$(O'_v)$ If $A$ and $B$ are any two disjoint closed subsets of $X$, then there exist two disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$.

It follows immediately that $(O_v)$ implies $(O'_v)$, for if $f$ is a continuous mapping of $X$ into $[0, 1]$ which is equal to $0$ on $A$ and to $1$ on $B$, then the open sets $\overline{f}^{-1}([0, 1/2[)$ and $\overline{f}^{-1}(]1/2, 1])$ contain $A$ and $B$ respectively and do not intersect.

To prove the converse, notice first that $(O'_v)$ is equivalent to the following axiom:

$(O''_v)$ Given any closed set $A$ and any open neighbourhood $V$ of $A$, there exists an open neighbourhood $W$ of $A$ such that $\overline{W} \subset V$.

If there is a continuous mapping $f : X \to [-1, +1]$ which is equal to $-1$ on $A$ and to $+1$ on $B$, and if we put $U(t) = \overline{f}^{-1}([-1, t[)$ for each $t \in [0, 1]$, then we have defined a family of open sets in $X$, indexed by $[0, 1]$, such that (i) $A \subset U(0)$, (ii) $B \subset \mathbf{C}U(1)$ and (iii) for each pair of real numbers $t, t'$ such that $0 \leq t < t' \leq 1$ we have

for $U(t)$ is contained in the closed set $\overline{f}^{-1}([-1, t])$. Conversely, suppose that we have defined a family $(U(t))$ of open sets $(0 \leq t \leq 1)$ with these three properties (i), (ii) and (iii). For each $x \in X$, put $g(x) = 1$ if $x \in \mathbf{C}U(1)$, and if $x \in U(1)$ let $g(x)$ be the greatest lower bound of the values of $t$ such that $x \in U(t)$. Clearly $0 \leq g(x) \leq 1$ for each $x \in X$, $g(x) = 0$ on $A$, $g(x) = 1$ on $B$; also $g$ is continuous on $X$, for if we put $g(x) = a$, we have $|g(y) - g(x)| \leq \varepsilon$ for all $y \in U(a + \varepsilon) \cap \overline{\mathbf{C}U}(a - \varepsilon)$, which is a neighbourhood of $x$ by (i) [with the conventions that $U(a + \varepsilon) = X$ if $a + \varepsilon > 1$, and $U(a - \varepsilon) = \varnothing$ if $a - \varepsilon < 0$].

Hence Theorem 1 will be proved if we can define a family $(U(t))$ of open sets satisfying conditions (i), (ii) and (iii) above; to do this we use Axiom $(\mathrm{O}'_V)$. Take $U(1) = \mathbf{C}B$; since $A \subset U(1)$ there exists an open set $U(o)$ such that $A \subset U(o)$ and $U(o) \subset U(1)$ by $(\mathrm{O}'_V)$. Suppose then that for each dyadic number $k/2^n$ ($k = 0, 1, \ldots, 2^n$) we have defined an open set $U(k/2^n)$, these sets being such that $\overline{U}(k/2^n) \subset U((k+1)/2^n)$ for $0 \leq k \leq 2^n - 1$. For each dyadic number

there exists by $(\mathrm{O}'_V)$ an open set $U((2k + 1)/2^{n+1})$ such that and Hence for each dyadic number $r$ such that $0 \leq r \leq 1$ we can define an open set $U(r)$, such that $A \subset U(o)$, $B \subset \mathbf{C}U(1)$, and for each pair of dyadic numbers $r, r'$ such that $0 \leq r < r' \leq 1$.

Now define, for each real number $t \in [0, 1]$,

by (2), this definition agrees with the preceding one for $t$ dyadic; also, if $0 \leq t < t' \leq 1$, then there exist two dyadic numbers $r, r'$ such that $t \leq r < r' \leq t'$; by (2) we have $U(t) \subset U(r)$, hence this proves (1) and therefore completes the proof.

Theorem 1 will enable us to show that two important categories of topological spaces are normal. In the first place:

A compact space satisfies axiom $(\mathrm{O}'_V)$, by Proposition 2 of Chapter I, § 9, no. 2.

As to locally compact spaces, every point of such a space has a compact neighbourhood, which is a normal subspace; but there are examples of locally compact spaces which are not normal (cf. Exercises 9 and 26, and § 5, Exercise 15).

Let $X$ be a metrizable space and let $d$ be a metric compatible with the topology of $X$. Let $A, B$ be two disjoint closed subsets of $X$; since the functions $d(x, A)$ and $d(x, B)$ are continuous, the set $U$ (resp. $V$) of points $x$ such that $d(x, A) < d(x, B)$ [resp. $d(x, B) < d(x, A)$] is open; clearly $A \subset U$ and $B \subset V$ and $U \cap V = \varnothing$, hence Axiom $(O'_V)$ is satisfied.

