Point-Set Topology — Chapter 2 Continuous functions

Chapter 2
Continuous functions

2.1. Continuous functions

In this section, $X$ and $Y$ will denote topological spaces.

Definition 2.1.1. §

A function $f : X \to Y$ is said to be continuous if $f^{- 1} (V)$ is open for every open subset $V$ of $Y$ .

Definition 2.1.2. §

We say that $f : X \to Y$ between topological spaces is continuous at a point $x \in X$ (or continuous at $x$ ) if for every open neighborhood $V$ of $f (x)$ , there exists an open neighborhood $U$ of $x$ such that $f (U) \subseteq V$ .

Remark 2.1.3. §

As every open subset is a union of open neighborhoods of its points, a function is continuous if and only if it is continuous at every point.

We can check continuity on basis elements of $Y$ .

Proposition 2.1.4. §

Let $B_{X}$ and $B_{Y}$ be fixed bases of the topologies on $X$ and $Y$ , respectively. Let $f : X \to Y$ be a function.

a.: A function $f : X \to Y$ is continuous if and only if $f^{- 1} (V)$ is open in $X$ for all $V \in B_{Y}$ .
b.: A function $f : X \to Y$ is continuous at $x \in X$ if and only if for every basic open neighborhood $V \in B_{Y}$ of $f (x)$ , there exists a basic open neighborhood $U \in B_{X}$ of $x$ with $f (U) \subseteq V$ ,

Proof.

If $f$ is continuous, then $f^{- 1} (V)$ is open for all $V \in B_{Y}$ by definition. Conversely, suppose $f^{- 1} (V)$ is open for all $V \in B_{Y}$ . As $B_{Y}$ is a base, for any open subset $W$ of $Y$ , there exists $𝒞 \subseteq B_{Y}$ such that $W = ⋃_{V \in 𝒞} V$ . Then $f^{- 1} (W) = ⋃_{V \in 𝒞} f^{- 1} (V)$ , so $f^{- 1} (W)$ is open as a union of open sets. Thus, we have proven part (a).

If $f$ is continuous at $x \in X$ and $V$ is a basic open neighborhood of $f (x)$ , then $f^{- 1} (V)$ contains an open neighborhood of $x$ , and such an open neighborhood contains a basic open neighborhood of $x$ . Conversely, if for every $V \in B_{Y}$ with $f (x) \in V$ , the set $f^{- 1} (V)$ contains a (basic) open neighborhood of $x$ , then for any open neighborhood $W$ of $f (x)$ , we have that $W$ contains some such $V$ . Thus, $f$ is continuous at $x$ . □

Here is the fundamental example, which states that a map between metric spaces is continuous with respect to their metric topologies if and only if it is continuous in the usual sense.

Proposition 2.1.5. §

Let $X$ and $Y$ be metric spaces, which we endow with their metric topologies. A function $f : X \to Y$ is continuous if and only if for every $x \in X$ and $𝜖 \in ℝ_{> 0}$ , there exists $δ \in ℝ_{> 0}$ such that $f (B (x, δ)) \subseteq B (f (x), 𝜖)$ .

Proof.

The sets $B_{X}$ and $B_{Y}$ of open balls of positive radius in $X$ and $Y$ , respectively, are bases for the metric topologies on $X$ and $Y$ . The result is therefore a direct consequence of the equivalence of Proposition 2.1.4(b). □

We give a few more examples.

Examples 2.1.6. §

a.: If $X$ is discrete, then any map $f : X \to Y$ is continuous.
b.: If $Y$ has the trivial topology, then any map $f : X \to Y$ is continuous.
c.: If the set $X$ has two topologies $𝒯_{X}$ and $𝒯_{X}^{'}$ with $𝒯_{X}^{'}$ finer than $𝒯_{X}$ and $f$ is continuous for $𝒯_{X}$ , then $f$ is continuous for $𝒯_{X}^{'}$ .
d.: If $Y$ has two topologies $𝒯_{Y}$ and $𝒯_{Y}^{'}$ with $𝒯_{Y}^{'}$ finer than $𝒯_{Y}$ and $f$ is continuous for $𝒯_{Y}^{'}$ , then it is continuous for $𝒯_{Y}$ .
e.: As a special case of the two latter examples, the identity map ${id}_{X} : X \to X$ given by ${id}_{X} (x) = x$ is continuous with potentially different topologies on the domain and codomain if and only if the topology of the domain is finer than the topology on the codomain.

We can also express continuity in terms of closed sets and closures.

Lemma 2.1.7. §

A function $f : X \to Y$ is continuous if and only if $f^{- 1} (B)$ is closed for every closed subset $B$ of $Y$ .

Proof.

We have $f^{- 1} (B) = X - f^{- 1} (B^{c})$ , so $f^{- 1} (B)$ is closed if and only if $f^{- 1} (B^{c})$ is open. The result follows since $B^{c}$ runs over the open sets of $Y$ as $B$ runs over the closed sets. □

Proposition 2.1.8. §

A function $f : X \to Y$ is continuous if and only if for every $A \subseteq X$ , we have $f (\bar{A}) \subseteq \bar{f (A)}$ .

Proof.

If $f$ is continuous and $A^{'} = f^{- 1} (\bar{f (A)})$ , then $A^{'}$ is closed and $A \subseteq f^{- 1} (f (A)) \subseteq A^{'}$ , so $\bar{A} \subseteq A^{'}$ . Thus

f (\bar{A}) \subseteq f (A^{'}) = f (f^{- 1} (\bar{f (A)})) \subseteq \bar{f (A)} .

