§ 1. Generation of a uniformity by a family of pseudometrics; uniformizable spaces

1. Generation of a uniformity by a family of pseudometrics; uniformizable spaces

1. Pseudometrics

Definition I. *If X is a set, a pseudometric on X is any mapping f of X × X into the interval [0, +∞] of the extended real line $\overline{\mathbf{R}}$ which satisfies the following conditions:

(EC_I) For all $x \in X, f(x, x) = 0$ . (EC{II}) For all $x \in X$ and all $y \in X, f(x, y) = f(y, x)$ (symmetry). (EC{III}) *For all $x, y, z$ in X,

f(x, y) \leq f(x, z) + f(z, y)

(triangle inequality).

Example 1

On real number space $\mathbf{R}^n$ , Euclidean distance (Chapter VI, § 2, no. 1) is a pseudometric.

Example 2

If X is any set, the function f defined on $X \times X$ by the conditions $f(x, x) = 0$ for all $x \in X, f(x, y) = +\infty$ if $x \neq y$ is a pseudometric on X.

Example 3

If X is any set and if g is any finite real-valued function defined on X, then the function f defined on $X \times X$ by $f(x, y) = |g(x) - g(y)|$ is a pseudometric on X.

Example 4

\* Let X be the set of all continuous mappings of the interval [0, 1] of $\mathbf{R}$ into $\mathbf{R}$ . If for each pair of elements $x, y$ of X we put

f(x, y) = \int_0^1 |x(t) - y(t)| \, dt,

then f is a pseudometric on X. \

Remarks: 1) Example 2 above shows that a pseudometric can take the value $+\infty$ for certain pairs of elements of $X$ . 2) If $f$ is a pseudometric on $X$ , we can in general have $f(x, y) = 0$ without $x = y$ , as Example 3 above shows (cf. § 2).

From the triangle inequality it follows that if $f(x, z)$ and $f(y, z)$ are finite then so is $f(x, y)$ ; moreover, in this case we have

f(x, z) \leq f(y, z) + f(x, y) \quad \text{and} \quad f(y, z) \leq f(x, z) + f(x, y),

so that (I)

|f(x, z) - f(y, z)| \leq f(x, y).

If $f$ is a pseudometric on $X$ , then so is $kf$ for any finite real number $k > 0$ . If $\{f_i\}_{i \in I}$ is any family of pseudometrics on $X$ , the sum $\sum_{i \in I} f_i$ is defined for all $x, y \in X \setminus X$ ; if $f$ denotes the value of this sum, then $f$ is a pseudometric on $X$ . Again, the supremum of the family $\{f_i\}$ (Chapter IV, § 5, no. 5) is a pseudometric on $X$ , for the relations $f_i(x, y) \leq f_i(x, z) + f_i(z, y)$ imply

\sup_{i \in I} f_i(x, y) \leq \sup_{i \in I} (f_i(x, z) + f_i(z, y)) \leq \sup_{i \in I} f_i(x, z) + \sup_{i \in I} f_i(z, y)

[Chapter IV, § 5, no. 7, formula (17)].

2. Definition of a uniformity by means of a family of pseudometrics

We have seen in Chapter VI, § 2, no. 3 that if, for each real number $s > 0$ , we denote by $U_s$ the set of all pairs $(x, y)$ of points of $\mathbf{R}^n$ whose Euclidean distance apart is $\leq s$ , then the $U_s$ form a fundamental system of entourages of the uniformity of $\mathbf{R}^n$ as $s$ runs through the set of real numbers $> 0$ .

