Introduction Basic Analysis Choice Sequences Conclusion All Functions are Continuous! A provocative introduction to constructive analysis Ryan Acosta Babb University of Warwick 2019 Introduction Basic Analysis Choice Sequences Conclusion A dissatisfying proof Exercise Find irrational numbers a and b such that ab is rational. \Solution". p p p 2 p 2 By cases! Either 2 2 or 2 2= . p Q Q p 2 p 2 2 . Just take a = b = 2. p Q p p 2 p 2 p 2 2= . Then take a = 2 and b = 2: Q p p 2 p p p 2 p 2 2 p 2 ab = 2 = 2 = 2 = 2: Sure ::: but we didn't find a and b! Introduction Basic Analysis Choice Sequences Conclusion The Law of Excluded Middle The \culprit" is The Law of Excluded Middle: ' _:' (LEM) Yields non-constructive proofs! What would mathematics without LEM look like? Introduction Basic Analysis Choice Sequences Conclusion Five things you wanted to know about constructive mathematics but were too afraid to ask 1 Does constructive mathematics accept contradictions? 2 Does constructive mathematics reject proof by contradiction? 3 Does constructive mathematics reject the Axiom of Choice? 4 Is constructive mathematics compatible with classical mathematics? 5 Is constructive mathematics \poorer" than classical mathematics? Is [your favourite theorem] still valid? Introduction Basic Analysis Choice Sequences Conclusion 1. Does constructive mathematics accept contradictions? NO! Constructive mathematics does not assume ' _:' in general. It does not accept ' ^ :' or :(' _:'). Compare to agnosticism about Axiom of Choice: we refrain from using it in proofs. Introduction Basic Analysis Choice Sequences Conclusion Decidability Constructivism does accept LEM in some cases: for decidable propositions. '(x) is decidable if there is a \general procedure" (\algorithm") to decide '(x) _:'(x) for each individual x in finitely may steps! Example: GC(n) ≡ 2n + 2 is the sum of two primes. Then: 8n 2 N(GC(n) _:GC(n)): (X) Just compute all prime numbers up to 2n + 2 and check each one! Compare the full Goldbach Conjecture: 8n 2 N(GC(n)) _ :8n 2 N(GC(n)) (7) Introduction Basic Analysis Choice Sequences Conclusion 2. Does constructive mathematics reject proof by contradiction? NO! Compare two types of \proof by contradiction": Proof of Negation Reductio Ad Absurdum ' :' . ? ? :' ' p OK! E.g. 2 2= Q. Not OK! Assumes LEM! Introduction Basic Analysis Choice Sequences Conclusion A closer look: Reductio Ad Absurdum :' . ? ::' Negation Introduction Rule Double' Negation Rule \Double Negation Rule" is equivalent to LEM! ' _:' () ::' ! ' Introduction Basic Analysis Choice Sequences Conclusion 3. Does constructive mathematics reject the Axiom of Choice? YES! Theorem (Diaconescu) The Axiom of Choice implies LEM. If we cannot assume LEM, then we must reject Choice! Introduction Basic Analysis Choice Sequences Conclusion Proof. 1 Pick a proposition ' and define: U := fx 2 f0; 1g :(x = 0)_'g;V := fx 2 f0; 1g :(x = 1)_'g 2 Choice function f s.t. f(U) 2 U and f(V ) 2 V 3 (f(U) = 0) _ ' and (f(V ) = 1) _ ' 4 (f(U) 6= f(V )) _ ' 5 Now, ' ! (U = V ), so ' ! (f(U) = f(V )) 6 (f(U) 6= f(V )) !:' 7 Hence ' _:' by (4) and (6). Introduction Basic Analysis Choice Sequences Conclusion 4. Is constructive mathematics compatible with classical mathematics? YES... andNO... It's complicated! YES: Rejecting LEM makes constructive mathematics a generalisation of classical mathematics: Constructivist Mathematics ( Classical Mathematics Compare: all fields are rings but not all rings are fields! NO: Some varieties of constructive mathematics make additional assumptions which are incompatible with classical mathematics. More on this later! Introduction Basic Analysis Choice Sequences Conclusion 5. Is constructive mathematics \poorer" than classical mathematics? It may seem that Constructive Mathematics \loses" something by avoiding LEM: many classical theorems become unprovable. Casualties include 1 Tychonov's Theorem. 2 Zorn's Lemma. 3 Intermediate Value Theorem. 4 Bolzano-Weierstraß Theorem. 