§ 2. Continuous functions · General Topology

2. Continuous functions

1. Continuous functions

Definition 1

A mapping $f$ of a topological space $X$ into a topological space $X'$ is said to be continuous at a point $x_0 \in X$ if, given any neighbourhood $V'$ of $f(x_0)$ in $X'$ , there is a neighbourhood $V$ of $x_0$ in $X$ such that the relation $x \in V$ implies $f(x) \in V'$ .

Definition 1 may be restated in the following more intuitive form: to say that $f$ is continuous at the point $x_0$ means that $f(x)$ is as near as we please to $f(x_0)$ whenever $x$ is sufficiently near $x_0$ .

The relation "for each $x \in V, f(x) \in V'$ " is equivalent to $f(V) \subset V'$ or again to $V \subset \overline{f}^{-1}(V')$ ; in view of the neighbourhood axiom ( $V_I$ ), we see that Definition 1 is equivalent to the following: $f : X \to X'$ is said to be continuous at the point $x_0$ if, for each neighbourhood $V'$ of $f(x_0)$ in $X'$ , $\overline{f}^{-1}(V')$ is a neighbourhood of $x_0$ in $X$ . Moreover, it is sufficient that $\overline{f}^{-1}(V')$ is a neighbourhood of $x_0$ for each neighbourhood $V'$ belonging to a fundamental system of neighbourhoods of $f(x_0)$ in $X'$ ( $§ 1$ , no. 3).

Proposition 1

Let $f$ be a mapping of a topological space $X$ into a topological space $X'$ . If $f$ is continuous at $x$ , and if $x$ lies in the closure of a subset $A$ of $X$ , then $f(x)$ lies in the closure of $f(A)$ .

Let $V'$ be a neighbourhood of $f(x)$ in $X'$ . Since $f$ is continuous at $x$ , $\overline{f}^{-1}(V')$ is a neighbourhood of $x$ in $X$ . Hence $\overline{f}^{-1}(V')$ meets $A$ , from which it follows that $V'$ meets $f(A)$ , and therefore $f(x)$ is in the closure of $f(A)$ .

Proposition 2

Let $X, X', X''$ be three topological spaces; let $f$ be a mapping of $X$ into $X'$ , continuous at $x \in X$ ; let $g$ be a mapping of $X'$ into $X''$ , continuous at $f(x)$ . Then the composition $h = g \circ f : X \to X''$ is continuous at $x$ .

Let $V''$ be a neighbourhood of $h(x) = g(f(x))$ in $X''$ . Since $g$ is continuous at $f(x)$ it follows that $\overline{g}^{-1}(V'')$ is a neighbourhood of $f(x)$ in $X'$ . But $f$ is continuous at $x$ , hence $\overline{f}^{-1}(\overline{g}^{-1}(V'')) = \overline{h}^{-1}(V'')$ is a neighbourhood of $x$ in $X$ , and therefore $h$ is continuous at $x$ .

Definition 2

A mapping of a topological space $X$ into a topological space $X'$ is said to be continuous on $X$ (or just continuous) if it is continuous at each point of $X$ .

Example 1

The identity mapping of a topological space $X$ onto itself is continuous.

Example 2

A constant map of a topological space into a topological space is continuous.

Example 3

Every mapping of a discrete space into a topological space is continuous.

Theorem 1

Let $f$ be a mapping of a topological space $X$ into a topological space $X'$ . Then the following statements are equivalent:

a) $f$ is continuous in $X$ .

b) For every subset $A$ of $X$ , $f(\overline{A}) \subset \overline{f(A)}$ .

c) The inverse image under $f$ of every closed subset of $X'$ is a closed subset of $X$ .

d) The inverse image under $f$ of every open subset of $X'$ is an open subset of $X$ .

We have already seen that a) implies b) (Proposition 1). To show that b) implies c), let $F'$ be a closed subset of $X'$ and let $F = \overline{f^{-1}(F')}$ ; then by hypothesis $f(\overline{F}) \subset \overline{f(F)} \subset \overline{F'} = F'$ , hence $\overline{F} \subset \overline{f^{-1}(F')} = F \subset \overline{F}$ , so that $F = \overline{F}$ and $F$ is closed. By virtue of the relation $\overline{f^{-1}(A')} = \overline{f^{-1}(CA')}$ for every subset $A'$ of $X'$ , c) implies d). Finally, suppose that d) is satisfied. Let $x$ be any point of $X$ and let $V'$ be any neighbourhood of $f(x)$ in $X'$ ; then there is an open set $A'$ in $X'$ for which

f(x) \in A' \subset V'

and hence $x \in \overline{f^{-1}(A')} \subset \overline{f^{-1}(V')}$ . By hypothesis, $\overline{f^{-1}(A')}$ is open in $X$ , so that $\overline{f^{-1}(V')}$ is a neighbourhood of $x$ in $X$ . Thus d) implies a).

