On the compatibility of unbroken recurrence with vacant persistent occupancy

Received: not recorded. Accepted: not recorded.

1. Definitions

Let H be a house and let n ∈ ℕ index its nights. Write P_n for the event that a party occurs in H on night n, and G_n for the set of beings present at it.

Say H is recurrent if a party occurs on every night,

R(H) = 1 ⟺ ∀n. P_n

and define its persistent occupancy as the set of beings present on every night without exception,

Ω(H) = ⋂_n∈ℕ G_n.

Call H non-retentive if no being present on one night is present on the next:

∀n. G_n ∩ G_n+1 = ∅.

2. Lemma

If H is non-retentive then Ω(H) = ∅.

Suppose x ∈ Ω(H). Then x ∈ G_n for every n, and in particular x ∈ G₁ and x ∈ G₂. Hence x ∈ G₁ ∩ G₂, which is empty by non-retention. So no such x exists.

3. Corollary

Recurrence does not entail occupancy. The conjunction

R(H) = 1  ∧  Ω(H) = ∅

is consistent, and is satisfied by any non-retentive recurrent house. Every night there is a party; no one attends twice; and there is no being of whom it may truthfully be said that they live there.

The temptation to infer an occupant from an unbroken record of parties is therefore an error of quantifier order and not of evidence. No further observation of H will correct it.

4. Remark on nomenclature

It is customary in this archive to name Ω(H) rather than merely to evaluate it, on the grounds that a set which is asked after nightly has, for practical purposes, a name.¹ The convention is not original to us.²

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¹ The nightly interrogative — is anyone home — is well formed and has, on the above, the determinate answer Ω(H). Practitioners report that possession of the answer does not abate the question.

² Hom. Od. ix. 366: Οὖτις ἐμοί γ᾽ ὄνομα. The name there given is the same object; the neighbours, hearing it correctly reported, went home.

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