Closure and subbasis Let $X$ be a topological space and $A \subset X$ with a subbasis $S$. Does it then hold that $x \in \overline{A}: \Leftrightarrow \forall s \in S: (x \in s \Rightarrow s \cap A \neq  \emptyset).$ This is certainly true if $S$ is a basis, but I suspect it is wrong for a subbasis, am I right?
 A: A subbasis $S$ for the standard topology on $\mathbb{R}$ is the collection of all the rays $(-\infty,a)$ and $(a,\infty)$ where $a \in \mathbb{R}$. Now, let 
\begin{align*}
A = \{0,1\} && x = 1/2.
\end{align*} 
Any ray $R$ containing $x$ contains either $0$ or $1$, i.e. $R \cap A \neq \varnothing$. Yet, $x$ does not belong to $\overline A$. 

Actually, if you think about it, you have the following:

Let $X$ be a topological space with subbasis $S$.  Then, the following are equivalent:
  
  
*
  
*$S$ is a basis.
  
*For every set $A \subset X$ and every point $x \in X$, one has $x \in \overline A$ if and only if every $U \in S$ such that $x \in U$ has $U \cap A \neq \varnothing$. 
  

I think you are already aware of the implication (1) implies (2). 
Suppose that (2) holds. We need to show that, given a point $x \in X$ and an open set $U \subset X$ such that $x \in U$, there exists $V \in S$ such that $x \in V \subset U$. Let $A$ be the complement of $U$, which is is closed. Since $x \in U$, we have $x \notin A = \overline A$. Thus, by (2), there exists some $V \in S$ such that $x \in V$ and $V \cap A = \varnothing$. In other words, $V \subset U$, which proves (1). 