It is easy to see that $(O''_V)$ implies $(O_V)$; for if $B$ and $C$ are two disjoint closed subsets of $X$, then the function which is equal to 0 on $B$ and equal to $i$ on $C$ is defined and continuous on the closed set $B \cup C$, hence by $(O''_{V})$ has a continuous extension $f$ to $X$. If $g = \inf (f^{+}, 1)$, then $g$ is continuous on $X$, takes its values in $[0, 1]$ and is equal to $o$ on $B$ and to $i$ on $C$.

Let us show conversely that $(O_{V})$ implies $(O''_{V})$. Since $\overline{\mathbf{R}}$ and the interval $[ -i, +i ]$ are homeomorphic, we need consider only the case where the continuous mapping $f : A \to \overline{\mathbf{R}}$ takes its values in $[ -i, +i ]$. We shall construct an extension $g$ of $f$ to $X$ by forming a sequence $(g_{n})$ of continuous functions on $X$, such that the sequence $(g_{n}(x))$ converges for all $x \in X$ to a point of the interval $[ -i, +i ]$; this limit will, by definition, be the value of $g$ at $x$, and it will follow from the choice of the $g_{n}$ that the function $g$ satisfies the required conditions.

The definition of the $g_{n}$ rests on the following lemma:

Let $H$ be the set of all $x \in A$ such that $-i \leq u(x) \leq -i/3$, and let $K$ be the set of all $x \in A$ such that $i/3 \leq u(x) \leq i$; $H$ and $K$ are closed in $A$, and therefore in $X$, and do not intersect; hence by $(O_{V})$ there is a continuous mapping $v$ of $X$ into $[ -i/3, +i/3 ]$ which is equal to $-i/3$ on $H$ and to $+i/3$ on $K$. The mapping $v$ satisfies the conditions of the lemma.

We now define the functions $g_{n}$ by induction. Applying the lemma with $u = f$, we define $g_{0}$ to be a continuous mapping of $X$ into $[ -i/3, +i/3 ]$ such that $|f(x) - g_{0}(x)| \leq 2/3$ for all $x \in A$. Suppose now that a continuous mapping $g_{n}$ of $X$ into the interval

has been defined, such that $|f(x) - g_{n}(x)| \leq (\frac{2}{3})^{n+1}$ for all $x \in A$. Applying the lemma to the function $u(x) = (\frac{2}{3})^{n+1} (f(x) - g_{n}(x))$, we see that there exists a continuous mapping $h_{n+1}$ of $X$ into the interval

such that

for all $x \in A$; the induction is completed by taking $g_{n+1} = g_{n} + h_{n+1}$, since this function satisfies the inequality $|g_{n+1}(x)| \leq 1 - (2/3)^{n+2}$ for all $x \in X$, by virtue of the definition of $h_{n+1}$.

From this definition it follows that, for $m \geq p$ and $n \geq p$, we have

at each point $x \in X$; hence the sequence $(g_n(x))$ is a Cauchy sequence for each $x \in X$, and therefore converges to a point $g(x)$ of the interval $[ -1, +1 ]$; and since $f(x) - g_n(x)$ tends to 0 for all $x \in A$ as $n \to \infty$, $g$ is an extension of $f$ to $X$. It remains therefore only to show that $g$ is continuous on $X$.

Now let $x$ be any point of $X$; then, given any $\varepsilon > 0$, there exists an integer $n_0$ such that $|g_m(y) - g_n(y)| \leq \varepsilon$ for all $y \in X$ and all $m \geq n_0$ and all $n \geq n_0$; hence, letting $m$ tend to $+\infty$, we have

Let $V$ be a neighbourhood of $x$ such that $|g_n(y) - g_n(x)| \leq \varepsilon$ for all $y \in V$; then, for each $y \in V$ we shall have

which shows that $g$ is continuous at $x$, and completes the proof of Theorem 2. (The last part of the proof uses, in a particular case, the idea of uniform convergence, which we shall study in general in Chapter X, no. 1.)

First consider the case in which $f(x) \geq 0$ for all $x \in A$; then there is a continuous extension $g_1$ of $f$ to $X$, taking its values in $[0, +\infty]$. If we put $B = \overline{g_1}(+\infty)$, then $B$ is closed and by hypothesis does not meet $A$; the function $h$ which is equal to $f$ on $A$ and to 0 on $B$ is therefore a continuous function on the closed set $A \cup B$. Let $g_2$ be a continuous extension of $h$ to $X$, again taking its values in $[0, +\infty]$; the function $g = \inf(g_1, g_2)$ is then a continuous extension of $f$ to $X$, whose values are $\geq 0$ and finite at every point of $X$.