Conversely, suppose that $f (\bar{A}) \subseteq \bar{f (A)}$ for all $A$ . Take $B$ to be a closed subset of $Y$ , and set $A = f^{- 1} (B)$ . Then $f (A) \subseteq B$ , and if $x \in \bar{A}$ , then $f (x) \in \bar{f (A)} \subseteq B$ . In other words, $\bar{A} \subseteq f^{- 1} (B) = A$ , so $f^{- 1} (B)$ is closed. Therefore, $f$ is continuous. □

Examples 2.1.9. §

a.: Constant functions are continuous.
b.: The inclusion map $ι_{A} : A \to X$ of a subspace $A$ of a space $X$ is continuous. That is, if $V$ is open in $X$ , then $ι^{- 1} (V) = A \cap V$ is open in $A$ in the subspace topology.
c.: A composition of continuous maps is continuous.
d.: If $f : X \to Y$ is a continuous function, then its restriction $f |_{A}$ to any subset $A$ of $X$ given by $f |_{A} (a) = f (a)$ for all $a \in A$ is continuous for the subspace topology on $A$ , since $f |_{A} = f \circ ι_{A}$ . (We say that $f$ restricts to $g = f |_{A}$ on $A$ , that $f$ extends $g$ to $X$ , and that $f$ is an extension of $g$ from $A$ to $X$ .)
e.: If $f : X \to Y$ is a continuous function, then the map $g : X \to f (X)$ is continuous for $g (x) = f (x)$ for all $x \in X$ , where $f (X)$ is endowed with the subspace topology from $Y$ . If $V \subseteq f (X)$ is open, then $V = U \cap f (X)$ for some open subset $U$ of $Y$ , so $g^{- 1} (V) = f^{- 1} (U)$ is open.

Definition 2.1.10. §

An open cover of a subset $A$ of a topological space $X$ is a collection $𝒰$ of open subsets of $X$ that covers $A$ .

Lemma 2.1.11. §

A function $f : X \to Y$ is continuous if and only if there is an open cover $𝒰$ of $X$ such that $f |_{U}$ is continuous for all $U \in 𝒰$ .

Proof.

We have seen that each $f |_{U}$ is continuous if $f$ is. On the other hand, suppose $f |_{U}$ is continuous for all $U \in 𝒰$ . If $V \subset Y$ is an open subset, then $f^{- 1} (V) \cap U = f |_{U}^{- 1} (V)$ is open. Thus

f^{- 1} (V) = ⋃_{U \in 𝒰} (f^{- 1} (V) \cap U)

is open, so $f$ is continuous. □

Lemma 2.1.12. §

If $f : X \to Y$ is a function, $A$ and $B$ are closed subspaces of $X$ such that $A \cup B = X$ , and $f |_{A}$ and $f |_{B}$ are continuous, then $f$ is continuous.

Proof.

Let $C \subseteq Y$ be closed. Then $f^{- 1} (C) = f |_{A}^{- 1} (C) \cup f |_{B}^{- 1} (C)$ , and ${(f |_{A})}^{- 1} (C)$ is closed in $A$ and ${(f |_{B})}^{- 1} (C)$ is closed in $B$ . Since $A$ and $B$ are closed in $X$ , the latter two inverse images are closed in $X$ , so $f^{- 1} (C)$ is closed as a union of two closed sets. □

Definition 2.1.13. §

A function $f : X \to Y$ is a homeomorphism if it is a continuous bijection and its inverse is continuous as well.

Definition 2.1.14. §

Two spaces $X$ and $Y$ are homeomorphic if there exists a homeomorphism $f : X \to Y$ .

We leave the following for the reader to verify.

Lemma 2.1.15. §

The relation $≃$ on any set of topological spaces given by $X ≃ Y$ if $X$ is homeomorphic to $Y$ is an equivalence relation.

Examples 2.1.16. §

a.: The function $f : ℝ \to ℝ$ given by $f (x) = x^{3}$ is a homeomorphism, as it is continuous with continuous inverse $f^{- 1} (x) = x^{1 ∕ 3}$ .
b.: The function $f : (0, 1) \to ℝ_{> 0}$ given by $f (x) = \frac{x}{1 - x}$ is a homeomorphism (where $(0, 1)$ has the subspace topology from $ℝ$ ).

Definition 2.1.17. §

A function $f : X \to Y$ is an embedding if it is a homeomorphism onto its image $f (X)$ with the subspace topology from $Y$ .

Definition 2.1.18. §

A function $f : X \to Y$ is an open map, or open, if $f (U)$ is open for every open subset $U$ of $X$ . A function $f : X \to Y$ is a closed map, or closed, if $f (U)$ is closed for every closed subset $U$ of $X$ .

Remark 2.1.19. §

If $f : X \to Y$ is a bijection, then $f$ is open if and only if $f$ is closed.

By definition, homeomorphisms are open maps; that is, they are the continuous, open bijections. However, in general, continuous bijections may not be open.

Examples 2.1.20. §

a.

Let $f : S^{1} \to ℝ^{2}$ be the natural embedding, where

S^{1} = {(x, y) \in ℝ^{2} ∣ x^{2} + y^{2} = 1} .

Then $S^{1}$ is open in itself, but $f (S^{1}) = S^{1}$ is not open in $ℝ^{2}$ .

b.

