Is nonempty intersection with every set in a base a sufficient condition to prove the closure of the set is the whole space? I've noticed in some proofs, if $A\subset B$ for $B$ a topological space, it seems that it is enough to show that $A$ intersects every set in a base of $B$ to prove that the closure $\overline{A}=B$. I noticed this for instance in the proof that the closure of $\mathbb{Q}$ is $\mathbb{R}$ in the lower limit topology.
Is this ultimately true in general, that if a subspace meets every subset of a base of the ambient space, that the closure of the subspace is the whole space? If so, why?
 A: Yes, it’s true. Suppose that $\langle X,\mathscr{T}\rangle$ is a space, $\mathscr{B}$ is a base for $\mathscr{T}$, and $D\subseteq X$ is such that $D \cap B \ne \varnothing$ for each $B \in \mathscr{B}$. Let $x \in X$ be arbitrary, and let $V$ be any open nbhd of $x$. By the definition of a base there is some $B\in\mathscr{B}$ such that $x\in B \subseteq V$. By hypothesis $D\cap B \ne \varnothing$, so $D \cap V \supseteq D\cap B \ne \varnothing$. Thus, every open nbhd of $x$ meets $D$, and therefore $x \in \operatorname{cl}D$. Since $x$ was an arbitrary point of $X$, $\operatorname{cl}D = X$.
A: $\bar{A}$, the closure of $A$, is equal to the intersection of all the closed sets containing $A$. Therefore, $\bar{A}^c$, the complement of $\bar{A}$ in $B$, is equal to the union of all the open sets that don't meet $A$. So showing $\bar{A}=B$ is equivalent to showing $\bar{A}^c=B^c=\varnothing$, i.e. there are no open sets in $B$ that don't meet $A$ (except $\varnothing$ itself, of course). But given a base for the topology of $B$, any open set in $B$ is a union of elements in the base, so to show $\bar{A}^c=\varnothing$, it suffices to show that every element of the base meets $A$. So, your conjecture is true!
