Proposition 1 of § 1, no. 3 shows that $\overline{d}$ is a finite pseudometric on $\hat{X}$, and that the uniformity defined by $\overline{d}$ on $\hat{X}$ is the uniformity obtained by completion; since this latter uniformity is Hausdorff, $\overline{d}$ is a metric. Whenever we consider the completion of a metric space $X$ as a metric space, it is always to be understood that the metric on $\hat{X}$ is that obtained by extending the metric on $X$ by continuity.

For $d(x, A) = 0$ expresses the fact that the ball $V_a(x)$ meets $A$ whatever the value of $a > 0$; and this is equivalent to $x \in \overline{A}$.

Let $x$ and $y$ be any two points of $X$; then given any $\varepsilon > 0$ there exists $z \in A$ such that $d(y, z) \leq d(y, A) + \varepsilon$, and therefore

by the triangle inequality.

A fortiori $d(x, A) \leq d(x, y) + d(y, A) + \varepsilon$, and since $\varepsilon$ is arbitrary it follows that $d(x, A) \leq d(x, y) + d(y, A)$. Similarly we have

so that whence the result follows.

Let $a$ be any point of $\overline{X}$; then for each $k > \omega(a; f)$ there exists an open neighbourhood $V$ of $a$ such that $\delta(f(V \cap X)) \leq k$; for each $x \in V \cap \overline{X}$, $V$ is a neighbourhood of $x$ and therefore

which shows that $\omega$ is upper semi-continuous at the point $a$.

In order that $\omega(x; f) = 0$ at a point $x \in \overline{X}$ it is necessary and sufficient that for each $\varepsilon > 0$ there should exist a neighbourhood $V$ of $x$ such that $f(V \cap X)$ is contained in a ball of radius $\varepsilon$; if $x \in X$, this condition expresses the fact that $f$ is continuous at the point $x$ (with respect to $X$); if $x \in \overline{X} \cap \mathbf{C}X$, the image under $f$ of the trace on $X$ of the neighbourhood filter of $x$ in $Y$ is a Cauchy filter base on $X'$; in particular:

The condition is necessary, for (with the notation of no. 2) the entourages $V_{1/n} (n \geq 1)$ form a base of the filter of entourages of the uniformity of a metric space.

The condition is sufficient, for, if it is satisfied, the uniformity under consideration is defined by a single pseudometric, by Proposition 2 of § 1, no. 4; since the uniformity is Hausdorff, this pseudometric is a metric.

For if $(f_n)$ is a sequence of pseudometrics defining such a structure, the filter of entourages is generated by the countable family of sets $\overline{f}_n^1([0, 1/m])$, where $m$ and $n$ each run through the set of integers $> 0$.

For such a space is Hausdorff and its uniformity has a countable fundamental system of entourages (Chapter II, § 2, no. 6).

More generally:

Let $d$ be a metric compatible with the topology of a metrizable space $X$. If $A$ is a closed subset of $X$, it is the intersection of the open sets $V_{1/n}(A)$ [the set of all $x \in X$ such that $d(x, A) < 1/n$; cf. Proposition 2]. The second part of the proposition follows by taking complements.

We know already, from Chapter I, § 7, no. 3, that the condition is sufficient. To see that it is necessary, consider a countable fundamental system $(V_n)$ of neighbourhoods of $x$ such that $V_{n+1} \subset V_n$ for each $n$. If $x$ lies in the closure of $A$ then each $V_n$ meets $A$, and if $x_n$ lies in $V_n \cap A$, the sequence $(x_n)$ converges to $x$(**).

From Proposition 8 we deduce:

Let $\hat{X}$ be the completion of $X$. If there is a point $x \in \hat{X}$ which does not belong to $X$, then there is a sequence $(x_n)$ of points of $X$ which converges to $x$, and this is a non-convergent Cauchy sequence in $X$.

The condition is necessary, from Chapter I, § 7, no. 4, Proposition 9, Corollary 1. To show that it is sufficient, consider the filter $\mathscr{B}'$ of neighbourhoods of $f(a)$ in $X'$; the hypothesis implies that $\overline{f}^{-1}(\mathscr{B}')$ is coarser than every elementary filter associated with a sequence which converges to $a$, that is to say every elementary filter which converges to $a$; but

() See e.g., J. Dieudonné, Notes de Tératopologie (I), Revue scientifique (Revue rose), 1939, p. 39. (*) This proposition can still be valid in certain spaces in which at least one point does not have any countable fundamental system of neighbourhoods; for example, the space obtained by compactifying an uncountable discrete space by adjoining a point at infinity (Chapter I, § 9, no. 8, Theorem 4).

the intersection of these filters is the neighbourhood filter of $a$ (Chapter I, § 6, no. 8, Proposition 11). Hence the result.

