Basis for a topology if every open set is union of some sets Let $(X, \mathcal{T})$ be a topological space, $\forall U$ open in $X$, $U= \bigcup B, B \in \mathcal {B}$, then $\mathcal{B}$ is a basis for a topology on $X$?
Check: Let $x \in X$, then $\exists U$ open in $X$ such that $x  \in U$ ($X$ is open in $X$) $\Rightarrow$  $x \in \bigcup B$ $\Rightarrow$ $\exists B \in \mathcal{B}$ such that $x \in B$.
Let $B_1,B_2 \in \mathcal{B}$. Since $B_1 = \bigcup B_1$, $B_2 = \bigcup B_2$ $\Rightarrow B_1, B_2$ are open in $X$, then $B_1 \cap B_2$ is open in $X$.
 A: The question is somewhat unclear. But I think the question is meant to be:
Suppose we have a collection of subsets $\mathcal{B}$ of a topological space $(X,\mathcal{T})$. Suppose that for every open set $U$ of $X$, there is some subcollection $\mathcal{B}_U \subseteq \mathcal{B}$ such that $U = \cup \mathcal{B}_U$. Can we then conclude that $\mathcal{B}$ is a base for the topology $\mathcal{T}$?
For this is it is of course necessary that all members of $\mathcal{B}$ are themselves open, as this is part of the definition of a base for a topology. So we need $\mathcal{B} \subseteq \mathcal{T}$. We cannot omit this condition, otherwise we could just take $\mathcal{B} = \{\{x\}: x \in X \}$ and this works for any topology on $X$ (any set is just the union of the singleton sets corresponding to its elements). At least the latter is at least a base for another, finer, topology, namely the discrete one. But even that (i.e. being a base for a finer topology) need not hold, as the example of $X = \mathbb{R}$, $\mathcal{T}$ is the usual topology and $\mathcal{B} = \{(a,b]: a,b \in \mathbb{R}, a < b \} \cup \{[a,b): a,b \in \mathbb{R}, a < b \}$ shows: these sets are not open, can form all open sets by unions, but themselves form no base for any topology (as we cannot find a member of $\mathcal{B}$ inside $\{a\} = [a,a+1) \cap (a-1,a]$ for any $a \in \mathbb{R}$).
If however all members of $\mathcal{B}$ are open, then yes, almost by definition $\mathcal{B}$ then forms a base for the topology: suppose $x \in O$ and $O$ is open. Then we write $O = \cup \mathcal{B}_O$ for some collection $\mathcal{B}_O \subseteq \mathcal{B}$. Then some $B \in \mathcal{B}_O$ must contain $x$, so $x \in B \subseteq O$. This condition is just a reformulation of the "every open set is a union of base elements" condition, if you think about it.
A "base" where the members need not themselves be open is a useful tool, though, and is called a network for the topology. It was introduced by Arhangel'skij to prove some theorems, but has proved to be a useful idea in general.