Remarks. 1) Let $\mathfrak{B}$ be a base (§ 1, no. 3) of the topology of $X'$ ; then for $f : X \to X'$ to be continuous, it is necessary and sufficient that $\overline{f^{-1}(U')}$ is open in $X$ for every $U' \in \mathfrak{B}$ .

Example

Let $a$ be any rational number. The mapping $x \to a + x$ of the rational line $\mathbf{Q}$ into itself is continuous on $\mathbf{Q}$ , for the inverse image under this mapping of an open interval $]b, c[$ is the open interval

]b - a, c - a[.

Likewise, the mapping $x \to ax$ is continuous on $\mathbf{Q}$ ; this is clear if $a = 0$ , for then $ax = 0$ for all $x$ ; if $a \neq 0$ then the inverse image under this mapping of the open interval $]b, c[$ is the open interval with end-points $b/a$ and $c/a$ .

2) The direct image of an open (resp. closed) set of $X$ under a continuous mapping $f : X \to X'$ is not necessarily open (resp. closed) in $X'$ (cf. § 5).

Example

\* The mapping $f : x \to 1/(1 + x^2)$ of $\mathbf{R}$ into itself is continuous, but $f(\mathbf{R})$ is the half-open interval $]0, 1]$ , which is neither open nor closed in $\mathbf{R}$ . \

Theorem 2

1) If $f : X \to X'$ and $g : X' \to X''$ are two continuous mappings, then $g \circ f : X \to X''$ is continuous.

2) For a bijection $f$ of a topological space $X$ onto a topological space $X'$ to be a homeomorphism, it is necessary and sufficient that $f$ and the inverse of $f$ are continuous (or, as is also said, that $f$ is bicontinuous).

The first assertion is an immediate consequence of Proposition 2; the second follows from Theorem 1, d) and the definition of a homeomorphism (§ 1, no. 1).

Remark 1

It is possible to have a continuous bijection of a topological space $X$ onto a topological space $X'$ which is not bicontinuous: for example, take $X'$ to be the rational line $\mathbf{Q}$ , and $X$ to be the set $\mathbf{Q}$ with the discrete topology; then the identity map $X \to X'$ is continuous but is not a homeomorphism.

Remark 2

To verify that a continuous bijection $f : X \to X'$ is a homeomorphism, it is enough to show that for each $x \in X$ and each neighbourhood $V$ of $x$ , $f(V)$ is a neighbourhood of $f(x)$ in $X'$ .

Remark 3

Let $X$ be a topological space, and for each $x \in X$ let $\mathfrak{B}(x)$ be the set of all neighbourhoods of $x$ . Let $x_0$ be a point of $X$ ; for each $x \in X$ , define a set $\mathfrak{B}_0(x)$ of subsets of $X$ as follows: $\mathfrak{B}_0(x_0) = \mathfrak{B}(x_0)$ , and if $x \neq x_0$ then $\mathfrak{B}_0(x)$ is to be the set of all subsets of $X$ which contain $x$ . It is immediately verified (§ 1, no. 2, Proposition 2) that the sets $\mathfrak{B}_0(x)$ are sets of neighbourhoods of points of $X$ for a topology on $X$ ; let $X_0$ denote the topological space thus obtained, and let $j : X_0 \to X$ denote the identity map, which is continuous but not in general bicontinuous. A mapping $f$ of $X$ into a topological space $X'$ is continuous at the point $x_0$ if and only if the composition $X_0 \xrightarrow{j} X \xrightarrow{f} X'$ is continuous on $X_0$ ; this follows immediately from the definitions.