Note that Propositions 8 and 10 are valid in any space $X$ in which every point has a countable fundamental system of neighbourhoods.

By transfer of structure we may assume that $a = 0$ and $b = 1$.

(i) Suppose first that $f = \varphi_A$, where $A$ is an open subset of $X$. Then the function $g_n$ defined by $g_n(x) = n \inf (d(x, X - A), 1/n)$ is continuous and $\geq 0$ on $X$; also $g_n(x) = f(x)$ when $x \in X - A$ and when

It follows immediately that $f = \sup_n (g_n)$.

(ii) In the general case consider, for each integer $n \geq 1$, the finite decreasing sequence of open sets

the function $g_n = \frac{1}{n} \sum_{k=1}^{n-1} \varphi_{A_k}$ is lower semi-continuous, and we have $0 \leq f(x) - g_n(x) \leq 1/n$ for all $n$; hence $f$ is the upper envelope of the sequence $(g_n)$. On the other hand, $g_n$ is a linear combination with positive coefficients of a finite number of characteristic functions of open sets and is therefore the upper envelope of a countable sequence $(h_{mn})_{m \geq 0}$ of continuous functions $\geq 0$, by (i); hence $f = \sup_{m,n} h_{mn}$. If we put we see that the sequence $(f_n)$ is an increasing sequence of continuous functions $\geq 0$, with $f$ as upper envelope, and which do not take the value 1, since $g_n \leq n - 1/n$.

a) X is of countable type. b) X has a countable dense subset. c) *X is homeomorphic to a subspace of the cube $\mathbf{I}^{\mathbf{N}}$, where $\mathbf{I}$ is the interval $[0, 1]$ in $\mathbf{R}$.

From the preceding remarks it is clear that c) implies a); a) implies b), for if $(U_n)$ is a countable base of the topology of X and $a_n$ is a point of $U_n$, the $a_n$ form a dense subset of X. Finally let us show that b) implies c). Let $(a_n)$ be a dense sequence of points of X, and for each $x \in X$ let $\varphi(x)$ be the point $(d(x, a_n))_{n \in \mathbf{N}}$ of $\mathbf{I}^{\mathbf{N}}$ ($d$ being a metric compatible with the topology of X, with respect to which the diameter of X is $\leq 1$). We shall show that $\varphi$ is a homeomorphism of X onto a subspace of $\mathbf{I}^{\mathbf{N}}$. Indeed, $\varphi$ is continuous, because each of the functions $x \to d(x, a_n)$ is continuous; and $\varphi$ is injective, because each point of X is the limit of some subsequence of $(a_n)$ (Proposition 8). Let B be a ball with centre $x_0$ and radius $r$ in X, and let $n$ be an integer such that

The image under $\varphi$ of the set W of points $x \in X$ such that

is by definition a neighbourhood of $\varphi(x_0)$ in $\varphi(X)$. But for each $x \in W$ we have $d(x, a_n) < d(x_0, a_n) + \frac{1}{3} r < \frac{2}{3} r$, whence

which shows that W is a neighbourhood of $x_0$ contained in B, and therefore that $\varphi$ is a homeomorphism of X onto $\varphi(X)$.

Note that for an arbitrary topological space X, property b) does not necessarily imply the existence of a countable base, even if X is compact and every point of X has a countable fundamental system of neighbourhoods (Exercise 13; cf. Chapter I, § 1, Exercise 7).

Let H be the subset of N consisting of indices n such that U_n is contained in at least one of the V_i; the sequence (U_n)_{n \in H} is a covering of X, because every point x \in X belongs to some V_i, and since V_i is open, there is an index n such that x \in U_n \subset V_i. Hence there is a mapping $\psi$ of H into I such that U_n \subset V_{\psi(n)} for each n \in H; taking J = $\psi(H)$, which is countable, the proposition is proved.

If we adjoin the hypothesis that X is complete we have a criterion for compactness of metric spaces.

From Proposition 14 we get a topological criterion for compactness, applicable to metrizable spaces:

Axiom (C) of Chapter I, § 9, no. 1 shows that the condition is necessary. To show sufficiency, let d be a metric compatible with the topology of X. We show first that the metric space X so defined is complete: every Cauchy sequence in X has a cluster point and is therefore convergent (Chapter II, § 3, no. 2, Proposition 5, Corollary 2); hence X is complete, by Proposition 9. Next we shall show that X is precompact; if this were not so, then by Proposition 14 there would exist a real number $\alpha > 0$ such that X could not be covered by any finite number of subsets of X of diameter $\leq \alpha$. We could then define by induction on n an infinite sequence $(x_n)$ of points of X by the condition $d(x_p, x_n) > \frac{1}{2} \alpha$ for all $p < n$; and such a sequence can have no cluster point, since every ball of radius $< \frac{1}{2} \alpha$ contains at most one point of the sequence.