Let $X$ be a set with two topologies $𝒯$ and $𝒯^{'}$ with $𝒯^{'}$ strictly finer than $𝒯$ . Then the identity map $f : X \to X$ with the domain having the topology $𝒯^{'}$ and the codomain having the topology $𝒯$ is continuous, but its inverse $f^{- 1}$ is not, so $f$ is not an open map. On the other hand, $f^{- 1}$ is an open map.

We next consider the behavior of continuous functions and sequences.

Proposition 2.1.21. §

Let $f : X \to Y$ be a function. If $f$ is continuous, then for every sequence ${(x_{n})}_{n \geq 1}$ in $X$ that converges to a point $x \in X$ , the sequence ${(f (x_{n}))}_{n \geq 1}$ converges to $f (x)$ . If $X$ is a metrizable space such that for every sequence ${(x_{n})}_{n \geq 1}$ in $X$ that converges to some $x \in X$ , then ${(f (x_{n}))}_{n \geq 1}$ converges to $f (x)$ , then $f$ is continuous.

Proof.

Suppose that $f$ is continuous. Let ${(x_{n})}_{n \geq 1}$ be a sequence in $X$ converging to $x$ . Let $V$ be an open neighborhood of $f (x)$ in $Y$ . Then $f^{- 1} (V)$ is an open neighborhood of $x \in X$ , so contains all $x_{n}$ for $n \geq N$ for some $N \geq 1$ . Thus $f (x_{n}) \in V$ for all such $n \geq N$ , and therefore the sequence of $f (x_{n})$ converges to $f (x)$ .

Now let be $X$ metrizable, and suppose that whenever ${(x_{n})}_{n \geq 1}$ converges to $x \in X$ , the sequence ${(f (x_{n}))}_{n \geq 1}$ converges to $f (x)$ . Let $A$ be a subset of $X$ and $x \in \bar{A}$ . It suffices by Proposition 2.1.8 to show that $f (x) \in \bar{f (A)}$ . For this, we need only exhibit a sequence ${(x_{n})}_{n \geq 1}$ in $A$ converging to $x$ . Then $f (x)$ is the limit of the $f (x_{n})$ . But such a sequence exists by the metrizability of $X$ and Proposition 1.6.21. □

Definition 2.1.22. §

Let ${(f_{n})}_{n \geq 1}$ be a sequence of functions from a set $X$ to a metric space $Y$ with metric $d_{Y}$ . The sequence ${(f_{n})}_{n \geq 1}$ converges uniformly to $f : X \to Y$ if for every $𝜖 > 0$ there exists an integer $N \geq 1$ such that $d_{Y} (f (x), f_{n} (x)) < 𝜖$ for all $n \geq N$ and $x \in X$ .

Remark 2.1.23. §

If $Y$ is a metrizable topological space, then the notion of uniform convergence of a sequence is independent of the choice of metric on $d_{Y}$ yielding the topology.

Proposition 2.1.24. §

Let $X$ be a topological space and $Y$ be a metrizable space. Let ${(f_{n})}_{n \geq 1}$ be a sequence of continuous functions $f_{n} : X \to Y$ that converges uniformly to a function $f : X \to Y$ . Then $f$ is continuous.

Proof.

Let $𝜖 > 0$ and $a \in X$ . Choose $N \geq 1$ such that $d_{Y} (f_{n} (x), f (x)) < \frac{𝜖}{3}$ for all $n \geq N$ and $x \in X$ . By the continuity of $f_{N}$ , we can find an open neighborhood $U$ of $a$ such that if $x \in U$ , then $d_{Y} (f_{N} (a), f_{N} (x)) < \frac{𝜖}{3}$ . For such $x \in X$ , we then have

d_{Y} (f (a), f (x)) \leq d_{Y} (f (a), f_{N} (a)) + d_{Y} (f_{N} (a), f_{N} (x)) + d_{Y} (f_{N} (x), f (x)) < 𝜖 .

□

2.2. Product spaces

Proposition 2.2.1. §

Let $I$ be an indexing set, and for each $i \in I$ , let $X_{i}$ be a topological space. Then the set $B$ of product sets $\prod_{i = 1}^{n} U_{i}$ with each $U_{i}$ open in $X_{i}$ and $U_{i} = X_{i}$ for all but finitely many $i \in I$ forms a base of a topology on $X = \prod_{i \in I} X_{i}$ .

Proof.

Let $U = \prod_{i \in I} U_{i}$ and $V = \prod_{i \in I} V_{i}$ be two sets in $B$ . Then

(\prod_{i \in I} U_{i}) \cap (\prod_{i \in I} V_{i}) = \prod_{i \in I} (U_{i} \cap V_{i}),

each $U_{i} \cap V_{i}$ is open in $X_{i}$ , and all but finitely many $U_{i} \cap V_{i}$ equal $X_{i}$ . Thus $U \cap V \in B$ . Since $X \in B$ as well, it follows that $B$ is a base. □

Definition 2.2.2. §

Let $I$ be an indexing set, and for each $i \in I$ , let $X_{i}$ be a topological space. The product topology on $X$ is the topology generated by the base of sets $\prod_{i \in I} U_{i}$ with each $U_{i}$ open in $X_{i}$ and all but finitely many $U_{i} = X_{i}$ .

