Let be a connected subspace of that is not an interval. Then there exist such that for some . Then may be partitioned into two disjoint open sets as follows

contradicting our requirement that be connected.

Conversely, let be an interval and for some inhabited disjoint open . Without loss of generality assume there exists such that . By the completeness of , the supremum exists, and , so either or , and from openness is either a subset of or . Both situations are a contradiction.