To pass to the general case, it is enough to remark that, if $f$ is finite and continuous on $A$, then so are $f^+$ and $f^-$; extending $f^+$ and $f^-$ to $X$ by finite continuous functions $g_1$ and $g_2$ respectively, we see that the function $g_1 - g_2$ is finite and continuous on $X$ and extends $f$.

Well-order the index set $I$ (Set Theory, Chapter III, § 2, no. 3, Theorem 1). We shall define a family $(B_i)_{i \in I}$ of open sets in $X$, by transfinite induction, such that (i) $\overline{B}_i \subset A_i$ for each $i \in I$; (ii) for each $i \in I$, the family formed by the $B_\lambda$ such that $\lambda \leq i$ and by the $A_\lambda$ such that $\lambda > i$ is an open covering of $Y$. Suppose that we have defined the $B_i$ for $i < \gamma$ so that (i) and (ii) are satisfied for all $i < \gamma$, and let us show that we can define $B_\gamma$ in such a way that (i) and (ii) are also satisfied for $i = \gamma$. Let us first show that the $B_i$ for which $i < \gamma$ and the $A_i$ for which $i \geq \gamma$ form a covering of $Y$. By hypothesis, for each $x \in Y$ there is only a finite number of indices $\lambda \in I$ such that $x \in A_\lambda$, say $\lambda_1 < \lambda_2 < \cdots < \lambda_n$; let $\lambda_h$ be the greatest of the $\lambda_i$ such that $\lambda_i < \gamma$; if $h < n$ we have $x \in A_{\lambda_n}$ and $\lambda_n \geq \gamma$, and if $h = n$ the inductive hypothesis shows that $x$ belongs to some $B_\lambda$ such that $\lambda \leq \lambda_n < \gamma$, and our assertion follows.

Now put $C = (\complement Y) \cup \left( \bigcup_{i < \gamma} B_i \right) \cup \left( \bigcup_{i > \gamma} A_i \right)$; $C$ is open, and from what has been said we have $\complement A_\gamma \subset C$; by virtue of Axiom $(O''_V)$ for normal spaces, there is therefore an open set $V$ such that $\complement A_\gamma \subset \overline{V} \subset V \subset C$. If we put $B_\gamma = \complement \overline{V}$, we have $\overline{B}_\gamma \subset \complement V \subset A_\gamma$ and $B_\gamma \cup C = X$, so that the $B_i$ such that $i \leq \gamma$ and the $A_i$ such that $i > \gamma$ cover $Y$.

By Theorem 3 there exists an open covering $(B_i)_{i \in I}$ of $X$ such that $\overline{B}_i \subset A_i$ for each $i \in I$, and it is clear that the covering $(B_i)$ is locally finite. By axiom $(O''_V)$, for each $i \in I$ there exists an open set $C_i$ such that $\overline{B}_i \subset C_i \subset \overline{C}_i \subset A_i$. By axiom $(O_V)$, for each $i \in I$ there exists a continuous mapping $g_i$ of $X$ into $[0,1]$, such that $g_i(x) = 1$ on $\overline{B}_i$ and such that the support of $\overline{g}_i$ is contained in $\overline{C}_i$, and therefore contained in $A_i$. Since $(B_i)$ is a covering of $X$, we have $\sum_{i \in I} g_i(x) > 0$ for each $x \in X$; if we put

for all $i \in I$ and all $x \in X$, then the $f_i$ form a continuous partition of unity subordinate to the covering $(A_i)$.

The family of sets consisting of $CF$ and the $A_i$ is a locally finite open covering of $X$. There is therefore a continuous partition of unity subordinate to this covering, consisting of a family $(f_i)_{i \in I}$ such that $\operatorname{Supp}(f_i) \subset A_i$ for each $i \in I$, and a function $g$ whose support is contained in the complement of $F$. The family $(f_i)$ clearly satisfies the required conditions.

This is a consequence of the following lemma:

Assuming the truth of this lemma for the moment, we can apply it to the case where $B$ consists of a single point, because $X$ is Hausdorff, and it shows then that $X$ is regular. We can then apply Lemma 2 again to any two disjoint closed subsets of $X$, and this shows that Axiom $(O'_V)$ is satisfied.

To prove the lemma, consider the open covering of $X$ consisting of $CA$ and the $V_x$, where $x \in A$; let $(T_i)_{i \in I}$ be a locally finite open refinement of this covering. Then, by definition, if $A \cap T_i \neq \emptyset$ there exists $x_i \in A$ such that $T_i \subset V_{x_i}$. Let $T$ be the open set which is the union of the $T_i$ which meet $A$, and let us show that there is an open neighbourhood $U$ of $B$ which does not meet $T$. For each $y \in B$ there is an open neighbourhood $S_y$ of $y$ which meets only a finite number of sets $T_i$; let $J$ be the finite subset of $I$ consisting of those indices $i$ such that $T_i$ meets both $S_y$ and $A$; if we put $U_y = S_y \cap \bigcap_{i \in J} W_{x_i}$, then $U_y$ is an open neighbourhood of $y$ which meets none of the $T_i$ which meet $A$, and hence $U_y \cap T = \emptyset$. Let $U = \bigcup_{y \in B} U_y$; then $U'$ is an open neighbourhood of $B$ which does not meet $T$, and the lemma is proved.

