Solution. Suppose \(A\) is open and let \(x\) be a limit point of \(A^{C}\). If \(x\in A\) then, \(A\) being open, some interval \((x-\varepsilon ,x+\varepsilon )\) lies wholly in \(A\), so that interval meets \(A^{C}\) nowhere — contradicting that \(x\) is a limit point of \(A^{C}\). Hence \(x\in A^{C}\), and \(A^{C}\) contains all its limit points, so it is closed.

Conversely, suppose \(A^{C}\) is closed and let \(x\in A\). Then \(x\notin A^{C}\), so \(x\) is not a limit point of \(A^{C}\) either, and therefore some \((x-\varepsilon ,x+\varepsilon )\) meets \(A^{C}\) only possibly at \(x\) itself — but \(x\notin A^{C}\), so it misses \(A^{C}\) entirely and lies in \(A\). Thus every point of \(A\) is interior and \(A\) is open.

Solution. No interval contains only rationals, since the irrationals are dense, so no rational is an interior point: \(\mathbb {Q}^{0}=\emptyset \) and \(\mathbb {Q}\) is not open. Every real is a limit of rationals, again by density, so every real is a limit point: \(\overline {\mathbb {Q}} = \mathbb {R}\). Since \(\overline {\mathbb {Q}}\neq \mathbb {Q}\), the set is not closed either.

A set may be neither open nor closed, and \(\mathbb {Q}\) is the standard witness. The words are not opposites — \(\emptyset \) and \(\mathbb {R}\) are both.