5 Trichotomy for (R; <),i.e 8x 2 R(x < 0 _ x = 0 _ x > 0). But we also gain some new theorems! More on this later. Introduction Basic Analysis Choice Sequences Conclusion Five things you wanted to know about constructive mathematics but were too afraid to ask 1 Does constructive mathematics accept contradictions?NO 2 Does constructive mathematics reject proof by contradiction?NO 3 Does constructive mathematics reject the Axiom of Choice? YES 4 Is constructive mathematics compatible with classical mathematics? Erm... 5 Is constructive mathematics \poorer" than classical mathematics? Is [your favourite theorem] still valid? Erm... 3/5 { not bad! Introduction Basic Analysis Choice Sequences Conclusion Building blocks of analysis We accept arithmetic on N and Z \as usual"! Atomic statements about integers, i.e. a = b and a < b, are decidable. In particular: 8x; y 2 Z(x < y _ x = y _ x > y): Arithmetic on Q can be reduced to arithmetic on Z! In particular: 8x; y 2 Q(x < y _ x = y _ x > y): N.B. Hereafter, quantifiers of the form 8n and 9n range over N. Introduction Basic Analysis Choice Sequences Conclusion Real number generators We introduce real numbers as \Cauchy sequences" over Q. Definition A real number generator (RNG) is a Cauchy sequence of rationals, i.e an (an) ⊂ Q such that 8k9n8p(jan+p − anj < 1=k): Definition Two RNG's (an) and (bn) coincide if 8k9n8p(jan+p − bn+pj < 1=k): We write:( an) ≈ (bn). We identify real numbers with RNGs modulo coincidence: a = b () (an) ≈ (bn) Introduction Basic Analysis Choice Sequences Conclusion A curious theorem Classically: Theorem If a 6= b is impossible, then a = b. Proof. SERIOUSLY?! Constructively things are little more subtle. Recall: a 6≈ b () a ≈ b !? Theorem If (an) 6≈ (bn) is impossible, then (an) ≈ (bn). Introduction Basic Analysis Choice Sequences Conclusion Proof (Heyting, 1966). 1 Fix k and let n be such that, for all p: jan+p − anj < 1=4k and jbn+p − bnj < 1=4k: 2 Suppose jan − bnj > 1=k. Then jan+p − bn+pj > 1=2k for all p 2 N: 3 Hence, (an) 6≈ (bn) { but this is impossible! 4 Thus, jan − bnj < 1=k. [an; bn 2 Q!] 5 It follows that jan+p − bn+pj < 2=k for all p 2 N: 6 Therefore, (an) ≈ (bn). Introduction Basic Analysis Choice Sequences Conclusion Is this necessary? YES! Constructively: ' 6() ::'! Example (Fundamental Theorem of \Algebra") Let ' ≡8 p 2 C[x]9z 2 C(p(z) = 0). A proof of ' is an \algorithm" that, given a polynomial p 2 C[x], yields (i) a z 2 C and (ii) a proof that p(z) = 0. A proof of ::' is an \algorithm" that given a proof of :' produces a contradiction. In general, a construction of a contradiction does not tell us how to construct a root z 2 C! In a sense, these are different facts { constructive mathematics is sensitive to this difference in a way that classical mathematics is not! Introduction Basic Analysis Choice Sequences Conclusion Logical fine-print Consider the three statements: 1 (an) ≈ (bn) () 8k9n8p(jan+p − bn+pj < 1=k) 2 (an) 6≈ (bn) () 8k9n8p(jan+p − bn+pj < 1=k) !? 3 :((an) 6≈ (bn)) () :8k9n8p(jan+p − bn+pj < 1=k) !? Classically, (1) and (3) are equivalent. Constructively, there is no a priori reason why (3) should yield the positive statement (1): the latter specifies the distance between two RNGs. This is because we lose the (classical) equivalence: :8x'(x) () 9x:'(x) due to the stronger constructive reading of 9. Introduction Basic Analysis Choice Sequences Conclusion Positive and negative notions Compare: \Negative inequality": :8k9n8p(jan+p − bn+pj < 1=k) \Positive inequality": 9k9n8p(jan+p − bn+pj > 1=k) The positive statement tells us exactly how \separated" the RNGs (an) and (bn) are, and is thus a stronger statement. Definition Two RNGs (an) and (bn) lie apart if 9k9n8p(jan+p − bn+pj > 1=k): We write:( an)#(bn), and also a # b. We already proved: :((an)#(bn)) () (an) ≈ (bn): Introduction Basic Analysis Choice Sequences Conclusion Non-decidability of equality Define the following RNG: ( 1=n if 8k 6 nGC(k); an := 1=m if 9m 6 n(:GC(m) ^ 8k < mGC(k)): This is a well-defined RNG. It is clear that a = 0 () the Goldbach Conjecture is proved. a # 0 () the Goldbach Conjecture is refuted. The only way to (constructively) decide if a = 0 or a # 0 is to have a proof or counterexample for the Goldbach Conjecture. Non-decidability of equality for R Constructively: R 6= fx : x = 0 _ x # 0g. Introduction Basic Analysis Choice Sequences Conclusion Proof that (an) is an RNG. 1 Fix k 2 N and take n = 2k > k. 2 For each l 6 n, decide whether GC(l) or :GC(l) holds. 3 If 8l 6 nGC(l), then an = 1=n and an+p 6 1=n, for all p, so jan+p − anj 6 1=n + 1=n < 1=k: 4 Otherwise, if m is the least number less than n such that :GC(m) holds, then an = an+p = 1=m for all p, so jan+p − anj = 0 < 1=k: 5 Therefore, 8p(jan+p − anj < 1=k). The proof only requires \algorithms" to find the right choices of n given k { no need to solve Goldbach! Introduction Basic Analysis Choice Sequences Conclusion Order relation for RNGs Definition Let a and b be reals. We say that a < b if 9k9n8p(bn+p − an+p > 1=k): We write a 6 b for :(b < a). We also let [a; b] := fx : a 6 x 6 bg: Warning a 6 b 6() (b < a _ a = b) It can be shown that: jx + yj 6 jxj + jyj .