Let d be a metric compatible with the topology of X. We shall show that the space $\overline{A}$ is compact, by applying the criterion of Proposition 15. Let $(x_n)$ be a sequence of points of $\overline{A}$; then for each index n there exists $y_n \in A$ such that $d(x_n, y_n) < 1/n$; the sequence $(y_n)$ has, by hypothesis, a cluster point $a \in X$, and a is also a cluster point of the sequence $x_i$, for if $y_m$ lies in the ball with centre $a$ and radius $1/n$, for some $m > n$, then $x_m$ lies in the ball with centre $a$ and radius $2/n$.

It should be remarked that Proposition 15 is not a consequence of the existence of a countable fundamental system of neighbourhoods at each point of $X$; there are examples of non-metrizable, non-compact spaces in which every point has a countable fundamental system of neighbourhoods and every infinite sequence of points has a cluster point (Exercise 15).

The condition is necessary. By Proposition 14, for each integer $n \geq 1$ there is a finite subset $A_n$ of $X$ such that the distance of $A_n$ from every point of $X$ is $\leq 1/n$; the countable set $A = \bigcup_n A_n$ is therefore dense in $X$, and the result follows from Proposition 12 of no. 8.

The condition is sufficient. Let $(U_n)$ be a countable base of the topology of $X$. Every neighbourhood of a point of the diagonal $\Delta$ of $X \times X$ therefore contains a neighbourhood of the form $U_n \times U_n$; applying the Borel-Lebesgue axiom to the compact subset $\Delta$ of $X \times X$, it follows that every neighbourhood of $\Delta$ contains a finite union of sets of the form $U_n \setminus U_n$, which is a neighbourhood of $\Delta$. Hence the neighbourhoods of $\Delta$ which are finite unions of sets of the form $U_n \times U_n$ form a fundamental system of entourages of the uniformity of $X$ (Chapter II, § 4, no. 1, Theorem 1), and the result therefore follows from Theorem 1 of no. 4.

a) The topology of $X$ has a countable base. b) $X'$ is metrizable. c) $X$ is metrizable and $\sigma$-compact.

$a \Longrightarrow b$: Let $U_n$ be a countable base of the topology of $X$. Each neighbourhood of a point $x \in X$ contains a compact neighbourhood of $x$, which in turn contains a neighbourhood of $x$ equal to some $U_n$. Hence the relatively compact $U_n$ form a base of the topology of $X$, and we may therefore suppose that all the $U_n$ are relatively compact. $X$ is therefore a countable union of compact sets $\overline{U}_n$, i.e. it is $\sigma$-compact; this implies that in $X'$ the point $\omega$ has a countable fundamental system $V_n$ of open neighbourhoods (Chapter I, § 9, no. 9, Proposition 15, Corollary 2). Hence each neighbourhood of a point $y \in X'$ contains either one of the $U_n$ or one of the $V_n$, which is a neighbourhood of $y$, and so the $U_n$ and the $V_n$ form a countable base of the topology of $X'$. Hence $X'$ is metrizable, by Proposition 16.

b) $\Longrightarrow c)$: If $X'$ is metrizable, then $\omega$ has a countable fundamental system of neighbourhoods, and therefore $X$ is $\sigma$-compact by Chapter I, § 9, no. 9, Proposition 15, Corollary 2.

c) $\Longrightarrow a)$: By hypothesis, there is an increasing sequence $(V_n)$ of relatively compact open sets which cover $X$ and are such that $\overline{V}_n \subset V_{n+1}$ (Chapter I, § 9, no. 9, Proposition 15). The subspace $\overline{V}_n$ is compact and metrizable and therefore has a countable base (Proposition 16), and therefore so does $V_n$. Let $(U_{mn})_{m \geq 1}$ be a base of the topology of $V_n$. For each $x \in X$ and each neighbourhood $W$ of $x$, there exists $n$ such that $x \in V_n$, hence there exists $m$ such that $x \in U_{mn} \subset V_n \cap W$. Hence the sets $U_{mn}$ ($m \geq 1, n \geq 1$) form a base of the topology of $X$.

Equivalently, if $f$ is a continuous mapping of a compact metrizable space $X$ into a Hausdorff space $X'$, then $f(X)$ is a metrizable subspace of $X'$ (Chapter I, § 9, no. 4, Theorem 2, Corollary 4).