More generally, let $f$ be a pseudometric on a set $X$ : for each $s > 0$ , let $U_s$ denote $f^{-1}([0, s])$ , and let us show that, as $s$ runs through the set of all real numbers $> 0$ , the $U_s$ form a fundamental system of a uniformity on $X$ . Axiom $U_1$ is satisfied by reason of $\mathrm{(EC_1)}$ : if $s = t$ , we have $U_s \subset U_t$ and therefore the $U_s$ satisfy $B_1$ : by $\mathrm{(EC_{11})}$ , we have $U_s \cup U_s$ and therefore $U_{2s}$ is satisfied; finally, by $\mathrm{(EC_{111})}$ , we have $U_s = U_{2s}$ , so that $U_{2s}$ is satisfied. (Cf. Chapter II, § 1, no. 1, Definition 2). Consequently we may make the following definition:

Definition 2

Given a pseudometric $f$ on a set $X$ , the uniformity defined by $f$ is the uniformity on $X$ which has as a fundamental system of entourages the family of sets $\overline{f}^{-1}([0, a])$ , where $a$ runs through the set of all real numbers $> 0$ .

Two pseudometrics on $X$ are said to be equivalent if they define the same uniformity.

Remark 1

If $(a_n)$ is any sequence of numbers $> 0$ and tending to 0, the $U_{a_n}$ form a fundamental system of entourages of the uniformity defined by $f$ .

Remark 2

The definition of a uniformity by a pseudometric $f$ consists in taking as a fundamental system of entourages the inverse image under $f$ of the neighbourhood filter of 0 in the subspace $[0, +\infty]$ of $\overline{\mathbf{R}}$ . Note that this procedure is quite analogous to that which allowed us to define the uniformities on a topological group (Chapter III, § 3, no. 1).

Let $f$ and $g$ be two pseudometrics on $X$ . From Definition 2 it follows that the uniformity defined by $f$ is coarser than the uniformity defined by $g$ if and only if, for each $a > 0$ there exists $b > 0$ such that the relation $g(x, y) \leq b$ implies $f(x, y) \leq a$ . A necessary and sufficient condition for $f$ and $g$ to be equivalent pseudometrics is that for each $a > 0$ there exists $b > 0$ such that $g(x, y) \leq b$ implies $f(x, y) \leq a$ , and $f(x, y) \leq b$ implies $g(x, y) \leq a$ .

In particular, if there exists a constant $k$ such that $f \leq kg$ , the uniformity defined by $f$ is coarser than the uniformity defined by $g$ .

Let $\varphi$ be a mapping of the interval $[0, +\infty]$ into itself, satisfying the following conditions: 1) $\varphi(0) = 0$ , and $\varphi$ is continuous at 0; 2) $\varphi$ is increasing in $[0, +\infty]$ and is strictly increasing in a neighbourhood of 0; 3) for all $u \geq 0$ and $v \geq 0$ , we have $\varphi(u + v) \leq \varphi(u) + \varphi(v)$ . Then if $f$ is any pseudometric on a set $X$ , the composition $g = \varphi \circ f$ is a pseudometric equivalent to $f$ .

The reader may easily verify that we may, for example, take $\varphi$ to be any one of the following functions:

\sqrt{u}, \quad \log (1 + u), \quad \frac{u}{1 + u}, \quad \inf (u, 1).

The last two examples show that there always exist bounded pseudometrics equivalent to any given pseudometric (finite or not).

Definition 3

If $(f_i)_{i \in I}$ is a family of pseudometrics on a set $X$ , then the least upper bound of the set of uniformities defined on $X$ by the pseudometrics $f_i$ is called the uniformity defined by the family $(f_i)$ .

Two families of pseudometrics on $X$ are said to be equivalent if they define the same uniformity on $X$ .

From the definition of the least upper bound of a set of uniformities (Chapter II, § 2, no. 5), the filter of entourages of the uniformity $U$ defined on $X$ by a family of pseudometrics $(f_i)_{i \in I}$ is the filter generated (Chapter I, § 6, no. 2) by the family of sets $f_i^{-1}([0, a])$ , where $i$ runs through $I$ and $a$ runs through the set of real numbers $> 0$ . In other words, we obtain a fundamental system of entourages of $\mathscr{U}$ by proceeding as follows: we take at random a finite number of indices $i_1, i_2, \ldots, i_n$ and, corresponding to each $i_k$ , a number $a_k > 0$ ; then we consider the set $V$ of pairs $(x, y) \in X \times X$ such that $f_{i_k}(x, y) \leq a_k$ for $1 \leq k \leq n$ ; these sets $V$ (for all possible choices of $n$ , the $i_k$ and the $a_k$ ) form a fundamental system of entourages for $\mathscr{U}$ . Moreover, we may restrict ourselves to the case in which all the $a_k$ are equal to the same number $a > 0$ , since the entourage consisting of all pairs $(x, y)$ such that

\sup_{1 \leq k \leq n} (f_{i_k}(x, y)) \leq \inf_{1 \leq k \leq n} a_k

is evidently contained in $V$ .