2. Comparison of topologies

Theorem 2 of no. 1 shows that we may take the continuous mappings as morphisms of topological structures (Set Theory, Chapter IV, § 2, no. 1); from now on, we shall assume that we have made this choice of morphisms. In accordance with the general definitions (Set Theory, Chapter IV, § 2, no. 2), this allows us to define an ordering on the set of topologies on a given set $X$ :

Definition 3

Given two topologies $\mathfrak{T}_1, \mathfrak{T}_2$ on the same set $X$ , we say that $\mathfrak{T}_1$ is finer than $\mathfrak{T}_2$ (and that $\mathfrak{T}_2$ is coarser than $\mathfrak{T}_1$ ) if, denoting by $X_i$ the set $X$ with the topology $\mathfrak{T}_i$ ( $i = 1, 2$ ), the identity mapping $X_1 \to X_2$ is continuous. If in addition $\mathfrak{T}_1 \neq \mathfrak{T}_2$ , we say that $\mathfrak{T}_1$ is strictly finer than $\mathfrak{T}_2$ (and that $\mathfrak{T}_2$ is strictly coarser than $\mathfrak{T}_1$ ).

Two topologies, one of which is finer than the other, are said to be comparable.

The criteria for a mapping to be continuous (no. 1, Definition 1 and Theorem 1) give the following proposition:

Proposition 3

Given two topologies $\mathfrak{T}_1, \mathfrak{T}_2$ on a set $X$ , the following statements are equivalent: a) $\mathfrak{T}_1$ is finer than $\mathfrak{T}_2$ . b) For each $x \in X$ , each neighbourhood of $x$ for $\mathfrak{T}_2$ is a neighbourhood of $x$ for $\mathfrak{T}_1$ . c) For each subset $A$ of $X$ , the closure of $A$ in the topology $\mathfrak{T}_1$ is contained in the closure of $A$ in the topology $\mathfrak{T}_2$ . d) Every subset of $X$ which is closed in $\mathfrak{T}_2$ is closed in $\mathfrak{T}_1$ . e) Every subset of $X$ which is open in $\mathfrak{T}_2$ is open in $\mathfrak{T}_1$ .

Example

\* In Hilbert space $H$ consisting of sequences $x = (x_n)$ of real numbers such that

\|x\|^2 = \sum_{n=0}^\infty x_n^2 < +\infty,

the neighbourhoods of a point

x_0

in the strong topology on

H

are the sets which contain an open ball

\|x - x_0\| < \alpha

centred at

x_0

; the neighbourhoods of

x_0

in the weak topology on

H

are the sets containing a set defined by a relation of the form

\sup_{1 \leq i \leq n} |(x - x_0|a_i)| \leq 1

, where the

a_i

are points of

H

and

(x|y) = \sum_{n=0}^\infty x_n y_n

x = (x_n)

and

y = (y_n)

. Now if

\beta = \sup_{1 \leq i \leq n} \|a_i\|

, the relation

\|x - x_0\| \leq 1 / \beta

implies that

|(x - x_0|a_i)| \leq \|x - x_0\| \cdot \|a_i\| \leq 1

for

1 \leq i \leq n

; hence the strong topology on

H

is finer than the weak topology. On the other hand, given any finite family

(a_i)_{1 \leq i \leq n}

of points of

H

, there are points

x

H

such that

(x - x_0|a_i) = 0

for

1 \leq i \leq n

and such that

\|x - x_0\|

is arbitrarily large; this shows that the strong topology is strictly finer than the weak topology. \

Remark 1

In the ordered set of all topologies on a set $X$ , the topology in which the only open sets are $\varnothing$ and $X$ is the coarsest and the discrete topology is the finest.

Remark 2

The finer the topology, the more open sets, closed sets and neighbourhoods; the finer the topology, the smaller (resp. the larger) the closure (resp. the interior) of a set; the finer the topology, the fewer dense sets.

Remark 3

If $f : X \to X'$ is a continuous mapping, it remains continuous if the topology of $X$ is replaced by a finer topology and the topology of $X'$ is replaced by a coarser topology (no. 1, Theorem 2). In other words, the finer the topology of $X$ and the coarser the topology of $X'$ , the more continuous mappings there are of $X$ into $X'$ .