Let $X$ be a compact metrizable space, and let $R$ be an equivalence relation on $X$ such that $X/R$ is Hausdorff. Then $X/R$ is compact (Chapter I, § 9, no. 4, Theorem 2), hence by Proposition 16 it is enough to show that the topology of $X/R$ has a countable base. To do this, we use the facts that $R$ is closed (Chapter I, § 10, no. 4, Proposition 8) and that the classes mod $R$ are compact. Let $\varphi$ be the canonical mapping of $X$ onto $X/R$, and let $(U_n)$ be a countable base of the topology of $X$. Let $z$ be any point of $X/R$ and let $V$ be a neighbourhood of $z$ in $X/R$; then $\overline{\varphi}^{-1}(V)$ is a neighbourhood in $X$ of the compact set $\overline{\varphi}^{-1}(z)$. If $x$ is any point of $\overline{\varphi}^{-1}(z)$, there is a set $U_n$ containing $x$ and contained in $\overline{\varphi}^{-1}(V)$, and therefore by the Borel-Lebesgue axiom there is a finite open covering $(U_{n_k})_{1 \leq k \leq r}$ of $\overline{\varphi}^{-1}(z)$ such that, if $W$ denotes $\bigcup_k U_{n_k}$, $W$ is a neighbourhood of $\overline{\varphi}^{-1}(z)$ contained in $\overline{\varphi}^{-1}(V)$. Since $R$ is closed, it follows that $\varphi(W)$ is a neighbourhood of $z$ in $X/R$, contained in $V$ (Chapter I, § 5, no. 4, Proposition 10). Let $\mathscr{B}$ denote the set of interiors of sets of the form $\varphi(W)$, where $W$ runs through the set $\mathscr{F}$ of all finite unions of sets of the form $U_n$; we have then shown that $\mathscr{B}$ is a base of the topology of $X/R$, and since $\mathfrak{F}$ is countable, so is $\mathscr{B}$.

Let $\mathscr{B}_1$ be the set of all open balls of radius $1/2$ in $X$. As $B$ runs through $\mathscr{B}_1$, the sets $\varphi(B)$ form an open covering of $K$, and therefore there exists a finite number of points $x_1, \ldots, x_m$ of $X$ such that the images under $\varphi$ of the open balls with radius $1/2$ and centre $x_i$ ($1 \leq i \leq m$) form an open covering of $K$. Let $H_1 = \{ x_1, \ldots, x_m \}$ and suppose that we have defined a finite set $H_i$, for $1 < i \leq n$, such that:

(i) $H_i \subset H_{i+1}$ and each point of $H_{i+1}$ is at a distance $< 1/2^i$ from $H_i$, for $1 \leq i \leq n - 1$;

(ii) the images under $\varphi$ of the open balls of radius $1/2^i$, with centres at the points of $H_i$, form an open covering of $K$, for $1 \leq i \leq n$.

Let $\mathscr{B}_{n+1}$ be the set of all open balls of radius $1/2^{n+1}$ whose centre $x$ is such that $d(x, H_n) < 1/2^n$ ($d$ being the metric on $X$). The properties of $H_n$ show that the sets $\varphi(B)$, for $B \in \mathscr{B}_{n+1}$, form an open covering of $K$; hence there is a finite set $L_{n+1} \subset X$ such that the images under $\varphi$ of the open balls of radius $1/2^{n+1}$ whose centre belongs to $L_{n+1}$ form an open covering of $K$. Taking $H_{n+1} = H_n \cup L_{n+1}$, we see that we can define inductively an infinite sequence $(H_n)$ of finite subsets of $X$ with properties (i) and (ii) above. Let $H = \bigcup_n H_n$, and let us show that $H$ is precompact. For each $p > 0$ and each point $z_{n+p} \in H_{n+p}$, there exists a sequence of points $z_{n+i} \in H_{n+i}$ ($0 \leq i \leq p - 1$) such that

it follows that $d(z_n, z_{n+p}) \leq \sum_{i=0}^{p-1} 1/2^{n+i} \leq 1/2^{n-1}$, and consequently $d(y, H_n) \leq 1/2^{n-1}$ for all $y \in H$, which proves our assertion. Since $X$ is complete, $\overline{H}$ is compact, hence $\varphi(\overline{H})$ is compact. Next, let us show that $K \subset \varphi(\overline{H})$. If $z \in K$, then by definition $d(\overline{H}_n, \overline{\varphi}(z)) \leq 1/2^n$ for all $n$, and therefore $d(\overline{H}, \overline{\varphi}(z)) = 0$; but $\overline{\varphi}(z)$ is closed and $\overline{H}$ is compact, so that this implies $\overline{H} \cap \overline{\varphi}(z) \neq \emptyset$ (no. 2, Remark following Proposition 3); hence the assertion. Thus if $K' = \overline{H} \cap \overline{\varphi}(K)$, then $K'$ is closed in $\overline{H}$ and therefore compact, and from what has been proved we have $\varphi(K') = K$.

Q.E.D.