Closure of a set

A point \(x\) lies in \(\overline{A}\) exactly when every open set containing \(x\) meets \(A\). That is why in usual \(\mathbb{R}\), \(\overline{(0,1)}=[0,1]\): the endpoints are added because any interval around them touches the set. Likewise \(\overline{\mathbb{Q}}=\mathbb{R}\), since every interval contains rationals.

Always \(A\subseteq\overline{A}\), and \(\overline{A}=A\) if and only if \(A\) is closed. Also \(\overline{\overline{A}}=\overline{A}\), \(A\subseteq B\) implies \(\overline{A}\subseteq\overline{B}\), and \(\overline{A\cup B}=\overline{A}\cup\overline{B}\). For intersections only \(\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}\) holds, and the inclusion can be strict: with \((0,1)\) and \((1,2)\) the left side is empty and the right side is \(\{1\}\).

The closure depends on the topology: in the discrete one \(\overline{A}=A\); in the trivial one every nonempty set has closure \(X\). It is tied to the interior by duality, \(X\setminus\overline{A}=\operatorname{int}(X\setminus A)\), and together they define the boundary. A common mistake is to think the closure only adds visible «edge» points: in a finite space it can add points in unintuitive ways.