William Lawvere and the paradox generating machine

This is an elementary introduction to Lawvere's fixed point theorem for cartesian closed categories. We begin by rewriting the conventional presentation of Cantor's diagonal argument, using a more functional approach. Then we show how Russell's paradox can be derived, and construct Curry's fixed point combinator (i.e. the Y combinator).

Transcript

1. William Lawvere & The Paradox Generating Machine Rongmin Lu FP-Syd 24 April 2019
2. Georg Cantor Bertrand Russell
3. Cantor’s Theorem There is no surjective function f : N → 2N. Proof. Such a function f would define an enumeration S1, S2, . . . , Sn, . . . of subsets of N: S1 = {1, 2, 3, 4, 5, . . .} S2 = {1, , 3, 4, 5, . . .} S3 = {1, 2, 3, 4, 5, . . .} S4 = {1, 2, 3, , 5, . . .} S5 = {1, 2, 3, 4, , . . .} . . . G = { , 2, , 4, 5, . . .} However, G cannot occur in the enumeration, since it differs from each Sn at the nth place. Hence, there is no such function f .
4. Cantor’s Theorem - The Statement There is no surjective function f : N → 2N. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}.
5. Cantor’s Theorem - The Statement There is no surjective function f : N → 2N. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N.
6. Cantor’s Theorem - The Statement There is no surjective function f : N → 2N. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N. ∀n ∈ N, |2n| = 2n.
7. Cantor’s Theorem - The Statement There is no surjective function f : N → 2N. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N. ∀n ∈ N, |2n| = 2n. There is a correspondence S ⊆ N ↔ g : N → 2
8. Cantor’s Theorem - The Statement There is no surjective function f : N → 2N. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N. ∀n ∈ N, |2n| = 2n. There is a correspondence S ⊆ N ↔ g(n) = 1 if n ∈ S 0 if n / ∈ S
9. Cantor’s Theorem - The Statement There is no surjective function f : N → (N → 2). 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N. ∀n ∈ N, |2n| = 2n. There is a correspondence S ⊆ N ↔ χS (n) = 1 if n ∈ S 0 if n / ∈ S
10. Cantor’s Theorem - The Statement There is no “surjective” function f : N × N → 2. 2 = {0, 1}, n = {0, 1, . . . , (n − 1)}. 2N is the power set of N. ∀n ∈ N, |2n| = 2n. There is a correspondence S ⊆ N ↔ χS (n) = 1 if n ∈ S 0 if n / ∈ S
11. Surjectivity Definition A function f : X → Y is surjective iff ∀y ∈ Y , ∃x ∈ X such that f (x) = y.
12. Surjectivity Definition A function ¯ f : X → Y X is surjective iff ∀y ∈ Y , ∃x ∈ X such that f (x) = y.
13. Surjectivity Definition A function ¯ f : X → Y X is surjective iff ∀g ∈ Y X , ∃x ∈ X such that ¯ f (x) = g.
14. Surjectivity Definition A function φ: X × X → Y is “surjective” iff ∀g ∈ Y X , ∃x ∈ X such that ¯ f (x) = g.
15. Surjectivity Definition A function φ: X × X → Y is “surjective” iff ∀g : X → Y , ∃x ∈ X such that ∀x ∈ X, φ(x , x) = g(x).
16. Surjectivity Definition A function φ: X × X → Y is not representable by some g : X → Y iff ∀g : X → Y , ∃x ∈ X such that ∀x ∈ X, φ(x , x) = g(x).
17. Surjectivity Definition A function φ: X × X → Y is not representable by some g : X → Y iff ∀x ∈ X, ∃x ∈ X such that φ(x , x) = g(x).
18. The Diagonal Argument — Power Set Version Suppose that there is such a function f . Then f defines an enumeration S1, S2, . . . , Sn, . . . of subsets of N: S1 = {1, 2, 3, 4, 5, . . .} S2 = {1, , 3, 4, 5, . . .} S3 = {1, 2, 3, 4, 5, . . .} S4 = {1, 2, 3, , 5, . . .} S5 = {1, 2, 3, 4, , . . .} . . . G = { , 2, , 4, 5, . . .} However, G cannot occur in the enumeration, since it differs from each Sn at the nth place. Hence, there is no such function f .
19. The Diagonal Argument — Uncountability Version Such a function f also defines an enumeration of infinite sequences of binary digits: S1 = (1 1 1 1 1 . . . ) S2 = (1 0 1 1 1 . . . ) S3 = (1 1 1 1 1 . . . ) S4 = (1 1 1 0 1 . . . ) S5 = (1 1 1 1 0 . . . ) . . . G = (0 1 0 1 1 . . . )
20. The Diagonal Argument n → f 1 2 3 4 5 . . . 1 1 1 1 1 1 . . . 2 1 0 1 1 1 . . . 3 1 1 1 1 1 . . . 4 1 1 1 0 1 . . . 5 1 1 1 1 0 . . . . . . . . . . . . . . . ... m ↓ f : N × N → 2, f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm
21. The Diagonal Argument — Constructing G How did we construct G? “f defines an enumeration”
22. The Diagonal Argument — Constructing G How did we construct G? “f defines an enumeration” G “differs from”
23. The Diagonal Argument — Constructing G How did we construct G? “f defines an enumeration” G “differs from” “each Sn at the nth place”
24. The Diagonal Argument — Constructing G How did we construct G? f : N × N → 2, f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm G “differs from” “each Sn at the nth place”
