Must $X$ be lindelöf if it has countable network? A network is like a base, except that its members need not be open sets.
A family $\mathcal N$ of subsets of a topological space $X$ is a network for $X$ if for every point $x\in X$ and any neighbourhood $U$ of $x$ there exists an $M \in \mathcal N$ such that $x\in M \subset U$.
Suppose $X$ has a countable network. Must $X$ be Lindelöf?

Yes. Let $\mathscr{N}$ be a countable network for $X$, and let $\mathscr{U}$ be an open cover of $X$. For each $x\in X$ choose $N_ x\in \mathscr{N}$ such that $x\in N_x\subseteq U$ for some $U\in\mathscr{U}$. Let $\mathscr{N}_0=\{N_x:x\in X\}$; clearly $\mathscr{N}_0$ is countable and covers$X$. $\mathscr{N}_0$ refines $\mathscr{U}$, so for each $N\in\mathscr{N}_0$ we can choose a $U_ N\in\mathscr{U}$ such that $N\subseteq U_N$. Then $\{U_N:N\in\mathscr{N}_0\}$ is a countable subcover of $\mathscr{U}$.
