , while the interior of $A$ is empty unless $A = X$ . So, any nonempty, proper subset $A$ of $X$ has boundary $X$ , whereas the boundary of $X$ is empty.

Example 1.4.14. §

Consider the three-point set $X = {a, b, c}$ with topology

𝒯 = {\emptyset, {a}, {a, b}, {a, c}, X} .

The closed sets of $𝒯$ are $∅$ , ${b}$ , ${c}$ , ${b, c}$ , and $X$ . Thus, the closure of ${a}$ is $X$ , while ${b}$ and ${c}$ are closed.

Remark 1.4.15. §

The taking of subspaces can change interiors and closures. For instance, the interior of the closed interval $[a, b]$ with $a < b$ in $ℝ$ is $(a, b)$ , but its interior in $[a, b]$ is $[a, b]$ , since $[a, b]$ is open in $[a, b]$ .

1.5. Limit points

Definition 1.5.1. §

A point $x$ in a topological space $X$ is called a limit point of a subset $A$ of $X$ if every (open) neighborhood of $x$ intersects $A$ in a point other than $x$ .

The following is a corollary of Proposition 1.4.4.

Lemma 1.5.2. §

The closure of a subset $A$ of a topological space $X$ is the union of $A$ and the set of its limit points.

Proof.

By Proposition 1.4.4, if $x \in \bar{A}$ , then every open neighborhood of $x$ intersects $A$ . If $x \notin A$ , these intersections cannot contain $x$ , so $x$ is a limit point. □

Since closed sets are their own closures, we have the following.

Corollary 1.5.3. §

Any closed subset $A$ of a topological space contains all of its limit points.

Remark 1.5.4. §

In Proposition 1.4.4, we may replace the condition on every open neighborhood with the same condition restricted to open neighborhoods in any base.

One might ask how the notion of a limit point compares to the notion of points to which sequences converge. For this, we need the following definition.

Definition 1.5.5. §

A sequence ${(x_{n})}_{n \geq 1}$ of points of a topological space $X$ converges to a point $x \in X$ if for every open neighborhood $U$ of $x$ , there exists $N \geq 1$ such that $x_{n} \in U$ for all $n \geq N$ . The point $x$ is said to be a limit of the sequence $x_{n}$ .

Remark 1.5.6. §

By definition, a limit of a convergent sequence ${(x_{n})}_{n \geq 1}$ that is not eventually constant is a limit point of the set ${x_{n} ∣ n \geq 1}$ . However, the latter set may have more then one limit point even if ${(x_{n})}_{n \geq 1}$ converges.

Example 1.5.7. §

If $X$ has the trivial topology, then every sequence in $X$ converges to every point of $X$ .

To avoid such pathologies as in the previous example, it is useful to put the following condition on a space.

Definition 1.5.8. §

A topological space $X$ is called is Hausdorff is for every two distinct points $a, b \in X$ , there exist open neighborhoods $U$ of $a$ and $V$ of $b$ such that $U \cap V = \emptyset$ .

More briefly, $X$ is Hausdorff if every two distinct points of $X$ have disjoint neighborhoods. In Hausdorff spaces, points are closed.

Lemma 1.5.9. §

In a Hausdorff space $X$ , every singleton set ${x}$ for $x \in X$ is closed.

Proof.

If $x \in X$ and $a \in X$ with $x \neq a$ , then there exists an open neighborhood $U_{a}$ of $a$ not containing $x$ . As the union of all such open sets $U_{a}$ is the complement of ${x}$ , the set ${x}$ is closed. □

The property of being Hausdorff is stronger than that of points being closed, however.

Example 1.5.10. §

In an infinite set $X$ with the finite complement topology, points are closed as the complement of open sets. However, any two nonempty open sets in $X$ intersect in all but finitely many elements of $X$ , so $X$ is not Hausdorff. Moreover, every non-repeating sequence ${(x_{n})}_{n \geq 1}$ in $X$ converges to every point of $X$ , as the reader should check using the fact that open sets have finite complements.

Even better, in Hausdorff spaces, every convergent sequence has a unique limit.

Proposition 1.5.11. §

Every convergent sequence in a Hausdorff space has a unique limit.