For each finite subset $H$ of $I$ , let $g_H$ denote the upper envelope of the family $(f_i)_{i \in H}$ . As $H$ runs through the set of all finite subsets of $I$ and $a$ runs through the set of real numbers $> 0$ , the sets $g_H^{-1}([0, a])$ form a fundamental system of entourages of the uniformity $\mathscr{U}$ . Now the $g_H$ are pseudometrics on $X$ (no. 1), and the upper envelope of a finite number of functions of the family $(g_H)$ belongs to this family, by definition; we express this property by saying that the family of pseudometrics $(g_H)$ is saturated. The family of pseudometrics $(g_H)$ is therefore equivalent to the family $(f_i)$ , and is said to be the family of pseudometrics obtained by saturating $(f_i)$ . From what has just been said it follows that we may always restrict ourselves to considering uniformities defined by saturated families of pseudometrics.

In the particular case where $I$ is a finite set, this argument shows that the uniformity defined by the family of pseudometrics $(f_i)_{i \in I}$ is also defined by the single pseudometric $g = \sup_{i \in I} f_i$ .

Let $\mathscr{U}, \mathscr{U}'$ be two uniformities on $X$ , defined respectively by two saturated families $(f_i)_{i \in I}, (g_x)_{x \in K}$ . Then $\mathscr{U}$ is coarser than $\mathscr{U}'$ if and only if, for each index $i \in I$ and each real number $a > 0$ , there is an index $x \in K$ and a number $b > 0$ such that the relation $g_x(x, y) \leq b$ implies $f_i(x, y) \leq a$ .

Example of a uniformity defined by a family of pseudometrics. Let $(f_i)_{i \in I}$ be an arbitrary family of (finite) real-valued functions defined on a set $X$ . Let $\mathscr{U}$ be the coarsest uniformity on $X$ with respect to which the $f_i$ are uniformly continuous (Chapter II, § 2, no. 3). Then it follows from the definition of the entourages of $\mathscr{U}$ (loc. cit.) that $\mathscr{U}$ is the uniformity defined on $X$ by the pseudometrics

g_i(x, y) = |f_i(x) - f_i(y)|.

3. Properties of uniformities defined by families of pseudometrics

Let $\mathscr{U}$ be a uniformity defined on a set $X$ by a family of finite pseudometrics $(f_i)$ . If we endow $X \times X$ with the uniformity which is the product of $\mathscr{U}$ by itself, then each of the real-valued functions $f_i$ is uniformly continuous on $X \times X$ ; for by (1) we have

|f_i(x, y) - f_i(x', y')| \leq f_i(x, x') + f_i(y, y'),

and therefore the relations $f_i(x, x') \leq \varepsilon/2, f_i(y, y') \leq \varepsilon/2$ imply

|f_i(x, y) - f_i(x', y')| \leq \varepsilon.

For $\mathscr{U}$ to be Hausdorff it is necessary and sufficient, from the definition of the entourages of $\mathscr{U}$ , that for each pair of distinct points $x, y$ of $X$ there is an index $i$ such that $f_i(x, y) \neq 0$ .