3. Initial topologies

Proposition 4

Let $X$ be a set, let $(Y_i)_{i \in I}$ be a family of topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $X$ into $Y_i$ . Let $\mathscr{S}$ be the set of subsets of $X$ of the form $\overline{f}_i^{-1}(U_i)$ ( $i \in I$ , $U_i$ open in $Y_i$ ), and let $\mathfrak{B}$ be the set of finite intersections of sets of $\mathscr{S}$ . Then $\mathfrak{B}$ is a base of a topology $\mathfrak{T}$ on $X$ which is the initial topological structure on $X$ for the family $(f_i)$ (Set Theory, Chapter IV, § 2, no. 3) and in particular is the coarsest topology on $X$ for which the mappings $f_i$ are continuous. More precisely, if $g$ is a mapping of a topological space $Z$ into $X$ , then $g$ is continuous at a point $z \in Z$ (X carrying the topology $\mathfrak{T}$ ) if and only if each of the functions $f_i \circ g$ is continuous at $z$ .

Let $\mathscr{D}$ be the set of all unions of sets belonging to $\mathfrak{B}$ ; clearly $\mathscr{D}$ satisfies axiom $(O_1)$ since formation of unions is associative; and $\mathscr{D}$ satisfies axiom $(O_{II})$ by reason of the definition of $\mathfrak{B}$ and the fact that finite intersection is distributive over arbitrary union [Set Theory, R § 4, formula (37)]. The set $\mathscr{D}$ is therefore the set of open subsets of $X$ for a topology $\mathfrak{T}$ of which $\mathfrak{B}$ is a base. We shall prove the last assertion of the proposition, which implies the others by reason of the general properties of initial structures (Set Theory, Chapter IV, § 2, no. 3, criterion CST 9). In the first place, the definition of $\mathscr{S}$ shows that the $f_i$ are continuous on $X$ (no. 1, Theorem 1); hence, if $g$ is continuous at $z$ , so are the mappings $f_i \circ g$ (no. 1, Proposition 2). Conversely, suppose that all the mappings $f_i \circ g$ are continuous at $z$ , and let $V$ be a neighbourhood of $g(z)$ in $X$ ; by definition, there is a finite subset $J$ of $I$ , and for each $i \in J$ an open subset $U_i$ of $Y_i$ such that $V$ contains the set $\bigcap_{i \in J} \overline{f}_i^{-1}(U_i)$ and $g(z)$ belongs to this set. It follows that

\overline{g}^{-1}(V) \supset \bigcap_{i \in J} \overline{g}^{-1}(\overline{f}_i^{-1}(U_i)),

and the hypothesis implies that each of the sets

\overline{g}^{-1}(\overline{f}_i^{-1}(U_i))

is a neighbourhood of

z

Z

; hence

\overline{g}^{-1}(V)

is also a neighbourhood of

z

Z

. This completes the proof.

Let $\mathfrak{B}_i$ be a base of the topology of $Y_i (i \in I)$ ; let $\mathscr{S}'$ denote the set of subsets of $X$ of the form $\overline{f}_i^{-1}(U_i)$ for $i \in I$ and $U_i \in \mathfrak{B}_i$ for each $i \in I$ ; if $\mathfrak{B}'$ is the set of finite intersections of sets of $\mathscr{S}'$ , it is evident that $\mathfrak{B}'$ is a base of the topology $\mathfrak{T}$ .

The general properties of initial structures (Set Theory, Chapter IV, § 2, no. 3, criterion CST 10) imply in particular the following transitivity property (the direct proof of which is quite straightforward):

Proposition 5

Let $X$ be a set, $(Z_i)_{i \in I}$ a family of topological spaces, $(J_\lambda)_{\lambda \in L}$ a partition of $I$ and $(Y_\lambda)_{\lambda \in L}$ a family of sets indexed by $L$ . Also for each $\lambda \in L$ let $h_\lambda$ be a mapping of $X$ into $Y_\lambda$ ; for each $\lambda \in L$ and each $i \in J_\lambda$ let $g_{i\lambda}$ be a mapping of $Y_\lambda$ into $Z_i$ , and put $f_i = g_{i\lambda} \circ h_\lambda$ . If each $Y_\lambda$ carries the coarsest topology for which the mappings $g_{i\lambda} (i \in J_\lambda)$ are continuous, then the coarsest topology on $X$ for which the $f_i$ are continuous is the same as the coarsest topology for which the $h_\lambda$ are continuous.