Proof.

If $x \in X$ is a limit of a convergent sequence ${(x_{n})}_{n \geq 1}$ in a Hausdorff space $X$ , then for any $y \in X - {x}$ , we have have disjoint open neighborhoods $U$ of $x$ and $V$ of $y$ . For sufficiently large $n$ , the point $x_{n}$ are all in $U$ , hence not in $V$ , and therefore $y$ is not a limit point of the sequence. □

The following is immediate from the definitions.

Lemma 1.5.12. §

Every subspace of a Hausdorff space is Hausdorff.

The following characterization of the Hausdorff property is often useful.

Lemma 1.5.13. §

A topological space $X$ is Hausdorff if and only if for every two distinct points $x, y \in X$ , there exists an open neighborhood $U$ of $x$ with $y \notin \bar{U}$ .

Proof.

Let $x$ and $y$ be distinct points of $X$ . We have disjoint open neighborhoods $U$ and $V$ of $x$ and $y$ , respectively, if and only if we have an open neighborhood $U$ of $x$ and a closed neighborhood $A$ containing $U$ and not containing $y$ . That is, we simply take $A$ and $V$ to be complements of each other. But such an $A$ exists if and only if we can take it to be the closure $\bar{U}$ , which is contained in any such $A$ , being the smallest closed set containing $U$ . □

1.6. Metric spaces

Metric spaces, and the open balls inside of them, provide fundamental examples of topological spaces. We review the definition here.

Definition 1.6.1. §

A metric on a set $X$ is a function $d : X \times X \to ℝ_{\geq 0}$ such that for all $a, b, c \in X$ , one has

i.: $d (a, b) = 0$ if and only if $a = b$ ,
ii.: $d (a, b) = d (b, a)$ , and
iii.: $d (a, c) \leq d (a, b) + d (b, c)$ .

Terminology 1.6.2. §

For a set $X$ , the condition $d (a, c) \leq d (a, b) + d (b, c)$ on $d : X \times X \to ℝ_{\geq 0}$ for all $a, b, c \in X$ is called the triangle inequality.

Definition 1.6.3. §

A pair $(X, d)$ consisting of a set $X$ and a metric $d$ on $X$ is called a metric space.

Notation 1.6.4. §

When the metric $d$ on a metric space $(X, d)$ is understood, we often write $X$ for the metric space.

Example 1.6.5. §

The set $ℝ^{n}$ is a metric space for the distance function $d : ℝ^{n} \times ℝ^{n} \to ℝ_{\geq 0}$ defined by the Euclidean metric

d (x, y) = \sqrt{\sum_{i = 1}^{n} {(x_{i} - y_{i})}^{2}}

for $x = {(x_{i})}_{i = 1}^{n}$ and $y = {(y_{i})}_{i = 1}^{n}$ . We remark that any three points $x$ , $y$ , and $z$ form the vertices of a triangle, and the distances between them are the lengths of the sides, so the triangle inequality reduces to the usual triangle inequality of Euclidean geometry.

Definition 1.6.6. §

In a metric space $(X, d)$ , the open ball of radius $𝜖 > 0$ about a point (or with center) $x \in X$ is the set

B (x, 𝜖) = {y \in X ∣ d (x, y) < 𝜖} .

The closed ball of radius $𝜖$ is

\bar{B} (x, 𝜖) = {y \in X ∣ d (x, y) \leq 𝜖} .

Proposition 1.6.7. §

The set of open balls in a metric space $X$ forms a base of a topology on $X$ , and the set of open balls with center $x$ is a base of open neighborhoods of $x$ in this topology.

Proof.

Clearly, the union of open balls in a metric space is $X$ , so by Theorem 1.3.5, it suffices to show that for any two open balls $U = B (x, 𝜖)$ and $V = B (y, δ)$ in $X$ and any point $z \in U \cap V$ , there exists an open ball containing $z$ and contained in $U \cap V$ . For this, choose any positive real number

ρ < \min {𝜖 - d (x, z), δ - d (y, z)} .

