Lindelöf spaces

(i) If the topology of $X$ admits a countable base, $X$ is a Lindelöf space.

(ii) Let $f : X \to Y$ be a continuous mapping from $X$ into a topological space $Y$ (not necessarily separated). If $X$ is a Lindelöf space, the same is true of the subspace $f(X)$ of $Y$.

(i) Let $\mathscr{R}$ be an open covering of $X$, $\mathscr{B}$ a countable base of the topology of $X$, $\mathscr{B}'$ the set of elements $A \in \mathscr{B}$ such that there exists an element of $\mathscr{R}$ containing $A$; $\mathscr{B}'$ is a countable covering of $X$, since every $x \in X$ is contained in a set $U \in \mathscr{R}$, and there exists then $A \in \mathscr{B}$ such that $x \in A \subset U$. There then exists a mapping $\Phi$ from $\mathscr{B}'$ into $\mathscr{R}$ such that one has $A \subset \Phi(A)$ for every $A \in \mathscr{B}'$; it is clear that $\Phi(\mathscr{B}')$ is a countable open covering of $X$ contained in $\mathscr{R}$.

(ii) Let $(V_\lambda)_{\lambda \in L}$ be a family of open sets in $Y$ forming a covering of $f(X)$. The sets $f^{-1}(V_\lambda)$ then form an open covering of $X$, and there exists consequently a countable subset $J$ of $L$ such that the open sets $f^{-1}(V_\lambda)$ for $\lambda \in J$ form a covering of $X$. But then the sets $V_\lambda$ for $\lambda \in J$ form a covering of $f(X)$, which proves that $f(X)$ is a Lindelöf space.

This follows immediately from the preceding prop. 1 and from the definition of sub-Lindelöf spaces (IX, p. 59, def. 2), since a Polish space admits a countable base.

Soit $X$ a regular Lindelöf space, and $\mathscr{R}$ an open covering of $X$. For every $x \in X$, let $U_x \in \mathscr{R}$ containing $X$, and let $V_x$ be an open neighbourhood of $x$ such that $V_x \subset U_x$. Since $X$ is a Lindelöf space, there exists a sequence $(x_n)_{n \geq 0}$ of points of $X$ such that the $V_{x_n}$ form a covering of $X$. Put then, for every integer $n \geq 0$, $W_n = U_{x_n} \cap \complement \left( \bigcup_{k < n} V_{x_k} \right)$. The sets $W_n$ are open and form a covering of $X$ finer than $\mathscr{R}$. On the other hand, this covering is locally finite, since every $x \in X$ belongs to some $V_{x_i}$ for at least one $i$; then $V_{x_i}$ is a neighbourhood of $x$, which meets no set $W_n$ for $n > i$.

Let us first show that b) implies a). Let $U$ be an open subspace of $X$, and let $(U_\lambda)_{\lambda \in L}$ be a family of open sets in $U$ (hence in $X$) covering $U$. The functions $\varphi_{U_\lambda}$ being lower semicontinuous in $X$, there exists a countable subset $J$ of $L$ such that the family $(\varphi_{U_\lambda})_{\lambda \in J}$ has the same upper envelope $\varphi_U$ as the family $(\varphi_{U_\lambda})_{\lambda \in L}$. Consequently, $U = \bigcup_{\lambda \in J} U_\lambda$, and $U$ is a Lindelöf space.

Let us now prove that a) implies b). Let $\mathscr{F}$ be a set of lower semicontinuous functions in $X$, and let $s$ be the upper envelope of $\mathscr{F}$. Let $D$ be a countable dense subset of $\overline{\mathbf{R}}$. For every function $f \in \mathscr{F}$ and every number $d \in D$, let $U_{f,d}$ be the open set of the $x \in X$ such that $f(x) > d$ (IV, p. 29). By assumption, there exists a countable subset $\mathscr{F}_d'$ of $\mathscr{F}$ such that

Let $\mathscr{F}' = \bigcup_{d \in D} \mathscr{F}'_d$, and denote by $s'$ the upper envelope of $\mathscr{F}'$. We obviously have $s \geqslant s'$; on the other hand, let $x$ be a point of $X$, and let $d$ be an element of $D$ such that $s(x) > d$; there exists a function $f \in \mathscr{F}$ such that $f(x) > d$; hence $x \in U_{f,d}$, and there exists a function $f' \in \mathscr{F}'_d$ such that $x \in U_{f',d}$. Consequently, we also have $s'(x) > d$, whence the inequality $s' \geqslant s$, and finally $s' = s$. Since $\mathscr{F}'$ is countable, the assertion is established for lower semi-continuous functions; the case of upper semi-continuous functions is reduced to it by changing the sign.

Indeed, $X \times X$ is a souslinian space (IX, p. 60, prop. 7); the same is true of every open subspace of $X \times X$ (IX, p. 59, prop. 5), which is therefore a Lindelöf space (IX, p. 76, corollary). To every function $h \in H$, associate the closed set $F_h$ of the pairs $(x, y) \in X \times X$ such that $h(x) = h(y)$. By virtue of prop. 3, there exists a countable subset $H'$ of $H$ such that one has $\bigcap_{h \in H'} F_h = \bigcap_{h \in H} F_h$. But by assumption the second member is the diagonal of $X \times X$; the same is therefore true of the first, and consequently $H'$ separates the points of $X$.

Let $X$ be a compact souslinian space, $I$ the interval $[0, 1]$ of $\mathbf{R}$, $H$ the set of continuous mappings of $X$ into $I$. Since $X$ is completely regular, $H$ separates the points of $X$, and cor. 1 implies the existence of a sequence $(f_n)$ of elements of $H$ which separates the points of $X$. But then the mapping $x \mapsto (f_n(x))_{n \in \mathbf{N}}$ of $X$ into $I^\mathbf{N}$, continuous and injective, is a homeomorphism of the compact space $X$ onto a subspace of $I^\mathbf{N}$; this proves that $X$ is metrizable.