Remark 2.2.3. §

The product topology on a finite product $\prod_{i = 1}^{n} X_{i}$ of topological spaces $X_{1}, \dots, X_{n}$ with $n \geq 1$ has a base $\prod_{i = 1}^{n} U_{i}$ of open sets with $U_{i}$ open in $X_{i}$ for $1 \leq i \leq n$ . I.e., the condition that all but finitely many $U_{i}$ be $X_{i}$ is vacuous, since ${1, \dots, n}$ is a finite set.

Definition 2.2.4. §

Let $I$ be an indexing set, and for each $i \in I$ , let $X_{i}$ be a topological space. The $j$ th projection map $π_{j} : \prod_{i \in I} X_{i} \to X_{j}$ for $j \in I$ is the function defined by $π_{j} ({(x_{i})}_{i \in I}) = x_{j}$ for ${(x_{i})}_{i \in I} \in \prod_{i \in I} X_{i}$ .

The following proposition explains the seemingly strange choice of the product topology on $X$ .

Proposition 2.2.5. §

Let $I$ be an indexing set, and for each $i \in I$ , let $X_{i}$ be a topological space. The product topology on $X = \prod_{i \in I} X_{i}$ is the coarsest topology on $X$ such that each projection map $π_{j} : X \to X_{j}$ with $j \in I$ is continuous.

Proof.

Let $i_{1}, \dots, i_{n} \in I$ be elements for some $n \geq 1$ , and for $1 \leq k \leq n$ , let $U_{i_{k}} \subset X_{i_{k}}$ be an open subset. Then

⋂_{k = 1}^{n} π_{k}^{- 1} (U_{i_{k}}) = \prod_{i \in I} V_{i}

with $V_{i} = X_{i}$ unless $i \in {i_{1}, \dots, i_{n}}$ and $V_{i}$ open in $X_{i}$ if so. The collection of sets $π_{j}^{- 1} (U_{j})$ with $j \in I$ and $U_{j} \subset X_{j}$ therefore forms a subbase for the product topology on $X$ . In other words, the product topology is the coarsest topology such that these sets are open in $X$ , which is to say such that every $π_{j}$ is continuous. □

Proposition 2.2.6. §

Let $X$ be a topological space, let $Y_{i}$ be topological spaces for each $i$ in an indexing set $I$ , and endow $Y = \prod_{i \in I} Y_{i}$ with the product topology. A function $f = {(f_{i})}_{i \in I} : X \to Y$ is continuous if and only if each $f_{i} : X \to Y_{i}$ is continuous.

Proof.

If $f$ is continuous, then each $f_{i} = π_{i} \circ f$ is continuous, where $π_{i} : Y \to Y_{i}$ is the projection map in the $i$ -coordinate. Conversely, suppose that each $f_{i}$ is continuous. For $J$ be a finite subset of $I$ , and for each $j \in J$ , let $V_{j}$ be an open subset of $Y_{j}$ . Let $V = \prod_{i \in I} V_{i}$ where we set $V_{i} = Y_{i}$ for $i \in I - J$ . Then

f^{- 1} (V) = ⋂_{i \in I} f_{i}^{- 1} (V_{i}) = ⋂_{j \in J} f_{j}^{- 1} (V_{j}),

which is open as a finite intersection of open sets. Therefore $f$ is continuous. □

Proposition 2.2.7. §

Let $I$ be an indexing set, and let $X_{i}$ be a Hausdorff topological space for each $i \in I$ . Then $X = \prod_{i \in I} X_{i}$ is Hausdorff in the product topology.

Proof.

Let $x = {(x_{i})}_{i \in I}$ and $y = {(y_{i})}_{i \in I}$ be distinct elements of $X$ , and let $j \in I$ be such that $x_{j} \neq y_{j}$ . Let $U_{j}$ and $V_{j}$ be disjoint open neighborhoods of $x_{j}$ and $y_{j}$ . Then $U = π_{j}^{- 1} (U_{j})$ and $V = π_{j}^{- 1} (V_{j})$ are open in $X$ in the product topology, and $U \cap V = \emptyset$ . □

Remark 2.2.8. §

From our definition of the product topology on a direct product $X = \prod_{i \in I} X_{i}$ of topological spaces, it is not hard to see that if $J$ is any finite subset of $I$ and $U$ is any open subset of $\prod_{j \in J} X_{j}$ in the product topology, then $W = U \times \prod_{i \in I - J} X_{i}$ is open in $X$ . However, not every open subset need have this form.

For instance, consider $ℝ^{I} = \prod_{i \in I} ℝ$ for any infinite set $I$ . Consider its open subsets

V_{j} = {{(x_{i})}_{i \in I} ∣ x_{j} \in (0, 1)}

for $j \in I$ . Then the union $V = ⋃_{j \in I} V_{j}$ is open in $X$ but is not a product of the stated form, since it consists of all tuples with at least one coordinate in $(0, 1)$ . To see this, consider an open set $W = U \times ℝ^{I - J}$ of the above form, where $U$ is an open set of $ℝ^{J}$ (in the coordinates corresponding to $J$ , and $ℝ^{I - J}$ is the product of $ℝ$ in all coordinates in $I - J$ ). If $W \subseteq V$ , then since no condition is imposed on the $i$ th coordinates of elements of $W$ for $i \notin J$ , we must have $W \subseteq ⋃_{j \in J} V_{j}$ . Take any $i \in I - J$ , and note that the element of $V$ which is $0$ in all coordinates but the $i$ th and $\frac{1}{2}$ in the $i$ th coordinate is not in $W$ . Thus $W \neq V$ .