In particular, if $\mathscr{U}$ is defined by a single pseudometric $f$ , then $\mathscr{U}$ is Hausdorff if and only if the relation $f(x, y) = 0$ implies $x = y$ (cf. § 2). If $\mathscr{U}$ is not Hausdorff, the intersection of all the entourages of $\mathscr{U}$ is the subset of $X \times X$ consisting of pairs $(x, y)$ such that $f_i(x, y) = 0$ for all $i$ ; this subset is the graph of an equivalence relation $R$ on $X$ , and the Hausdorff uniformity associated with $\mathscr{U}$ is defined on $X/R$ (cf. Chapter II, § 3, no. 8). It is then easily verified that the functions $f_i$ are compatible (in $x$ and in $y$ ) with the relation $R$ (Set Theory, R, § 5, no. 7) and that the functions $\overline{f_i}$ , obtained from $f_i$ by passing to the quotient (with respect to $x$ and $y$ ), are pseudometrics on $X/R$ which define the Hausdorff uniformity associated with $\mathscr{U}$ (cf. § 2, no. 1).

If $A$ is a non-empty subset of $X$ , the restriction to $A \times A$ of a pseudometric on $X$ is clearly a pseudometric on $A$ . The uniformity induced by $\mathscr{U}$ on $A$ is clearly that defined by the family of restrictions to $A \times A$ of the pseudometrics $f_i$ .

Let us now look at the completion of the uniform space $X$ when $\mathscr{U}$ is Hausdorff.

Proposition 1

Let $X$ be a Hausdorff uniform space whose uniformity $\mathscr{U}$ is defined by a family of finite pseudometrics* $(f_i)$ , and let $\hat{X}$ be the completion of $X$ . Then the functions $f_i$ can be extended by continuity to $\hat{X} \times \hat{X}$ ; the extended functions $\bar{f}_i$ are finite pseudometrics on $\hat{X} \times \hat{X}$ , and the family $(f_i)$ defines the uniformity of $\hat{X}$ .

First, the $f_i$ can be extended by continuity to $\hat{X} \times \hat{X}$ , because they are uniformly continuous on $X \times X$ ; and the extended functions $\bar{f}_i$ are uniformly continuous on $\hat{X} \times \hat{X}$ (Chapter II, § 3, no. 6, Theorem 2); moreover, they are pseudometrics on $\hat{X}$ by virtue of the principle of extension of inequalities (Chapter IV, § 5, no. 2, Theorem 1). Let $\mathscr{U}_1$ denote the uniformity on $\hat{X}$ obtained by completion, and let $\mathscr{U}_2$ denote the uniformity defined by the family of pseudometrics $(\bar{f}_i)$ . Then $\mathscr{U}_2$ is coarser than $\mathscr{U}_1$ ; for each $\bar{f}_i$ is uniformly continuous on $\hat{X} \times \hat{X}$ with respect to $\mathscr{U}_1$ , and hence for each $a > 0$ there exists an entourage $V$ of $\mathscr{U}_1$ such that, whenever $(x, y) \in V$ , we have $|\bar{f}_i(x, y) - \bar{f}_i(x, x)| \leq a$ , that is [since $\bar{f}_i(x, x) = 0$ ], $V \subset \bar{f}_i([0, a])$ ; hence every entourage of $\mathscr{U}_2$ is an entourage of $\mathscr{U}_1$ . On the other hand, $\mathscr{U}_1$ and $\mathscr{U}_2$ induce the same uniformity $\mathscr{U}$ on $X$ . As $\hat{X}$ is complete with respect to $\mathscr{U}_1$ , it follows that $\mathscr{U}_1$ and $\mathscr{U}_2$ are identical (Chapter II, § 3, no. 7, Proposition 14).

4. Construction of a family of pseudometrics defining a uniformity

The significance of defining a uniformity by means of a family of pseudometrics lies in the fact that all uniformities can be so obtained. Namely:

Theorem 1

Given a uniformity $\mathscr{U}$ on a set $X$ , there is a family of pseudometrics on $X$ such that the uniformity defined by this family is identical with $\mathscr{U}$ .

