In $\mathbb{R}$ the interiors of $\mathbb{Q}$ and of its complement are both empty while the interior of their union is everything

Refuted: that $\operatorname{int}(A \cup B) = \operatorname{int}(A) \cup \operatorname{int}(B)$. Claim 3 of Interior commutes with finite intersections and closure with finite unions, while the two reverse combinations are inclusions only and both fail for infinite families; the space is the disjoint union of interior, boundary and exterior asserts only the inclusion $\operatorname{int}(A) \cup \operatorname{int}(B) \subseteq \operatorname{int}(A \cup B)$, and the gap can be the whole space.

Witness. In $\mathbb{R}$ with its usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded) take $A := \mathbb{Q}$, the image of the rationals in $\mathbb{R}$, and $B := \mathbb{R} \setminus \mathbb{Q}$. Then

$\operatorname{int}(A) = \operatorname{int}(B) = \varnothing, \qquad \operatorname{int}(A \cup B) = \operatorname{int}(\mathbb{R}) = \mathbb{R} .$

• The dual failure is the same witness read through complements. Since $\mathbb{R} \setminus \operatorname{int}(S) = \overline{\mathbb{R} \setminus S}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space), the computation above says $\overline{\mathbb{Q}} = \overline{\mathbb{R} \setminus \mathbb{Q}} = \mathbb{R}$ while $\overline{\mathbb{Q} \cap (\mathbb{R} \setminus \mathbb{Q})} = \overline{\varnothing} = \varnothing$; so the same pair witnesses the strictness of $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$.

• Both sets are dense and neither is open. A set with empty interior and full closure is codense and dense at once (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets); $\mathbb{Q}$ is the standard example, and its complement is another. Neither is nowhere dense, their closures being everything.

• The cofinite witness on the general page is the same phenomenon with less machinery. There $A = \{p\}$ and $B = X \setminus \{p\}$ already give a strict inclusion (Interior commutes with finite intersections and closure with finite unions, while the two reverse combinations are inclusions only and both fail for infinite families; the space is the disjoint union of interior, boundary and exterior); the present pair is recorded because it is the failure a reader is most likely to have met, and because it needs the uncountability of intervals rather than a finiteness count.