Example 1

Inverse image of a topology. Let $X$ be a set, $Y$ a topological space, $f$ a mapping of $X$ into $Y$ ; the coarsest topology $\mathfrak{T}$ on $X$ for which $f$ is continuous is called the inverse image under $f$ of the topology of $Y$ . It follows from Proposition 4 and the formulae for the inverse image of a union and an intersection [Set Theory, R, § 4, formulae (34) and (46)] that the open (resp. closed) sets in the topology $\mathfrak{T}$ are the inverse images under $f$ of the open (resp. closed) sets of $Y$ ; consequently, for each $x \in X$ , the sets $f^{-1}(W)$ , where $W$ runs through a fundamental system of neighbourhoods of $f(x)$ in $Y$ , form a fundamental system of neighbourhoods of $x$ in the topology $\mathfrak{T}$ . In § 3 we shall study, under the name of induced topology, the particular case in which $X$ is a subset of $Y$ and $f$ is the canonical injection $X \to Y$ ; $X$ , with the induced topology, is then called a subspace of $Y$ .

For a mapping $f$ of a topological space $X$ into a topological space $X'$ to be continuous it is necessary and sufficient that the topology of $X$ is finer than the inverse image under $f$ of the topology of $X'$ .

Example 2

Least upper bound of a set of topologies. Every family $(\mathfrak{T}_i)_{i \in I}$ of topologies on a set $X$ has a least upper bound $\mathfrak{T}$ in the ordered set of all topologies on $X$ , i.e. there exists a topology on $X$ which is coarsest among all the topologies on $X$ which are finer than each of the $\mathfrak{T}_i$ . To see this we may apply Proposition 4, taking $Y_i$ to be the set $X$ with the topology $\mathfrak{T}_i$ , and $f_i$ to be the identity map $X \to Y_i$ ; $\mathfrak{T}$ is the coarsest topology for which all the mappings $f_i$ are continuous.

Let $\mathfrak{S}$ be an arbitrary set of subsets of a set $X$ ; amongst the topologies $\mathfrak{T}$ on $X$ for which the sets of $\mathfrak{S}$ are open, there is a topology $\mathfrak{T}_0$ which is coarser than all the others and which is called the topology generated by $\mathfrak{S}$ . For each set $U \in \mathfrak{S}$ let $\mathfrak{T}_U$ be the topology whose open sets are $\varnothing, U$ and $X$ [it is clear that this set of subsets of $X$ satisfies (O_I) and (O_{II})]; then $\mathfrak{T}_0$ is just the least upper bound of the topologies $\mathfrak{T}_U$ . By Proposition 4, if $\mathfrak{B}$ is the set of finite intersections of sets belonging to $\mathfrak{S}$ , then $\mathfrak{B}$ is a base of the topology $\mathfrak{T}_0$ . We say that $\mathfrak{S}$ is a subbase of $\mathfrak{T}_0$ .

Example 3

Product topology. Let $(X_i)_{i \in I}$ be a family of topological spaces. The coarsest topology on the product set $X = \prod_{i \in I} X_i$ for which the projections $\mathrm{pr}_i : X \to X_i$ are continuous mappings is called the product of the topologies of the $X_i$ ; we shall study it in more detail in

4. Final topologies

Proposition 6

Let $X$ be a set, let $(Y_i)_{i \in I}$ be a family of topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $Y_i$ into $X$ . Let $\mathscr{D}$ be the set of subsets $U$ of $X$ such that $\overline{f}_i^{-1}(U)$ is open in $Y_i$ for each $i \in I$ ; then $\mathscr{D}$ is the set of open subsets of $X$ in a topology $\mathfrak{T}$ on $X$ which is the final structure on $X$ for the family $(f_i)$ (Set Theory, Chapter IV, § 2, no. 5), and in particular $\mathfrak{T}$ is the finest topology on $X$ for which the mappings $f_i$ are continuous. In other words, if $g$ is a mapping of $X$ into a topological space $Z$ , then $g$ is continuous ( $X$ carrying the topology $\mathfrak{T}$ ) if and only if each of the mappings $g \circ f_i$ is continuous.

It is immediately verified that $\mathscr{D}$ satisfies the axioms $(O_I)$ and $(O_{II})$ . [Set Theory, R, § 4, formulae (34) and (46)]. We shall prove the last assertion of the proposition, which implies the other assertions by reason of the general properties of final structures (Set Theory, Chapter IV, § 2, no. 5, criterion CST 18). It is clear that the $f_i$ are continuous in the topology $\mathfrak{T}$ , by the definition of $\mathscr{D}$ (no. 1, Theorem 1); hence if $g$ is continuous, so is each mapping $g \circ f_i$ (no. 1, Theorem 2). Conversely, suppose that each $g \circ f_i$ is continuous, and let $V$ be an open set in $Z$ ; by hypothesis, $\overline{f}_i^{-1}(\overline{g}^{-1}(V))$ is open in $Y_i$ for each $i \in I$ ; hence $\overline{g}^{-1}(V) \in \mathscr{D}$ , and the proof is complete.