Remark 2.2.9. §

Just to speak of an element of an arbitrary product $A = \prod_{i \in I} A_{i}$ for an uncountable set $I$ and nonempty sets $A_{i}$ , we run into the issue of needing to make uncountably many choices. If we cannot, we may not be able to say that that $A$ contains a single element! That one can in fact do this is equivalent to an axiom of set theory, called the axiom of choice. It states that, given a collection of disjoint sets ${A_{i} ∣ i \in I}$ for some indexing set $I$ , there exists a function $f : I \to ∐_{i \in I} A_{i}$ to the disjoint union of the sets $A_{i}$ such that $f (i) \in A_{i}$ for each $i \in I$ . In other words, we can pick one element from each set $A_{i}$ .

There is a more simply described topology on a product that agrees with the product topology for finite products but is rather finer for infinite products. It might constitute a first guess at a natural topology on the product.

Definition 2.2.10. §

Let $X_{i}$ be a topological space for each $i$ in an indexing set $I$ , and set $X = \prod_{i \in I} X_{i}$ . The box topology on $X$ is the topology generated by the base consisting of products $\prod_{i \in I} U_{i}$ of open sets $U_{i} \subseteq X_{i}$ for each $i \in I$ .

Remark 2.2.11. §

Suppose that $X$ is an infinite product of spaces that do not have the trivial topology. Then the box topology on $X$ is strictly finer than the product topology.

Proposition 2.2.12. §

Let $I$ be an indexing set. Let $X_{i}$ be a topological space and let $A_{i}$ be a subset of $X_{i}$ for each $i \in I$ . The closure of $A = \prod_{i \in I} A_{i}$ in the box or product topology on $\prod_{i \in I} X_{i}$ is $\prod_{i \in I} \bar{A_{i}}$ .

Proof.

If $a = {(a_{i})}_{i \in I} \in \prod_{i \in I} X_{i}$ and $U = \prod_{i \in I} U_{i}$ is an open neighborhood of $a$ in the product (i.e., all but finitely many $U_{i} = X_{i}$ ) or box topologies, then $U \cap A = \prod_{i \in I} (U_{i} \cap A_{i})$ is nonempty if and only if each $U_{i} \cap A_{i}$ is nonempty. So, if $a \in \bar{A_{i}}$ for all $i \in I$ , then $a$ lies in the closure of $\prod_{i \in I} A_{i}$ . Conversely, if $a$ lies in the latter closure for all possible choices of $U$ , then $a \in \bar{A_{i}}$ for each $i \in I$ . □

The following tells us when a product of maps is continuous.

Proposition 2.2.13. §

For each $i$ in an indexing set $I$ , let $f_{i} : X_{i} \to Y_{i}$ be a function between topological spaces. Set $X = \prod_{i \in I} X_{i}$ and $Y = \prod_{i \in I} Y_{i}$ and $f = {(f_{i})}_{i \in I} : X \to Y$ . Then $f$ is continuous with respect to the product (resp., box) topology on $X$ and $Y$ if and only if each $f_{i}$ is continuous.

Proof.

Let $J$ be a finite subset of $I$ , and let $V_{j}$ be open in $Y_{j}$ for each $j \in J$ . Let $V = \prod_{i \in I} V_{i}$ where $V_{i} = X_{i}$ for $i \in I - J$ . Then $f^{- 1} (V) = \prod_{i \in I} f_{i}^{- 1} (V_{i})$ is open if and only if $f_{j}^{- 1} (V_{j})$ is open for every $j \in J$ . Thus in the product topology, $f$ is continuous if and only if each $f_{j}$ is continuous. In the box topology, we may simply replace $J$ by $I$ in the above argument. □

Example 2.2.14. §

Consider $ℝ^{I} = \prod_{i \in I} ℝ$ . The metric topology for the uniform metric given by

d (x, y) = \sup {\min {| x_{i} - y_{i} |, 1} ∣ i \in I},

is called the uniform topology. The reason for the name is as follows: the space of functions $f : I \to ℝ$ from a set $I$ to $ℝ$ is in bijection with $ℝ^{I}$ via the map which takes a function $f$ to ${(f (i))}_{i \in I}$ . A sequence of functions ${(f_{n})}_{n \geq 1}$ with $f_{n} : I \to ℝ$ converges uniformly to some $f : I \to ℝ$ if and only if the corresponding sequence in $ℝ^{I}$ converges to ${(f (i))}_{i \in I}$ in the uniform topology.

Example 2.2.15. §

If $I$ is finite, the product, uniform, and box topologies on $ℝ^{I}$ are all simply the Euclidean topology. If $I$ is infinite, then the box topology is strictly finer than the uniform topology, which is strictly finer than the product topology, as we next explain.

The product topology on $ℝ^{I}$ has a basis of open neighborhoods of $0$ consisting of a product $P_{𝜖}$ of open intervals $(- 𝜖, 𝜖)$ in finitely many coordinates and $ℝ$ in the others. The uniform topology has a basis of open balls $B_{𝜖} = B (𝜖, 0)$ of radius $𝜖 < 1$ . We have $B_{𝜖} \subset P_{𝜖}$ , but $B_{𝜖}$ contains no $P_{δ}$ . The box topology has a basis of open neighborhoods consisting of products of open intervals centered at $0$ of lengths depending on the coordinates. Inside $B_{𝜖}$ , we have the product of open intervals $(- \frac{𝜖}{2}, \frac{𝜖}{2})$ in every coordinate, which is open in the box topology. On the other hand, a product of open intervals centered at $0$ the infinimum of the lengths of which is $0$ contains no $B_{𝜖}$ .