There exist normal spaces which are not paracompact (Exercise 19).

Let $(U_\lambda)_{\lambda \in L}$ be a locally finite open covering of $X$ which refines the covering $(A_i)_{i \in I}$; then there is a mapping $\varphi : L \to I$ such that $U_\lambda \subset A_{\varphi(\lambda)}$ for each $\lambda \in L$. By Propositions 3 and 4, there exists a continuous partition of unity $(g_\lambda)_{\lambda \in L}$ subordinate to $(U_\lambda)$. For each $i \in I$, put

this sum is defined and continuous since the supports of the $g_\lambda$ form a locally finite covering; moreover, the union $B_i$ of the supports of the $g_\lambda$ such that $\varphi(\lambda) = i$ is closed, by Proposition 4 of Chapter I, § 1, no. 6, and is contained in $A_i$. Since we have $f_i(x) = 0$ whenever $x \in \mathbf{C}B_i$, the support of $f_i$ is contained in $B_i$, and therefore in $A_i$. On the other hand, the family $B_i$ is locally finite, because for each $x \in X$ there is a neighbourhood $V$ of $x$ and a finite subset $H$ of $I$ such that $V \cap U_\lambda = \emptyset$ for all $\lambda \notin H$, and it follows therefore that $V \cap B_i = \emptyset$ for all $i \notin \varphi(H)$. Finally, for each $x \in X$ we have

and the proof is complete.

By Proposition 16 of Chapter I, § 9, no. 10, any closed subspace of X is paracompact; the corollary therefore follows from Proposition 4 and Axiom $(O_v')$.

The theorem is a consequence of the following four lemmas.

Let $d$ be a metric on X compatible with its topology. For each $\alpha \in A$ and each integer $n$, let $F_{n\alpha}$ denote the set of all $x \in U_\alpha$ such that $d(x, X - U_\alpha) \geq 2^{-n}$. Since $X - U_\alpha$ is closed, we have $U_\alpha = \bigcup_n F_{n\alpha}$.

Well-order the set $A$; for each $\alpha \in A$ and each integer $n$, let $G_{n\alpha}$ be the set of all $x \in F_{n\alpha}$ such that $x \notin F_{n+1,\beta}$ for all $\beta < \alpha$, and let $V_{n\alpha}$ be the set of all $y \in X$ such that $d(y, G_{n\alpha}) > 2^{-n-3}$. $V_{n\alpha}$ is clearly an open set; on the other hand, $V_{n\alpha} \subset U_\alpha$, because for each $y \in V_{n\alpha}$ there exists $x \in G_{n\alpha}$ such that $d(x, y) \leq 2^{-n-1}$, and since $x \in F_{n\alpha}$, we have

so that $y \in U_\alpha$. Furthermore, for each $x \in X$ let $\alpha$ be the smallest index in $A$ such that $x \in U_\alpha$; then there is an integer $n$ such that $x \in F_{n\alpha}$, and the definition of $\alpha$ implies that $x \in G_{n\alpha}$, so that $x \in V_{n\alpha}$. This shows that if we put $\mathfrak{S}_n = (V_{n\alpha})_{\alpha \in A}$, then $\mathfrak{S} = \bigcup_n \mathfrak{S}_n$ is an open covering of $X$ which refines $\mathscr{R}$; thus it remains to be shown that each of the families $\mathfrak{S}_n$ is locally finite. To this end we shall first show that $d(G_{n\alpha}, G_{n\beta}) \geq 2^{-n-1}$ if $\alpha \neq \beta$. Suppose that $\beta < \alpha$; then if $x \in G_{n\alpha}$ and $y \in F_{n\beta}$ we have $x \notin F_{n+1,\beta}$ by definition, hence $d(x, X - U_\beta) < 2^{-n-1}$ and $d(y, X - U_\beta) \geq 2^{-n}$, and therefore $d(x, y) \geq 2^{-n-1}$; since $G_{n\beta} \subset F_{n\beta}$, the assertion follows. From this it follows immediately, using the definition of $V_{n\alpha}$ and $V_{n\beta}$, that $d(V_{n\alpha}, V_{n\beta}) \geq 2^{-n-2}$. From this last inequality we deduce that, for each $z \in X$, the open ball with centre $z$ and radius $2^{-n-3}$ meets at most one set of the family $\mathfrak{S}_n$; hence $\mathfrak{S}_n$ is a locally finite family, and the proof is complete.