Then $B (z, ρ) \subseteq U \cap V$ by the triangle inequality. That is, if $d (z, w) < ρ$ , then

d (w, x) \leq d (w, z) + d (z, x) < ρ + d (x, z) < 𝜖,

and similarly $d (w, y) \leq d (w, z) + d (z, y) < δ$ . Thus the set of open balls in $X$ form a base.

If $x \in X$ and $U$ is any open neighborhood of $x$ , then it contains some basic open neighborhood $B (y, 𝜖)$ of $x$ , i.e., with $d (x, y) < 𝜖$ . Then $B (x, 𝜖 - d (x, y)) \subseteq U$ as in the above argument, so the balls centered at $x$ form a base of open neighborhoods of $x$ . □

Definition 1.6.8. §

The metric topology on a set $X$ induced by a metric $d$ on $X$ is the topology on $X$ generated by the set of open balls under $d$ .

Lemma 1.6.9. §

Let $X$ be a metric space. Then every closed ball $\bar{B} (x, 𝜖)$ is closed in the metric topology.

Proof.

It suffices to show that the complement of $\bar{B} (x, 𝜖)$ is open. If $y \notin \bar{B} (x, 𝜖)$ , then $δ = d (x, y) - 𝜖 > 0$ , and $B (y, δ)$ and $\bar{B} (x, 𝜖)$ are disjoint by the triangle inequality. That is, if $z \in B (y, δ)$ , then $d (x, z) + δ > d (x, z) + d (z, y) \geq d (x, y)$ , so $d (x, z) > 𝜖$ . Thus, $z \notin \bar{B} (x, 𝜖)$ . □

Two different metrics on a set $X$ can have the same metric topology. Take the following example.

Example 1.6.10. §

Consider the Euclidean metric $d$ on $ℝ^{n}$ and the box metric $d^{'}$ on $ℝ^{n}$ defined by

d^{'} (x, y) = \max {| x_{i} - y_{i} | ∣ 1 \leq i \leq n}

for $x = {(x_{i})}_{i = 1}^{n}, y = {(y_{i})}_{i = 1}^{n} \in ℝ^{n}$ . The reader should check that $d^{'} (x, y)$ is in fact a metric.

We have bases $B (x, 𝜖)$ and $B^{'} (x, 𝜖)$ of open balls about a point $x$ with respect to these respective metrics. For any $y \in ℝ^{n}$ , we have

\max {| x_{i} - y_{i} |^{2} ∣ 1 \leq i \leq n} \leq \sum_{i = 1}^{n} {(x_{i} - y_{i})}^{2} \leq n \max {| x_{i} - y_{i} |^{2} ∣ 1 \leq i \leq n},

B (x, 𝜖) \subseteq B^{'} (x, 𝜖) \subseteq B (x, \sqrt{n} 𝜖)

for all $𝜖 > 0$ . Thus, the two metric topologies coincide.

We will often consider a metric space as a topological space endowed with the metric topology. We can then examine the topological properties of metric spaces.

Proposition 1.6.11. §

Metric spaces are Hausdorff.

Proof.

Let $X$ be a metric space, and suppose that $x, y \in X$ are distinct points. Let $𝜖 = \frac{1}{2} d (x, y)$ . Then $B (x, 𝜖)$ and $B (y, 𝜖)$ are disjoint by the triangle inequality. □

Definition 1.6.12. §

A topological space $(X, 𝒯)$ is metrizable if there exists a metric $d$ on $X$ such that the metric topology induced by $d$ is the topology $𝒯$ on $X$ .

The following is a direct corollary of Proposition 1.6.11.

Corollary 1.6.13. §

If $X$ is a metrizable topological space, then $X$ is Hausdorff.

Discrete spaces are metric spaces as well.

Definition 1.6.14. §

Let $X$ be a set. The discrete metric $d$ on $X$ is defined by

d (x, y) = {\begin{matrix} 1 & if x \neq y \\ 0 & if x = y . \end{matrix}

We have the following.

Lemma 1.6.15. §

The discrete metric on a set $X$ is a metric, and the topology induced by this metric is the discrete topology.

Proof.

That $d$ is a metric is straightforward. Since $B (x, \frac{1}{2}) = {x}$ for all $x \in X$ , singleton sets are open in this topology, so $X$ is discrete. □

The following example shows that even metric spaces can defy our intuition from Euclidean geometry.