Proposition 2.2.16. §

Let $J$ be an indexing set. If $J$ is infinite, then $ℝ^{J}$ with the box topology is not metrizable. The space $ℝ^{J}$ with the product topology is metrizable if and only if $J$ is countable.

Proof.

For the first statement, it suffices to consider the case that $J$ is countable, which we can then take to be the set of positive integers. Consider the subset $U = ℝ_{> 0}^{J}$ of $ℝ^{J}$ in the box topology. The $0 \in \bar{U}$ , as the reader can check. However, if ${(a_{n})}_{n \geq 1}$ is a sequence in $U$ and $a_{n, m}$ is its $m$ th coordinate, then the product $\prod_{n = 1}^{\infty} (- a_{n, n}, a_{n, n})$ is an open neighborhood of $0$ that does not contain any $a_{n}$ . By Proposition 1.6.21, the space $ℝ^{J}$ with the box topology is not metrizable.

Now, take the product topology on $ℝ^{J}$ with $J$ uncountable, and consider the set

A = {{(x_{j})}_{j \in J} ∣ x_{j} = 1 for all but finitely many j \in J} .

Then $0 \in \bar{A}$ , since any product of open intervals $(- 𝜖, 𝜖)$ in finitely many coordinates and $ℝ$ in the others clearly intersects $A$ . On other other hand, if ${(a_{n})}_{n \geq 1}$ is a sequence in $A$ , let $I$ be the subset of $J$ consisting of those $j \in J$ such that the $j$ -coordinate of some $a_{n}$ is not $1$ . This is a countable set, so $J \neq I$ . But then there exists $j \in J$ such that the $j$ -coordinate of $a_{n}$ is $1$ for all $n \geq 1$ , which means in particular the neighborhood of $0$ that is $ℝ$ in all coordinates but the $j$ -coordinate and $(- \frac{1}{2}, \frac{1}{2})$ in the $j$ -coordinate does not contain any $a_{n}$ .

For the product topology on $ℝ^{J}$ with $J = ℤ_{\geq 1}$ , the reader may check that we have a metric $d$ defined by

d (x, y) = \sup {\frac{1}{n} \min {| x_{n} - y_{n} |, 1} ∣ n \geq 1} .

□

2.3. Quotient spaces

Recall that we say that a function $f : X \to Y$ is surjective (or a surjection) if $f (X) = Y$ .

Definition 2.3.1. §

A quotient map $π : X \to Y$ of topological spaces $X$ and $Y$ is a surjection such that a subset $V$ of $Y$ is open if and only if $π^{- 1} (V)$ is open in $X$ .

The following is easily verified.

Lemma 2.3.2. §

A surjective map $π : X \to Y$ is a quotient map if and only if a subset $B$ of $Y$ is closed if and only if $π^{- 1} (B)$ is closed in $X$ .

Examples 2.3.3. §

Let $X$ be a topological space.

a.: The identity map ${id}_{X} : X \to X$ is a quotient map.
b.: Any constant map from a nonempty $X$ is a quotient map onto its singleton image.
c.: Let $X = \prod_{i \in I} X_{i}$ be a product of topological spaces. Then the projection maps $π_{i} : X \to X_{i}$ are quotient maps, as the reader can check.

Example 2.3.4. §

Let $S^{1}$ denote the unit open circle in $ℂ$ . Consider the map $f : ℝ \to S^{1}$ given by $f (x) = e^{2 𝜋𝑖𝑥}$ . Then $f$ is a quotient map.

The product map $F : ℝ^{2} \to S^{1} \times S^{1}$ given by $F (x, y) = (f (x), f (y))$ is a quotient map as well, realizing a torus (which has the shape of the surface of a donut) as a quotient of the plane. In fact, if we restrict this function to the square $X = {[0, 1]}^{2}$ , the resulting map $F : X \to S^{1} \times S^{1}$ is also a quotient map. It satisfies $F (0, y) = F (1, y)$ and $F (x, 0) = F (x, 1)$ for all $x, y \in [0, 1]$ . In particular,

F (0, 0) = F (1, 0) = F (0, 1) = F (1, 1) = (1, 1) .

One may think of the quotient map as identifying the left and right sides of the square with each other and the top and bottom sides of the square with each other, which in the process identifies the four corners with each other.

The following example illustrates that quotient maps need not be open maps.

Example 2.3.5. §

Let $X$ be the subspace of $ℝ^{2}$ consisting of points with $y$ -coordinate $0$ or $1$ . A subset of $X$ is open if and only if its intersection with each of the two lines in $X$ is open in the Euclidean topology.

Define $f : X \to ℝ$ by

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

Then $f$ is a quotient map. Let $U$ be the open subset of $X$ consisting of those $(x, y)$ with $y = 1$ , or with $y = 0$ and $x = 0$ . Then $f (U) = [0, \infty)$ , so $f$ is not open.

Now consider the restriction $f |_{U} : U \to ℝ$ . It is a continuous surjection. However, $f |_{U}^{- 1} ([0, \infty)) = {(x, 1) ∣ x \in ℝ}$ , but $[0, \infty)$ is not open, so $f |_{U}$ is not a quotient map.