For each entourage $V$ of the uniformity $\mathscr{U}$ , define inductively a sequence of symmetric entourages $(U_n)$ such that $U_1 \subset V$ and $U_{n+1} \subset U_n$ for all $n \geq 1$ . The sequence $(U_n)$ is a fundamental system of entourages of a uniformity $\mathscr{U}_V$ coarser than $\mathscr{U}$ ; moreover, it is clear that $\mathscr{U}$ is the least upper bound of all the uniformities $U_V$ as $V$ runs through the filter of entourages of $\mathscr{U}$ . Theorem 1 is therefore a consequence of the following proposition:

Proposition 2

If a uniformity $\mathscr{U}$ on $X$ has a countable fundamental system of entourages, then there is a pseudometric $f$ on $X$ such that $\mathscr{U}$ is identical with the uniformity defined by $f$ .

Let $(V_n)$ be a countable fundamental system of entourages of $\mathscr{U}$ . Define inductively a sequence $(U_n)$ of symmetric entourages of $\mathscr{U}$ such that $U_1 \subset V_1$ and

\bigcup_{n+1}^3 \subset U_n \cap V_n \quad \text{for} \quad n \geq 1.

Clearly

(U_n)

is another fundamental system of entourages of

\mathscr{U}

, and we have in particular

\bigcup_{n+1}^3 \subset U_n

for

n \geq 1

. We define a real-valued function

g

X \times X

as follows:

g(x, y) = 0

(x, y) \in U_n

for all

n

;

g(x, y) = 2^{-k}

(x, y) \in U_n

for

1 \leq n \leq k

, but

(x, y) \notin U_{k+1}

;

g(x, y) = 1

(x, y) \notin U_1

. The function

g

is symmetric and positive, and we have

g(x, x) = 0

for all

x \in X

. Put

f(x, y) = \inf \sum_{i=0}^{p-1} g(z_i, z_{i+1}),

the greatest lower bound being taken over the set of all finite sequences

(z_i)_{0 \leq i \leq p}

(

p

arbitrary) such that

z_0 = x

and

z_p = y

. We shall show that

f

is a pseudometric which satisfies the inequalities

\frac{1}{2} g(x, y) \leq f(x, y) \leq g(x, y).

It follows immediately from the definition that

f

is symmetric and positive and satisfies the triangle inequality. Also it is clear that

f(x, y) \leq g(x, y)

, hence

f(x, x) = 0

for all

x \in X

, and therefore

f

is a pseudometric. To prove the left-hand half of the inequalities (2), let us show by induction on

p

that, for every finite sequence

(z_i)_{0 \leq i \leq p}

p + 1

points of

X

such that

z_0 = x

and

z_p = y

, we have

\sum_{i=0}^{p-1} g(z_i, z_{i+1}) \geq \frac{1}{2} g(x, y).

This is clear if

p = 1

. Put

a = \sum_{i=0}^{p-1} g(z_i, z_{i+1})

; the inequality (3) is true if

a \geq 1/2

, because

g(x, y) \leq 1

. Suppose then that

a < 1/2

, and let

h

be the largest of the indices

q

such that

\sum_{i < q} g(z_i, z_{i+1}) \leq \frac{a}{2};

we have then

\sum_{i < h} g(z_i, z_{i+1}) \leq a/2

and

\sum_{i < h+1} g(z_i, z_{i+1}) > a/2

, whence

\sum_{i > h} g(z_i, z_{i+1}) \leq \frac{a}{2}.

By the inductive hypothesis we have $g(x, z_h) \leq a$ and $g(z_{h+1}, y) \leq a$ ; on the other hand it is clear that $g(z_h, z_{h+1}) \leq a$ . Let $k$ be the smallest integer $> 0$ such that $2^{-k} \leq a$ ; then $k \geq 2$ , and $(x, z_h) \in U_k, (z_h, z_{h+1}) \in U_k, (z_{h+1}, y) \in U_k$ by the definition of $g$ ; hence $(x, y) \in U_k \subset U_{k-1}^3$ , which implies that $g(x, y) \leq 2^{1-k} \leq 2a$ .