25. The Diagonal Argument — Constructing G How did we construct G? f : N × N → 2, f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm t : 2 → 2, t(n) = 1 n = 0 0 n = 1 Note that t has no fixed points. “each Sn at the nth place”
26. The Diagonal Argument — Constructing G How did we construct G? f : N × N → 2, f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm t : 2 → 2, t(n) = 1 n = 0 0 n = 1 Note that t has no fixed points. ∆: N → N × N is the diagonal map.
27. The Diagonal Argument — Constructing G How did we construct G? f : N × N → 2, f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm t : 2 → 2, t(n) = 1 n = 0 0 n = 1 Note that t has no fixed points. ∆: N → N × N is the diagonal map. We can compose these to get a function g : N → 2: N ∆ − − − − → N × N f − − − − → 2 t − − − − → 2
28. The Diagonal Argument — The Diagram We get the following diagram, which commutes: N × N f − − − − → 2 ∆   t N g − − − − → 2 Note that g = χ(G) is the characteristic function of the set G = {n ∈ N|n / ∈ Sn}, f : N × N → 2 is f (n, m) = 1 n ∈ Sm 0 n / ∈ Sm , and t has no fixed points.
29. The Diagonal Argument N × N f − − − − → 2 ∆   t N g − − − − → 2 Note that g is the characteristic function of the set G = {n ∈ N|n / ∈ Sn}. Observe that ∀n ∈ N, f (−, n) = g(−). Otherwise, if f (−, n0) = g(−) for some n0 ∈ N, then f (n0, n0) = g(n0) = t(f (n0, n0)). This shows that t has a fixed point, which is a contradiction.
30. Russell’s Paradox Substitute Sets for N and define f : Sets × Sets → 2 to be f (S, T) = 1 S ∈ T 0 S / ∈ T . Russell’s paradox arises by noting that we cannot construct a similar function g : Sets → 2, as that would lead us to conclude that, for some S ∈ Sets, f (S, S) = g(S) = t(f (S, S)), i.e. that S ∈ S and S / ∈ S at the same time.
31. Grelling’s Paradox Call an adjective “heterological” if it does not describe itself.
32. Grelling’s Paradox Call an adjective “heterological” if it does not describe itself. “English” is English, but “French” is not French.
33. Grelling’s Paradox Call an adjective “heterological” if it does not describe itself. “English” is English, but “French” is not French. “Short” is not short and “long” is not long.
34. Grelling’s Paradox Call an adjective “heterological” if it does not describe itself. “English” is English, but “French” is not French. “Short” is not short and “long” is not long. Is “heterological” heterological?
35. Grelling’s Paradox Call an adjective “heterological” if it does not describe itself. “English” is English, but “French” is not French. “Short” is not short and “long” is not long. Is “heterological” heterological? Adj × Adj f − − − − → 2 ∆   t Adj g − − − − → 2
36. The Y Combinator The diagonal argument can be used positively.
37. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t.
38. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x).
39. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A).
40. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A). Construction of Y : Define f (x, y) := x(y) as the composition of terms x and y.
41. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A). Construction of Y : Define f (x, y) := x(y) as the composition of terms x and y. Then t(f (x, x)) = t(x(x)) = A(x) = f (A, x).
42. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A). Construction of Y : Define f (x, y) := x(y) as the composition of terms x and y. Then t(f (x, x)) = t(x(x)) = A(x) = f (A, x). Let A := t(x(x)).
43. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A). Construction of Y : Define f (x, y) := x(y) as the composition of terms x and y. Then t(f (x, x)) = t(x(x)) = A(x) = f (A, x). Let A := t(x(x)). Then Y = f (A, A) = A(A) = (t(x(x)))(t(x(x))).
44. The Y Combinator The diagonal argument can be used positively. We want to find fix t, the fixed point for a function t. We need to find terms f and A such that for any term x, t(f (x, x)) = f (A, x). Then the fixed point is Y := f (A, A). Construction of Y : Define f (x, y) := x(y) as the composition of terms x and y. Then t(f (x, x)) = t(x(x)) = A(x) = f (A, x). Let A := t(x(x)). Then Y = f (A, A) = A(A) = (t(x(x)))(t(x(x))). In lambda calculus, Y = λt.(λx.t(xx))(λx.t(xx)).
45. References William Lawvere, “Diagonal Arguments and Cartesian Closed Categories”, Lecture Notes in Mathematics vol. 92 (1969), pp. 134–145. William Lawvere, “Diagonal Arguments and Cartesian Closed Categories”, Reprints in Theory and Applications of Categories, 15 (2006), pp. 1–13. Noson Yanofsky, “A Universal Approach to Self-Referential Paradoxes, Incompleteness and Fixed Points”, arXiv:math/0305282.