Interior of a set

A point \(x\) lies in \(\operatorname{int}A\) exactly when \(A\) is a neighborhood of \(x\). Thus, in \(\mathbb{R}\) with the usual topology, \(\operatorname{int}[0,1]=(0,1)\): the endpoints are lost because no open interval around them fits inside \([0,1]\). On the other hand \(\operatorname{int}\mathbb{Q}=\varnothing\), since every interval contains irrationals.

One always has \(\operatorname{int}A\subseteq A\), and \(\operatorname{int}A=A\) if and only if \(A\) is open. Also \(\operatorname{int}(\operatorname{int}A)=\operatorname{int}A\), \(\operatorname{int}(A\cap B)=\operatorname{int}A\cap\operatorname{int}B\), and \(A\subseteq B\) implies \(\operatorname{int}A\subseteq\operatorname{int}B\). For unions only the inclusion \(\operatorname{int}A\cup\operatorname{int}B\subseteq\operatorname{int}(A\cup B)\) holds, and it can be strict.

The interior depends on the topology: the same \(A\) can have a large interior in one topology and an empty one in another. In the trivial topology every proper subset has empty interior; in the discrete one, \(\operatorname{int}A=A\) for all \(A\). It is the dual of the closure, and together they describe the boundary of a set.