Lemma 2.3.6. §

Let $X$ be a topological space, and let $π : X \to Y$ be a surjective function onto a set $Y$ . There exists a unique topology on $Y$ such that $π$ is a quotient map.

Proof.

The set of subsets of $Y$ such that $π^{- 1} (V)$ is open is easily seen to be a topology on $Y$ , since $π^{- 1}$ commutes with the taking of intersections and unions. Moreover, this is the only topology such that $π$ is a quotient map, since $π$ being a quotient map means exactly that the $V \subseteq Y$ such that $π^{- 1} (V)$ is open are the open sets. □

Given Lemma 2.3.6, we may make the following definition.

Definition 2.3.7. §

Let $X$ be a topological space and $π : X \to Y$ a surjective function to a set $Y$ . The quotient topology on $Y$ is the unique topology on $Y$ such that $π$ is a quotient map.

Definition 2.3.8. §

If $π : X \to Y$ is a quotient map between topological spaces, then $Y$ is said to be a quotient space of $X$ .

Example 2.3.9. §

Let $Y = {a, b}$ be the two-point set. Consider the function $f : ℝ \to Y$ given by $f (x) = a$ if $x < 0$ and $f (x) = b$ if $x \geq 0$ . The quotient topology on $Y$ is contains the sets $∅$ , ${a}$ and ${a, b}$ but not ${b}$ , as $f^{- 1} (b)$ is not open. Then $Y$ is not Hausdorff, though $ℝ$ is.

Proposition 2.3.10. §

Let $π : X \to Y$ be a quotient map. Let $B$ be a subset of $Y$ and $A = π^{- 1} (B)$ . Consider the map $p : A \to π (A)$ given by $p (a) = π (a)$ for $a \in A$ .

a.: If $A$ is open or closed in $X$ , then $p$ is a quotient map.
b.: If $π$ is open or closed, then $p$ is a quotient map.

Proof.

By definition, $p$ is continuous and surjective. If $V$ is a subset of $π (A)$ , then $p^{- 1} (V) = π^{- 1} (V)$ is open (resp., closed) in $X$ if and only if $V$ is open (resp., closed) in $X$ . If $A$ is open (resp., closed) in $X$ , then $p^{- 1} (V)$ is open (resp., closed) in $X$ if and only if $p^{- 1} (V)$ is open (resp., closed) in $A$ , and therefore, $p$ is a quotient map. Thus, we have part a.

As for part b, suppose that $π$ is open. If $V \subseteq B$ is such that $p^{- 1} (V)$ is open in $A$ , then $π^{- 1} (V) = p^{- 1} (V) = U \cap A$ for some open $U$ in $X$ , and

V = π (π^{- 1} (V)) = π (U \cap A) = π (U) \cap B,

the last step as $π^{- 1} (A) = B$ . Since $π$ is open, $π (U)$ is open in $Y$ , and therefore $π (U) \cap B$ is open in $B$ under the subspace topology. That is, $V$ is open in $B$ , so $p$ is a quotient map. The argument in the case that $π$ is closed proceeds in the same manner, replacing “open” by “closed” everywhere. □

The quotient map has a certain universality property, as expressed in the following theorem.

Theorem 2.3.11. §

Let $π : X \to Y$ be a quotient map, and let $f : X \to Z$ be any continuous map such that $f$ is constant on $π^{- 1} ({y})$ for all $y \in Y$ . Then there exists a unique function $g : Y \to Z$ such that $f = g \circ π$ , and $g$ is continuous.

Proof.

The function $g$ is determined by $g (y) = f (x)$ where $π (x) = y$ , which is well-defined by the constancy of $f$ on the nonempty set $π^{- 1} ({y})$ . Let $W$ be open in $Z$ , and note that $π^{- 1} (g^{- 1} (W)) = f^{- 1} (W)$ . The set $f^{- 1} (W)$ is open in $X$ , so as $π$ is a quotient map, $g^{- 1} (W)$ is open in $Y$ . Thus, $g$ is continuous. □

Example 2.3.12. §

Let $X$ be a topological space, and let $A$ be a subset. Let $X ∕ A = (X - A) ∐ {*}$ as sets, and define $π : X \to X ∕ A$ by

π (x) = {\begin{matrix} x & if x \in X - A \\ * & if a \in A . \end{matrix}

We give $Y$ the quotient topology. This is the space given by collapsing $A$ to a point. It may not be Hausdorff even if $X$ is: for instance, if we take $X = ℝ$ and $A = (0, \infty)$ , then $0$ is in every open neighborhood of $*$ in $Y$ . However, $Y$ will be Hausdorff if for every $x \notin A$ , there exist disjoint open sets $U$ containing $x$ and $V$ containing $A$ .

Example 2.3.13. §

Let $X$ be a topological space. The cone $C (X)$ on $X$ is the quotient space of (the cylinder) $X \times [0, 1]$ in which we collapse $X \times {0}$ to a point. If $X = S^{1}$ , this is homeomorphic to a cone in $ℝ^{3}$ , e.g.,

D = {(𝑟𝑥, 𝑟𝑦, r) ∣ r \in [0, 1], (x, y) \in S^{1}},

where $S^{1}$ is the unit circle in $ℝ^{2}$ . Here, the homeomorphism $D \to C (S^{1})$ takes $(𝑟𝑥, 𝑟𝑦, r)$ to the image of $((x, y), r) \in X \times [0, 1]$ in $C (S^{1})$ .