Hence the inequalities (2) are proved; they show that, for each $a > 0$ , the set $f^{-1}([0, a])$ contains $U_k$ for each index $k$ such that $2^{-k} < a$ , and conversely that each $U_k$ contains the set $f^{-1}([0, 2^{-k-1}])$ ; hence the sets $f^{-1}([0, a])$ form a fundamental system of entourages of the structure $\mathscr{U}$ .

Q.E.D.

Remark

A uniformity $\mathscr{U}$ on $X$ is defined by the family $\Phi$ of all pseudometrics on $X$ which are uniformly continuous on $X \times X$ . For clearly the uniformity defined by the family $\Phi$ is coarser than $\mathscr{U}$ ; conversely, Theorem 1 shows that there is a subfamily of $\Phi$ which defines the uniformity $\mathscr{U}$ and therefore the uniformity defined by $\Phi$ is finer than $\mathscr{U}$ .

5. Uniformizable spaces

In Chapter II, § 4, no. 1, we posed the problem of characterizing uniformizable topological spaces. The solution is given by the following theorem:

Theorem 2

A topological space* $X$ *is uniformizable if and only if it satisfies the following axiom:

(OIV) Given any point $x_0 \in X$ and any neighbourhood $V$ of $x_0$ , there exists a continuous real-valued function on $X$ which takes its values in $[0, 1]$ , is equal to $0$ at $x_0$ , and is equal to $1$ on $C_V$ .

The condition is necessary. For if there is a uniformity on $X$ compatible with the topology of $X$ , then by Theorem 1 this uniformity can be defined by a family $(f_i)$ of pseudometrics on $X$ , and we may assume with no loss of generality that this family is saturated (no. 2). From the definition of the entourages of the uniformity defined by such a family of pseudometrics, there is a pseudometric $f_\alpha$ of the family $(f_i)$ , and a number $a > 0$ , such that $f_\alpha(x_0, x) \geq a$ for all $x \in C_V$ . It follows that the function $g(x) = \inf \left( 1, \frac{1}{a} f_\alpha(x_0, x) \right)$ satisfies all the conditions laid down in (OIV).

The condition is sufficient. For let $\Phi$ be the set of all continuous mappings of $X$ into $[0, 1]$ . Axiom (OIV) shows that the coarsest uniformity with respect to which all the functions belonging to $\Phi$ are uniformly continuous is compatible with the topology of $X$ (Chapter II, § 2, no. 3).

Definition 4

A topological space is said to be completely regular if it is uniformizable and Hausdorff.

Equivalently, in view of Theorem 2, a space is completely regular if it satisfies axioms (H) [cf. Chapter I, § 8, no. 1, Proposition 1] and (O_{IV}).

Remark

Axiom (O_{IV}) implies (O_{III}) (cf. Chapter I, § 8, no. 4), for if V is a neighbourhood of x_0 and if f is a continuous real-valued function on X with values in [0, 1], such that f(x_0) = 0, f(x) = 1 for all x \in \overline{V}, then the set \overline{f}([0, 1/2]) is a closed neighbourhood of x_0 contained in V. In particular, every completely regular space is regular (which justifies the terminology). But there are examples of regular spaces which are not completely regular (), so that (O{III}) does not imply (O_{IV}).

Every compact space is completely regular (Chapter II, § 4, no. 1, Theorem 1) and therefore so is every subspace of a compact space. We can now complete this proposition by proving its converse:

Proposition 3

A topological space X is completely regular if and only if it is homeomorphic to a subspace of a compact space.

Consider the coarsest uniformity on X with respect to which all continuous mappings of X into [0, 1] are uniformly continuous; we have already used this uniformity in the proof of Theorem 2, where we saw that it is compatible with the topology of X if X is uniformizable. Furthermore, this uniformity is a structure of a precompact space, by the compactness of the interval [0, 1] and Proposition 3 of Chapter II, § 4, no. 2. If X is Hausdorff, the completion of X with respect to this is therefore compact, and the proposition is proved.

