Martin Escardo, 20 Feb 2012. We give a negative answer to a question posed by Altenkirch, Anberrée and Li. They asked whether for every definable type X in Martin-Lof type theory, it is the case that for any two provably distinct elements x₀, x₁ : X, there is a function p: X→₂ and a proof d: p x₀ ≠ p x₁. Here ₂ is the type of binary numerals, or booleans if you like, but I am not telling you which of ₀ and ₁ is to be regarded as true or false. If one thinks of ₂-valued maps as characteristic functions of clopen sets in a topological view of types, then their question amounts to asking whether the definable types are totally separated, that is, whether the clopens separate the points. See Johnstone's book "Stone Spaces" for some information about this notion - e.g. for compact spaces, it agrees with total disconnectedness (the connected components are the singletons) and zero-dimensionality (the clopens form a base of the topology), but in general the three notions don't agree. We give an example of a type X whose total separatedness implies WLPO (the weak limited principle of omniscience). The proof works by constructing two elements x₀ and x₁ of X, and a discontinuous function ℕ∞→₂ from any hypothetical p:X→₂ with p x₀ ≠ p x₁, and then reducing discontinuity to WLPO. Our proof postulates extensionality, which entitles us to use UIP on the type ℕ∞. Without the postulate there are fewer closed terms of type X→₂, and their question was for closed terms X, x₀,x₁:X, and d:x₀≠x₁, and so the negative answer also works in the absence of extensionality and of UIP for ℕ∞. But assuming extensionality we get a stronger result, which is not restricted to closed terms. \begin{code} {-# OPTIONS --without-K #-} module FailureOfTotalSeparatedness where open import Two open import GenericConvergentSequence open import SetsAndFunctions open import Equality open import BasicDiscontinuityTaboo open import WLPO open import Extensionality open import CurryHoward open import HSets \end{code} The idea of the following construction is that we replace ∞ in ℕ∞ by two copies ∞₀ and ∞₁, which are different but not distinguishable by maps into ₂, unless WLPO holds. (We can use the Cantor space (ℕ→₂) or the Baire space (ℕ → ℕ), or many other types instead of ℕ∞, with ∞ replaced by any fixed element. But I think the proposed construction gives a more transparent argument.) \begin{code} X : Set X = Σ \(u : ℕ∞) → u ≡ ∞ → ₂ ∞₀ : X ∞₀ = (∞ , λ r → ₀) ∞₁ : X ∞₁ = (∞ , λ r → ₁) \end{code} The elements ∞₀ and ∞₁ look different: \begin{code} naive : π₁ ∞₀ refl ≡ ₀ ∧ π₁ ∞₁ refl ≡ ₁ naive = ∧-intro refl refl \end{code} But there is no function p : X → ₂ such that p x = π₁ x refl, because π₀ x may be different from ∞, in which case π₁ x is the function with empty graph, and so it can't be applied to anything, and certainly not to refl. In fact, the definition p : X → ₂ p x = π₁ x refl doesn't type check, and hence we haven't distinguished ∞₀ and ∞₁ by applying the same function to them. This is clearly seen when the implicit arguments are made explicit. No matter how hard we try to find such a function, we won't succeed: \begin{code} failure : ∀(p : X → ₂) → p ∞₀ ≢ p ∞₁ → WLPO failure p = disagreement-taboo p₀ p₁ lemma where p₀ : ℕ∞ → ₂ p₀ u = p(u , λ r → ₀) p₁ : ℕ∞ → ₂ p₁ u = p(u , λ r → ₁) lemma : ∀ n → p₀(under n) ≡ p₁(under n) lemma n = cong (λ h → p(under n , h)) (funext claim) where claim : ∀(r : under n ≡ ∞) → (λ r → ₀) r ≡ (λ r → ₁) r claim s = ⊥-elim(∞-is-not-ℕ n (sym s)) \end{code} Precisely because one cannot construct maps from X into ₂ that distinguish ∞₀ and ∞₁, it is a bit tricky to prove that they are indeed different: \begin{code} ∞₀-and-∞₁-different : ∞₀ ≢ ∞₁ ∞₀-and-∞₁-different r = zero-is-not-one claim₃ where p : ∞ ≡ ∞ p = cong π₀ r claim₀ : subst {ℕ∞} {λ u → u ≡ ∞ → ₂} p (λ p → ₀) ≡ (λ p → ₁) claim₀ = Σ-≡-lemma ∞₀ ∞₁ r claim₁ : subst {ℕ∞} {λ r → r ≡ ∞ → ₂} p (λ p → ₀) refl ≡ ₁ claim₁ = cong (λ f → f refl) claim₀ fact : refl ≡ p fact = ℕ∞-hset refl p claim₂ : ₀ ≡ subst {ℕ∞} {λ r → r ≡ ∞ → ₂} p (λ p → ₀) refl claim₂ = cong (λ p → subst {ℕ∞} {λ r → r ≡ ∞ → ₂} p (λ p → ₀) refl) fact claim₃ : ₀ ≡ ₁ claim₃ = trans claim₂ claim₁ \end{code}