Corollary

Under the hypotheses of Proposition 6, a subset $F$ of $X$ is closed in the topology $\mathfrak{T}$ if and only if $\overline{f}_i^{-1}(F)$ is closed in $Y_i$ for each $i \in I$ .

This follows from the definition of the open sets in the topology $\mathfrak{T}$ by taking complements.

The general properties of final structures (Set Theory, Chapter IV, § 2, no. 5, criterion CST 19) imply the following transitivity property (the direct proof of which is also straightforward):

Proposition 7

Let $X$ be a set, $(Z_i)_{i \in I}$ a family of topological spaces, $(J_\lambda)_{\lambda \in L}$ a partition of $I$ , and $(Y_\lambda)_{\lambda \in L}$ a family of sets indexed by $L$ . Also, for each $\lambda \in L$ let $h_\lambda$ be a mapping of $Y_\lambda$ into $X$ ; for each $\lambda \in L$ and each $i \in J_\lambda$ let $g_{\lambda i}$ be a mapping of $Z_i$ into $Y_\lambda$ , and put $f_i = h_\lambda \circ g_{\lambda i}$ . If each $Y_\lambda$ carries the finest topology for which the mappings $g_{\lambda i} (i \in J_\lambda)$ are continuous, then the finest topology on $X$ for which the $f_i$ are continuous is the same as the finest topology for which the $h_\lambda$ are continuous.

Examples

1) Quotient topology. Let $X$ be a topological space, $R$ an equivalence relation on $X$ , $Y = X/R$ the quotient set of $X$ with respect to the relation $R$ , $\varphi : X \to Y$ the canonical mapping. The finest topology on $Y$ for which $\varphi$ is continuous is called the quotient of the topology of $X$ by the relation $R$ ; we shall study it in more detail in § 3.

2) Greatest lower bound of a set of topologies. Every family $(\mathfrak{T}_i)_{i \in I}$ of topologies on a set $X$ has a greatest lower bound $\mathfrak{T}$ in the set of all topologies on $X$ , i.e. $\mathfrak{T}$ is the finest of all the topologies on $X$ which are coarser than each of the $\mathfrak{T}_i$ . To see this we may apply Proposition 6, taking $Y_i$ to be the set $X$ with the topology $\mathfrak{T}_i$ , and $f_i$ to be the identity mapping $Y_i \to X$ . If $\mathscr{D}_i$ is the set of subsets of $X$ which are open in the topology $\mathfrak{T}_i$ , then the set $\bigcap_{i \in I} \mathscr{D}_i$ is the set of subsets of $X$ which are open in $\mathfrak{T}$ . $\mathfrak{T}$ is also called the intersection of the topologies $\mathfrak{T}_i$ .

3) Sum of topological spaces. Let $(X_i)_{i \in I}$ be a family of topological spaces, $X$ the set which is the sum of the $X_i$ (Set Theory, Chapter II, § 4, no. 8, Definition 8); for each $i \in I$ , let $j_i$ be the canonical (injective) mapping of $X_i$ into $X$ . The finest topology $\mathfrak{T}$ on $X$ for which the mappings $j_i$ are continuous is called the sum of the topologies of the $X_i$ , and $X$ with this topology is said to be the sum of the topological spaces $X_i$ . Let us identify each of the $X_i$ with a subset of $X$ by means of $j_i$ ; then a set $A \subset X$ is open (resp. closed) in the topology $\mathfrak{T}$ if and only if each of the sets $A \cap X_i$ is open (resp. closed) in $X_i$ . In particular, each of the $X_i$ is both open and closed.

The following proposition generalizes the situation of Example 3:

Proposition 8

Let $X$ be a set, $(X_\lambda)_{\lambda \in L}$ a family of subsets of $X$ . Suppose each $X_\lambda$ carries a topology $\mathfrak{T}_\lambda$ such that, for each pair of indices $(\lambda, \mu)$ : 1) $X_\lambda \cap X_\mu$ is open (resp. closed) in each of the topologies $\mathfrak{T}_\lambda, \mathfrak{T}_\mu$ . 2) The topologies induced on $X_\lambda \cap X_\mu$ by $\mathfrak{T}_\lambda$ and $\mathfrak{T}_\mu$ coincide. Let $\mathfrak{T}$ be the finest topology on $X$ for which the injections $j_\lambda : X_\lambda \to X$ are continuous. Then, for each $\lambda \in L$ , $X_\lambda$ is open (resp. closed) in $X$ in the topology $\mathfrak{T}$ , and the topology induced by $\mathfrak{T}$ on $X_\lambda$ coincides with $\mathfrak{T}_\lambda$ .

