Proposition 3

Let $X$ be a topological space (not necessarily separated). The following properties are equivalent: a) Every open subspace of $X$ is a Lindelöf space. b) Every set of lower (resp. upper) semicontinuous functions in $X$ contains a countable subset which admits the same upper (resp. lower) envelope.

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.