# Clarification on definition of closure

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.

• A closed set is defined as the complement of an open set. An important property of $\overline A$ is that $\overline A$ is closed , and $\overline A$ is the common intersection of all the closed sets that have $A$ as a subset. May 10, 2019 at 15:50

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 $$(1,\infty)$$).

• Thank you. Is this equivalent to saying that $\overline{A}$ is the intersection of all open sets containing $A$?
– user88923
May 10, 2019 at 11:13
• No, because otherwise the closure of an open set $A$ would be $A$ again. However, $\overline A$ is the intersection of all closed sets containing $A$. May 10, 2019 at 11:15

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.