No Set Carries Exactly Three Dense Linear Orders without Endpoints: A Proof in a Weak Zermelo Theory without Choice or Replacement

Abstract

Let Z_sep be the theory consisting of Extensionality, Pairing, Infinity, Union, Power Set, and the full Separation schema. Neither Choice, Replacement, Foundation, nor any form of Countable Choice is assumed. For a standard finite n ≥ 1, the notation s(X) = n abbreviates a first-order formula saying that X carries n, but not n + 1, pairwise nonisomorphic dense linear orders without endpoints. We prove Z_sep ⊢ ¬∃X (s(X) = 3). The first part is a countable-benchmark argument. If ω ↪ X for a DLO carrier X, a choice-free monotone-subsequence construction produces a countable set B ⊆ X whose complement is again a DLO. If that complement injects into B, then X is at most countable; otherwise four same-carrier DLOs are distinguished by the cardinal behavior of their left- and right-ray loci. For the second part, exact three supplies a self-dual DLO type. A nontrivial increasing automorphism immediately gives an injection ω ↪ X. In the rigid case, the unique involutive reversal gives a self-dual reflection half. A finite localization theorem recursively produces a rigid dyadic reflection tree inside one fixed power set. Its center set is at most countable. If the half is not at most countable, one point outside all centers determines a branch, and successive local reflections form an injective ω-sequence. Both alternatives contradict exact three. The result applies to every carrier and therefore gives a negative answer to Shelah's exact-three question for models of Th(Q,<) on one underlying set. The general finite-spectrum problem is not resolved here.