If \(A\) is a connected subspace of \(X\), does it follow that \(\operatorname{Int} A\) and \(\operatorname{Bd} A\) are connected? Does the converse hold? Justify your answers.
Solution
Neither the interior nor the boundary of a connected set need be connected.
For the interior, let \(A\) be the union of two closed disks in \(\mathbb R^2\) tangent at one point. The set \(A\) is connected, but its interior is the disjoint union of the two open disks, hence is disconnected.
For the boundary, take the closed annulus \[ A=\{(x,y):1\le x^2+y^2\le4\}. \] It is connected, while its boundary is the disjoint union of the two circles \(x^2+y^2=1\) and \(x^2+y^2=4\).
The converse also fails. Take \(A=\mathbb Q\subset\mathbb R\). Then \(A\) is disconnected, but \[ \operatorname{Int}A=\varnothing \] is connected, and since \(\mathbb Q\) is dense, \[ \operatorname{Bd}A=\mathbb R, \] which is connected.