Because $\mathrm{Set}$ itself is a type, a large type of small types, it must have a secret topology. What is it? There are a number of ways of approaching topology. The most popular one is via open sets. For some spaces, one can instead use convergent sequences, and this approach is more convenient in our situation. It turns out that the topology of the universe $\mathrm{Set}$ is indiscrete: every sequence of types converges to any type! I apply this to deduce that $\mathrm{Set}$ satisfies the conclusion of Rice’s Theorem: it has no non-trivial, extensional, decidable property.

To see how this works, check:

• A short paper with the proofs in mathematical vernacular, and further discussion of the intuitions, motivations and consequences.
• Literate proofs in Agda of the universe indiscreteness theorem and Rice’s Theorem for the universe.
• Agda proofs of related facts.

The Agda pages can be navigated be clicking at any (defined) symbol or word, in particular by clicking at the imported module names.

In the last year or so eff has matured considerably:

• It now looks and feels like OCaml, so you won’t have to learn yet another syntax.
• It has static typing with parametric polymorphism and type inference.
• Eff now clearly separates three basic concepts: effect types, effect instances, and handlers.
• How eff works is explained in our paper on Programming with Algebraic Effects and Handlers.
• We moved the source code to GitHub, so go ahead and fork it!

# fun a b c -> (a b) (fun d -> d c) ;;
- : ('a -> (('b -> 'c) -> 'c) -> 'd) -> 'a -> 'b -> 'd = <fun>

# fun a b c d e e' f g h i j k l m n o o' o'' o''' p q r r' s t u u' v w x y z ->
    q u i c k b r o w n f o' x j u' m p s o'' v e r' t h e' l a z y d o''' g;;
  - : 'a -> 'b -> 'c -> 'd -> 'e -> 'f -> 'g -> 'h -> 'i -> 'j ->
    'k -> 'l -> 'm -> 'n -> 'o -> 'p -> 'q -> 'r -> 's -> 't ->
    ('u -> 'j -> 'c -> 'l -> 'b -> 'v -> 'p -> 'w -> 'o -> 'g ->
     'q -> 'x -> 'k -> 'y -> 'n -> 't -> 'z -> 'r -> 'a1 -> 'e ->
     'b1 -> 'c1 -> 'i -> 'f -> 'm -> 'a -> 'd1 -> 'e1 -> 'd -> 's
     -> 'h -> 'f1) -> 'v -> 'b1 -> 'z -> 'c1 -> 'u -> 'y -> 'a1
     -> 'w -> 'x -> 'e1 -> 'd1 -> 'f1 = <fun>

What is the easiest way to see that this really is the case?

A related question is this (I am sure people have thought about it): how big can a type of a typeable $\lambda$-term be? For example, the Ackermann function can be typed as follows, although the type prevents it from doing the right thing in a typed setting:

# let one = fun f x -> f x ;;
val one : ('a -> 'b) -> 'a -> 'b =
# let suc = fun n f x -> n f (f x) ;;
val suc : (('a -> 'b) -> 'b -> 'c) -> ('a -> 'b) -> 'a -> 'c =
# let ack = fun m -> m (fun f n -> n f (f one)) suc ;;
val ack :
  ((((('a -> 'b) -> 'a -> 'b) -> 'c) ->
   (((('a -> 'b) -> 'a -> 'b) -> 'c) -> 'c -> 'd) -> 'd) ->
   ((('e -> 'f) -> 'f -> 'g) -> ('e -> 'f) -> 'e -> 'g) -> 'h) -> 'h = <fun>

That’s one mean type there! Can it be “explained”? Hmm, why does ack compute the Ackermann function in the untyped $\lambda$-calculus?