# Every open set is the union of net of increasing open sets

I'm struggling to find a solution to the following problem:

Let $(X,\mathcal{T})$ be an arbitrary topological space and let $\mathcal{U}$ be an class of subsets of $X$, i.e. $\mathcal{U}\subseteq\mathcal{P}(X)$, that contains a base $\mathcal{B}$ of the topology $\mathcal{T}$ and that is closed with respect to finite intersections. Show that every open set $O\in\mathcal{T}$ can be represented in the form of the union of a net of increasing open sets $U_i$ that are finite unions of sets in $\mathcal{U}$.

I don't really have an idea how to show this. As far as I see, since $O$ is open, there exists for every $x\in O$ a neighbourhood $U_x\in\mathcal{U}(x)$ such that $x\in U_x\subseteq O$. Further, since $U$ is a neighbourhood of $x$, there exists an open set $O_x\in\mathcal{T}$ with $x\in O_x\subseteq U_x\subseteq O$. So, $O$ can be represented in the form $$O=\bigcup_{x\in O}O_x.$$ On the other hand, since $\mathcal{B}$ is a base of $\mathcal{T}$, there exists a subfamily $\mathcal{B}_0\subseteq\mathcal{B}$ such that $$O=\bigcup_{B_0\in\mathcal{B}_0}B_0.$$ I don't know how to contruct from this a net of increasing sets whose members are finite unions of members of $\mathcal{U}$.

Any help would be highly appreciated...thank you very much in advance!

Greetings, Florian

• What is a net of open sets ? – Bernard Jul 24 '15 at 13:56
• do you know what a net is? – Jorge Fernández Hidalgo Jul 24 '15 at 13:59
• What does "a net of increasing sets" mean? A map $i \mapsto U_i$ from a directed set $I$ satisfying $i \leqslant j \implies U_i \subset U_j$? – Daniel Fischer Jul 24 '15 at 13:59
• @Bernard: A net in $\mathcal{T}$ is a family of elements $(O_i)_{i\in I}$ in $\mathcal{T}$ indexed by a directed set $I$. Have a look at this: en.wikipedia.org/wiki/Net_(mathematics) – Florian Jul 24 '15 at 14:02
• Given $O$, consider $I = \bigl\{ \mathscr{F} \subset \mathcal{B} : \mathscr{F}\text{ is finite}, \bigl(\forall F\in \mathscr{F}\bigr)\bigl(F\subset O\bigr)\bigr\}$. – Daniel Fischer Jul 24 '15 at 14:07