In view of Proposition 6 and its corollary, it is enough to show that for each $\lambda$ and each subset $A_\lambda$ of $X_\lambda$ the following statements are equivalent: (i) $A_\lambda$ is open (resp. closed) in the topology $\mathfrak{T}_\lambda$ . (ii) For each $\mu \in L$ , $A_\lambda \cap X_\mu$ is open (resp. closed) in the topology $\mathfrak{T}_\mu$ .

It is clear that (ii) implies (i) by taking $\mu = \lambda$ . Conversely, if (i) is satisfied, then $A_\lambda \cap X_\mu$ is open (resp. closed) in $X_\lambda \cap X_\mu$ for the topology $\mathfrak{T}_{\lambda \mu}$ induced on $X_{\lambda} \cap X_{\mu}$ by $\mathfrak{T}_{\lambda}$ ; but $\mathfrak{T}_{\lambda \mu}$ is also the topology induced on $X_{\lambda} \cap X_{\mu}$ by $\mathfrak{T}_{\mu}$ ; hence $A_{\lambda} \cap X_{\mu}$ is also the intersection of $X_{\lambda} \cap X_{\mu}$ with a subset $B_{\mu}$ of $X_{\mu}$ which is open (resp. closed) in the topology $\mathfrak{T}_{\mu}$ ; since $X_{\lambda} \cap X_{\mu}$ is open (resp. closed) in $\mathfrak{T}_{\mu}$ , so is $A_{\lambda} \cap X_{\mu}$ . This completes the proof.

We remark that if the union of the $X_{\lambda}$ is different from $X$ , then the topology induced by $\mathfrak{T}$ on $X - \left( \bigcup_{\lambda \in L} X_{\lambda} \right)$ is discrete. For if $x \in X$ belongs to none of the $X_{\lambda}$ , then $\{ x \} \cap X_{\lambda} = \varnothing$ is open in each topology $\mathfrak{T}_{\lambda}$ and therefore $\{ x \}$ is open in the topology $\mathfrak{T}$ .

5. Pasting together of topological spaces

Let $(X_{\lambda})_{\lambda \in L}$ be a family of sets, and let $X$ be the set which is the sum of the $X_{\lambda}$ (Set Theory, Chapter II, § 4, no. 8, Definition 8); we shall identify each $X_{\lambda}$ with a subset of $X$ by means of the canonical injection $j_{\lambda} : X_{\lambda} \to X$ .

Let $R$ be an equivalence relation on $X$ such that each equivalence class of $R$ has at most one element in each $X_{\lambda}$ ; for each pair of indices $(\lambda, \mu)$ let $A_{\lambda \mu}$ be the subset of $X_{\lambda}$ consisting of the elements $x$ for which there is an element $y \in X_{\mu}$ which belongs to the equivalence class of $X$ . Clearly to each $x \in A_{\lambda \mu}$ corresponds a unique $y \in A_{\mu \lambda}$ which is congruent to $x$ mod $R$ ; the mappings $h_{\mu \lambda} : A_{\lambda \mu} \to A_{\mu \lambda}$ so defined satisfy the following conditions:

(i) For each $\lambda \in L$ , $h_{\lambda \lambda}$ is the identity mapping of $A_{\lambda \lambda} = X_{\lambda}$ .

(ii) For each triple of indices $(\lambda, \mu, \nu)$ of $L$ and each $x \in A_{\lambda \mu} \cap A_{\lambda \nu}$ , we have $h_{\mu \lambda}(x) \in A_{\mu \nu}$ and

h_{\nu \lambda}(x) = h_{\nu \mu}(h_{\mu \lambda}(x)).

