Lindelöff Lemma

October 24, 2025

Every open subset of the real numbers is a disjoint union of at most a countable number of open intervals.

Let $O$ be an open set. Then every point of $O$ is an interior point of $O$ . If $O = \emptyset$ then the set is open and is a disjoint union of zero open intervals. Assume that $O$ is nonempty. For any $x \in O$ the collection $\mathcal{I}(x)$ of all open intervals containing $x$ and contained in $O$ is nonempty. Since $x\in \cap \mathcal{I}(x)$ is nonempty $J_x = \cup \mathcal{I}(x)$ is an open interval, and in fact the largest open interval containing $x$ and contained in $O$ . If $J_x \cap J_y$ is nonempty, then $J_x \cup J_y$ is an open interval containing $x,y$ and contained in $O$ and so $J_x, J_y \subseteq J_x \cup J_y \subseteq J_x,J_y$ and so $J_x = J_y$ . Hence the collection $\mathcal{J} = \{J_x : x \in O\}$ is a disjoint collection. Clearly, $O = \cup_{x \in O} J_x$ . Since the collection is disjoint and since the rationals are a dense subset of the reals, every open interval in $\mathcal{J}$ contains a distinct rational number. Hence the elements of the collection can be labeled by distinct rational numbers. Since any subset that is distinctly labeled by a subset of the rational numbers is at most countable, the conclusion follows.