We can therefore say that a completely regular space can be embedded in a compact space. It is often convenient to present this result in the following way:

In general, a cube is a topological space K^I, the product of a family of topological spaces each identical with a compact interval K of \mathbf{R}, indexed by an arbitrary set I. If I is finite and has n elements, we recover the notion of an n-dimensional closed cube, which was defined in Chapter VI, § 1, no. 1. A cube is a compact space (Chapter I, § 9, no. 5, Theorem 3).

Proposition 4

If a topological space X is completely regular, it is homeomorphic to a subspace of a cube.

Let $(f_i)_{i \in I}$ denote the family of all continuous mappings of X into K = [0, 1], and let g denote the mapping $x \mapsto (f_i(x))$ of X into

(*) See A. Tychonoff, Math. Ann., 102, (1930), p. 553.

K¹. If $x, y$ are any two distinct points of $X$ , it follows from axioms (H) and (O_IV) that there is an index $i$ such that $f_i(x) \neq f_i(y)$ , and therefore $g$ is a one-to-one mapping of $X$ into $K^1$ . Moreover, it is immediate that $g$ is an isomorphism of the coarsest uniformity on $X$ for which all the $f_i$ are uniformly continuous, onto the uniformity induced on $g(X)$ by the product uniformity of $K^1$ ; a fortiori, $g$ is a homeomorphism of $X$ onto $g(X)$ .

6. Semi-continuous functions on a uniformizable space

In Chapter IV, § 6, no. 2, Corollary to Theorem 4, we showed that the upper envelope of a family of continuous real-valued functions on a topological space is a lower semi-continuous function. If the space is uniformizable, there is a converse to this proposition:

Proposition 5

In order that every lower semi-continuous real-valued function $f$ (finite or not) on a topological space $X$ should be the upper envelope of the continuous real-valued functions on $X$ (finite or not) which are $\leq f$ , it is necessary and sufficient that $X$ be uniformizable.

The condition is necessary. Let $x_0$ be any point of $X$ and let $V$ be any open neighbourhood of $x_0$ ; then the characteristic function $\varphi_V$ of the set $V$ is lower semi-continuous (Chapter IV, § 6, no. 2, Corollary to Proposition 1); by hypothesis, there is therefore a continuous real-valued function $g$ on $X$ such that $g \leq \varphi_V$ and $g(x_0) = a > 0$ . The continuous function $\inf \left( 1, \frac{1}{a} g^+ \right)$ takes its values in $[0, 1]$ , is equal to 0 in $C_V$ , and equal to 1 at $x_0$ . Hence (Theorem 2) $X$ is uniformizable.

The condition is sufficient. Consider first the case in which $f$ takes its values in $[-1, +1]$ . We have to show that, for each $x_0 \in X$ and each number $a < f(x_0)$ , there is a continuous real-valued function $g$ on $X$ such that $g \leq f$ and $g(x_0) \geq a$ . If $a \leq -1$ , we may take $g$ to be the constant $-1$ . If $-1 < a < f(x_0)$ , there is a neighbourhood $V$ of $x_0$ such that $f(x) \geq a$ for all $x \in V$ . Since $X$ is uniformizable, there is a continuous real-valued function $h$ on $X$ , with values in $[0, 1]$ , such that $h(x_0) = 0$ and $h(x) = 1$ for all $x \in C_V$ . We may then take $g(x) = a - (a + 1)h(x)$ , and we have a continuous function satisfying the stated conditions. Note that this function takes its values in $[-1, +1]$ .

The general case follows by transfer of structure; for there is a strictly increasing homeomorphism of $[-1, +1]$ onto $\overline{\mathbf{R}}$ (Chapter IV, § 4, no. 2, Proposition 2), and the definition of a semi-continuous function involves only the order structure and the topology of $\overline{\mathbf{R}}$ .

Remark

In the above proof we see that the function $g$ does not take the value $+1$ . By transfer of structure it follows that every lower semi-continuous real-valued function $f$ on the uniformizable space $X$ is the upper envelope of the continuous real-valued functions $g \leq f$ on $X$ which do not take the value $+\infty$ .

Exercises

See the exercises for § 1.