2.4. Disjoint unions

To give maps to a direct product of sets is to give a collection of maps to the individual sets, and if these sets are topological spaces, then with the product topology, we have seen in Proposition 2.2.13 that to give a continuous map to the product of topological spaces is to give a collection of continuous maps to the individual spaces. However, to give a function from a product of sets to a set is not the same as given a collection of maps from the individual set to the set. One might ask if there’s another set that does this, and the answer is yes, the disjoint union. In fact, when the individual sets are topological spaces, we can put a topology on the disjoint union so that we can make the same connection with continuous maps.

Definition 2.4.1. §

The disjoint union of a collection ${A_{i} ∣ i \in I}$ of sets is the set $∐_{i \in I} A_{i}$ that contains the $A_{i}$ as mutually disjoint subsets and is equal to the union $⋃_{i \in I} A_{i}$ .

Remark 2.4.2. §

To give a function $f : ∐_{i \in I} A_{i} \to B$ from a disjoint union of sets $A_{i}$ to a set $B$ is exactly to give a collection of maps $f_{i} : A_{i} \to B$ . These satisfy $f_{i} (a_{i}) = f (a_{i})$ for all $i \in I$ , so $f$ determines the $f_{i}$ and conversely.

Definition 2.4.3. §

Let ${X_{i} ∣ i \in I}$ be a collection of topological spaces. The disjoint union $∐_{i \in I} X_{i}$ of the spaces $X_{i}$ is the topological space with underlying set indicated disjoint union and with the topology under which a subset $U$ is open if and only if $U \cap X_{i}$ is open for all $i \in I$ .

Remark 2.4.4. §

Under this definition, we have continuous inclusion maps $ι_{i} : X_{i} ↪ ∐_{i \in I} X_{i}$ given by $ι_{i} (a_{i}) = a_{i}$ for $a_{i} \in X_{i}$ .

Proposition 2.4.5. §

Let $f_{i} : X_{i} \to Y$ be a collection of functions from topological spaces $X_{i}$ to a topological space $Y$ . The map $f : ∐_{i \in I} X_{i} \to Y$ that restricts to $f_{i}$ on $X_{i}$ for all $i$ is continuous if and only if every $f_{i}$ is continuous.

Proof.

Let $V$ be an open subset of $Y$ . Then $f^{- 1} (V) \cap X_{i} = f_{i}^{- 1} (V)$ , so $f^{- 1} (V)$ is open if and only if $f_{i}^{- 1} (V)$ is open for all $i$ . □

Lemma 2.4.6. §

Let $X = ∐_{i \in I} X_{i}$ be a disjoint union of topological spaces $X_{i}$ . Then each $X_{i}$ is open and closed in $X$ .

Proof.

We have $X_{i} \cap X_{j} = \emptyset$ if $i \neq j$ and $X_{i} \cap X_{i} = X_{i}$ , so $X_{i}$ is open in $X$ . As for its complement, we have

X - X_{i} = ⋃_{j \in I - {j}} X_{j},

which is open as a union of open sets. □

Example 2.4.7. §

Any union $A$ of the two parallel lines in the plane $ℝ^{2}$ is homeomorphic to $ℝ∐ℝ$ .

This example generalizes considerably.

Lemma 2.4.8. §

Let ${A_{i} ∣ i \in I}$ be a collection of subspaces of a space $X$ . Then the continuous map $∐_{i \in I} A_{i} \to X$ with restriction to $A_{i}$ the embedding of $A_{i}$ as a subspace of $X$ is a homeomorphism if the $A_{i}$ are mutually disjoint with union $X$ and every $A_{i}$ is open (and therefore closed) in $X$ .

Proof.

The continuous map $f$ in the statement is onto if and only if $⋃_{i \in I} A_{i} = X$ and is one-to-one if and only if $A_{i} \cap A_{j} = \emptyset$ for all $i \neq j$ . If $f$ is open, then every $X_{i}$ is open in $X$ (and then closed as well, since $A_{i}$ is the complement of the union of the $A_{j}$ for $j \neq i$ ). Conversely, if every $A_{i}$ is both open and closed in $X$ and $U$ is open in $∐_{i \in I} A_{i}$ , then $f (U) = ⋃_{i \in I} (U \cap A_{i})$ is open as a union of open sets. □

Lemma 2.4.8 justifies the following terminology.

Terminology 2.4.9. §

If $X$ is a topological space and ${A_{i} ∣ i \in I}$ is a collection of disjoint open subsets with union $X$ , then we say that $X$ is the disjoint union of the subspaces $A_{i}$ and write $X = ∐_{i \in I} A_{i}$ .

Examples 2.4.10. §

a.: The subspace $X = {(x, y) \in ℝ^{2} ∣ y \in {0, 1}}$ of $ℝ^{2}$ is the disjoint union of the lines $y = 0$ and $y = 1$ in the plane.
b.: Any discrete spaces is the disjoint union of its singleton subsets.

Example 2.4.11. §

Let ${X_{i} ∣ i \in I}$ be a collection of topological spaces, and choose $x_{i} \in X_{i}$ for each $i \in I$ . We have a quotient of $X = ∐_{i \in I} X_{i}$ given by collapsing the subset $A = {x_{i} ∣ i \in I}$ to a point. This will be a Hausdorff space if $X$ is. This space is called a one-point union of the spaces $X_{i}$ . In the case, for instance, that $X = S^{1} ∐ S^{1} ∐ \dots ∐ S^{1}$ , we get a finite collection of circles joined at a point.