Clarification on definition of closure

Question

Let $X$ be a topological space and $A\subset X$, $$\overline{A}=\{x\in X \mid \forall U\text{ open with }x\in U,U\cap A\neq \emptyset\},$$ where $\overline{A}$ is the closure of $A$ in $X$. I didn't understand the language, does this say that elements of all open sets whose intersection with $A$ is nonempty? Then that would be $X$ always, which is wrong. Please help me understand.

Answer 1

No, it is not about all open subsets of $X$. It's about those open subsets of $X$ which contain $x$. The closure of $A$ is the set of those $x\in X$ such that every open set containing $x$ intersects $A$. For instance, in $\mathbb R$, with its usual topology, $2\notin\overline{(0,1)}$ because there are open sets containing $2$ which do not intersect $(0,1)$ (such as, for instance $(1,\infty)$).

Answer 2

It's a criterion for the point $x$: let $\mathcal{O}_x$ be the set of all open sets of $X$ that contain $x$, the open neighbourhoods of $x$.

So $x \in \overline{A}$ iff all $O \in \mathcal{O}_x$ intersect $A$. The only open sets that are considered to see whether $x$ is in the closure of $A$ are those that contain $x$, all other open sets are irrelevant for that $x$. In metric topologies, if $x \in O$ and $O$ is open means there is some $\varepsilon>0$ such that $B(x,\varepsilon) \subseteq O$. Also, $B(x,\varepsilon)$ is itself an open set containing $x$ in the metric topology. So then we can reformulate it as

$$\overline{A} = \{x \in X: \forall \varepsilon>0: B(x,\varepsilon)\cap A\neq \emptyset\}$$

which is a very common thing to see.