Conversely, suppose that for each pair of indices $(\lambda, \mu)$ we are given a subset $A_{\lambda \mu}$ of $X_{\lambda}$ and a mapping $h_{\mu \lambda} : A_{\lambda \mu} \to A_{\mu \lambda}$ satisfying the conditions (i) and (ii) above. It follows first of all from (ii) applied to the triples $(\lambda, \mu, \lambda)$ and $(\mu, \lambda, \mu)$ that $h_{\lambda \mu} \circ h_{\mu \lambda}$ (resp. $h_{\mu \lambda} \circ h_{\lambda \mu}$ ) is the restriction of $h_{\lambda \lambda}$ (resp. $h_{\mu \mu}$ ) to $A_{\lambda \mu}$ (resp. $A_{\mu \lambda}$ ); hence we deduce from (i) that $h_{\lambda \mu}$ and $h_{\mu \lambda}$ are bijections which are inverses of each other. Now let $R \{ x, y \}$ be the relation "there exist $\lambda, \mu$ such that $x \in A_{\lambda \mu}$ , $y \in A_{\mu \lambda}$ and $y = h_{\mu \lambda}(x)$ ". It follows from (i) and from what precedes that $R$ is reflexive and symmetric; on the other hand, if $x \in A_{\lambda \mu}$ ,

y = h_{\mu \lambda}(x) \in A_{\mu \lambda} \cap A_{\mu \nu} \quad \text{and} \quad z = h_{\nu \mu}(y),

then also $x = h_{\lambda \mu}(y)$ and therefore, by (ii), $x \in A_{\lambda \mu} \cap A_{\lambda \nu}$ ; the relation (i) thus shows that $R$ is transitive, and therefore $R$ is an equivalence relation on $X$ . It follows also from (i) and from the definition of $R$ that each equivalence class mod $R$ has at most one element in each of the sets $X_\lambda$ , and that $A_{\lambda \mu}$ is the set of all $x \in X_\lambda$ for which there is an element $y \in X_\mu$ congruent to $x$ mod $R$ . We say that the quotient set $X/R$ is obtained by pasting together the $X_\lambda$ along the $A_{\lambda \mu}$ by means of the bijections $h_{\mu \lambda}$ . If $\varphi : X \to X/R$ is the canonical mapping, the restriction of $\varphi$ to each $X_\lambda$ is a bijection of $X_\lambda$ onto $\varphi(X_\lambda)$ .

Now suppose that each $X_\lambda$ is a topological space, and let $\mathfrak{T}_\lambda$ be its topology. Let $\mathfrak{T}$ be the finest topology on the set $X/R$ for which the mappings $\varphi \circ j_\lambda$ are continuous; $\mathfrak{T}$ is the quotient by $R$ of the topology on $X$ which is the sum of the topologies $\mathfrak{T}_\lambda$ . We say that the topological space $X/R$ (with the topology $\mathfrak{T}$ ) is obtained by pasting together the topological spaces $X_\lambda$ along the $A_{\lambda \mu}$ by means of the bijections $h_{\mu \lambda}$ . The open (resp. closed) subsets of $X/R$ are thus the canonical images of the subsets $B$ of $X$ which are saturated with respect to $R$ and are such that $B \cap X_\lambda$ is open (resp. closed) in $X_\lambda$ for each $\lambda \in L$ .

Since the restriction of $\varphi$ to each $X_\lambda$ is a bijection onto the subset $X'_\lambda = \varphi(X_\lambda)$ of $X/R$ , we can transport the topology $\mathfrak{T}_\lambda$ to $X'_\lambda$ by means of this bijection, so that $X'_\lambda$ carries a topology $\mathfrak{T}'_\lambda$ ; and the topology $\mathfrak{T}$ on $X/R$ is the finest for which the canonical injections $X'_\lambda \to X/R$ are continuous. In general, the topology induced by $\mathfrak{T}$ on $X'_\lambda$ is coarser than $\mathfrak{T}'_\lambda$ , but not identical with the latter; even if the $h_{\mu \lambda}$ are homeomorphisms (§ 3, Exercise 15). However, it follows from no. 4, Proposition 8 that, with the preceding notation:

Proposition 9

Suppose that the* $h_{\mu \lambda}$ are homeomorphisms and that each $A_{\lambda \mu}$ is open (resp. closed) in $X_\lambda$ ; then each $\varphi(X_\lambda)$ is open (resp. closed) in $X/R$ and the restriction of $\varphi$ to $X_\lambda$ is a homeomorphism of $X_\lambda$ onto the subspace $\varphi(X_\lambda)$ of $X/R$ .

Exercises

See the exercises for § 2.