Example 1.6.16. §

While in any metric space, the closure of the open ball $B (x, 𝜖)$ is contained in the closed ball $\bar{B} (x, 𝜖)$ , the closure of $B (x, 𝜖)$ can in fact be smaller. For instance, if $(X, d)$ is a metric space with the discrete metric $d$ , then the set $B (x, 1) = {x}$ is both open and closed (as are all subsets of $X$ ), while $\bar{B} (x, 1) = X$ .

Definition 1.6.17. §

A subset $A$ of a metric space $X$ is bounded if there exists $N > 0$ such that $d (x, y) \leq N$ for all $x, y \in A$ .

This notion of boundedness is not a topological one.

Lemma 1.6.18. §

If $(X, d)$ is a metric space, then the function $d^{'} : X \times X \to ℝ_{\geq 0}$ given by

d^{'} (x, y) = \min {d (x, y), 1}

is a metric on $X$ , and the metric topologies on $X$ from $d$ and $d^{'}$ are the same.

Proof.

Let $x, y, z \in X$ . If $d (x, y) \leq 1$ , then

d^{'} (x, y) = d (x, y) \leq \min {d (x, z) + d (z, y), 1 + d (z, y), d (x, z) + 1, 1 + 1} = d^{'} (x, z) + d^{'} (z, y),

by the triangle inequality for the first of the terms in the set and the fact that $d (x, y) \leq 1$ for the others. If $d (x, y) \geq 1$ , then

d^{'} (x, y) = 1 \leq \min {d (x, z) + d (z, y), 1 + d (z, y), d (x, z) + 1, 1 + 1} = d^{'} (x, z) + d^{'} (z, y)

by the triangle inequality for the first term in the set (in that $1 \leq d (x, y) \leq d (x, z) + d (z, y)$ ) and the fact that the other terms are clearly at least one. Thus, $d^{'}$ satisfies the triangle inequality, and it clearly satisfies the other two conditions for being a metric.

The open balls of radius less than $1$ form a base of any metric topology, as any open ball contains one of these. Since these sets of open balls coincide for $d$ and $d^{'}$ , these metrics induce the same topology on $X$ . □

Example 1.6.19. §

Under the Euclidean metric, $ℝ$ is not bounded, but it is with respect to the metric $d^{'} (x, y) = \min {| x - y |, 1}$ on $ℝ$ . Nevertheless, both of these metrics induces the Euclidean topology.

Sequences in metric spaces behave as one might expect.

Proposition 1.6.20. §

Let $(X, d)$ be a metric space. A sequence ${(x_{n})}_{n \geq 1}$ in $X$ converges to $a \in X$ if and only if $\lim_{n \to \infty} d (a, x_{n}) = 0$ .

Proof.

If $U$ is an open neighborhood of $a$ , then $U$ contains some open ball $B (a, 𝜖)$ . If $\lim_{n \to \infty} d (a, x_{n}) = 0$ , then there exists $N \geq 1$ such that $d (a, x_{n}) < 𝜖$ for all $n \geq N$ , so $x_{n} \in U$ . Conversely, if ${(x_{n})}_{n \geq 1}$ converges to $a$ , then for any $𝜖$ , there exists $N \geq 1$ such that $x_{n} \in B (a, 𝜖)$ for $n \geq N$ , which is to say $d (a, x_{n}) < 𝜖$ . Thus, the limit $\lim_{n \to \infty} d (a, x_{n})$ is $0$ . □

Proposition 1.6.21. §

Let $X$ be a metrizable space, let $A \subseteq X$ , and let $x \in X$ . Then $x \in \bar{A}$ if and only if it is the limit of a convergent sequence of elements of $A$ .

Proof.

We need only see that any $x \in \bar{A}$ is such a limit. Fix a metric $d$ on $X$ so that we may consider open balls in $X$ . If $x \in \bar{A}$ , then for any $n \geq 1$ , there exists $x_{n} \in B (x, \frac{1}{n}) \cap A$ by definition of $\bar{A}$ . Then the $x_{n}$ converge to $x$ by Proposition 1.6.20. □