Logic with Elements of Set Theory Relations & cardinality January 2018 Let ∼ be a relation on X2 := X × X, X 6= ?. We say that ∼ is reflexive, if x ∼ x, for all x ∈ X, irrreflexive, if x 6∼ x, for all x ∈ X, symmetric, if x ∼ y implies that y ∼ x, for all x, y ∈ X, asymmetric, if x ∼ y implies that y 6∼ x, for all x, y ∈ X, antisymmetric, if (x ∼ y ∧ y ∼ x) implies that x = y, for all x, y ∈ X, connex, (or total) if (x ∼ y ∨ y ∼ x), for all x, y ∈ X, semi-connex, if (x ∼ y ∨ y ∼ x), for all x, y ∈ X such that x 6= y, serial, if for all x ∈ X, x ∼ y for some y ∈ X, transitive, if (x ∼ y ∧ y ∼ z) implies that x ∼ z, for all x, y, z ∈ X. Note that the question about antisymmetry can be answered by checking if (x ∼ y ∧ x 6= y) implies that y 6∼ x. Semi-connexity can be restated as (x ∼ y ∨ y ∼ x ∨ x = y) for all x, y ∈ X. An equivalence relation is a relation that is simultaneously reflexive, symmetric and transitive, and then for any x ∈ X the set [x]∼ := {y ∈ X : x ∼ y} is called x’s equivalence class under ∼. The collection of equivalence classes under a given ∼ has the following properties: 1. [x]∼ 6= ?, for all x ∈ X, S 2. x∈X [x]∼ = X, 3. [x]∼ ∩ [y]∼ = ? or [x]∼ = [y]∼ for all x, y ∈ X. Any collection satisfying these conditions is called a partition of X.A partial order is a relation which is reflexive, antisymmetric and transitive. A complete partial order is a total order. Exercise 0. Hilbert’s hotel. Hilbert’s hotel is a hotel with infinitely many single rooms numbered with natural numbers. Assume that every room is occupied. Prove that it is possible to find. 1. at least one vacancy. 4. as many vacancies as rooms in the hotel. 2. at least two vacancies. 5. twice as many vacancies as rooms. 3. any desired finite number of vacancies. Let X be a universal set, i.e. such that all the sets discussed today will be assumed to be subsets of. We say that A and B are equinumerous or that they have the same cardinality, when there is a bijection f : A → B. The relation of equinumerosity is a relation of equivalence in 2X . Its equivalence classes are called the cardinal numbers. The cardinal number of N is denoted by ℵ0 or by ω or ω0 and every set equinumerous to N is called denumerable or countably infinite. The elements of each countably infinite set can be put into a sequence with no repetitions. The cardinal number of [0, 1] is denoted by c or Ω and any set equinumerous to [0, 1] is said to be a set of power of the continuum. 1 A set that is either finite or countably infinite is called countable or at most countable. The elements of such a set can be put into a sequence (maybe with repetitions). A set which is not countable is called uncountable or uncountably infinite or nondenumerable. By the famous Cantor-Schröder-Bernstein theorem, A and B are equinumerous, iff there exist injections f : A → B and g : B → A, or—which is equivalent—surjections φ : A → B and ψ : B → A. Another very important and useful result of Georg Cantor says that A and 2A are never equinumerous. Precisely: there is no surjection from A onto 2A. If there is an injection f : A → B, we say that the cardinality of A is smaller than or equal to that of B and strictly smaller, if that injection is never a bijection. Thus, the Cantor theorem can be expressed by saying that the cardinality of A is strictly smaller than that of 2A. For a set A its cardinality is denoted by card A or |A| or A¯ or # A. Natural numbers can be defined to be the cardinalities of finite sets, 0 being the cardinality of ?. Operations on cardinal numbers: card A + card B := card((A × {0}) ∪ (B × {1})), or when A ∩ B = ?, card A + card B = card(A ∪ B). card A · card B := card(A × B), (card A)card B := card{f : B → A}. When at least one of m, n is infinite: max{m, n} = m + n = m · n. In particular m2 = m, or more general mn = m, when n < m and m is infinite. Moreover, m m m if 2 6 n 6 2 , then n = 2 . Cardinal numbers can be subtracted only when subtrahend is strictly smaller than the minuend (m < n) and the result is n − m = n, when n is infinite. The continuum hypothesis (CH) is a statement that ω ω if ω 6 m 6 2 then m = ω or m = 2 . This cannot be proved nor disproved using the standard set of axioms (= the sentences one doesn’t prove, ω but takes for granted). If we denote the smallest uncountable cardinal by ω1, then CH is saying that ω1 = 2 . 1. For each of the following prove or disprove each of the properties from reflexivity through transitivity. If you find that the relation is an equivalence, derive an explicit formula for [x]∼ if possible, and find the number of equivalence classes and of the elements of each class. 1. X 6= ?: 2 (a) x ∼ y, for all x, y ∈ X (i.e. ∼ = X ), (c) x ∼ y, only when x = x (i.e. ∼ = idX ), (b) x ∼ y, for no x, y ∈ X (i.e. ∼ = ?), (d) x ∼ y, when f(x) = f(y) for a fixed f : X → Y . 2. X = R: (a) x ∼ y, when |x − y| < 1, (k) x ∼ y, when x · y < 1, (b) x ∼ y, when x − y < 1, (l) x ∼ y, when x · y = 0 or x · y = 1, (c) x ∼ y, when |x − y| > 1, (m) x ∼ y, when x − y ∈ Z, (d) x ∼ y, when x · y = 0, (n) x ∼ y, when bxc = byc, (e) x ∼ y, when x · y 6= 0, (o) x ∼ y, when b2xc = b2yc, (f) x ∼ y, when x · y > 0, (p) x ∼ y, when bxc 6= byc, (g) x ∼ y, when x · y > 0, (q) x ∼ y, when bxc 6= by + 1/2c, (h) x ∼ y, when x · y < 0, (r) x ∼ y, when bxc = dye, (i) x ∼ y, when x · y = 1, (s) x ∼ y, when mx = ny, for some m, n ∈ N, (j) x ∼ y, when x · y 6= 1, (t) x ∼ y, when mx = ny, for some m, n ∈ Z. 2 3. Let A, B ⊂ X and x ∼ y, when x ∈ A ∧ y ∈ B or x ∈ B ∧ y ∈ A. Consider the following cases: A ∩ B = ?, A ∩ B 6= ?, A ∪ B = X, A ∪ B 6= X, A = B 6= X, A = X \ B. n p o a c 4. X = q : p, q ∈ Z, q 6= 0 : b ∼ d , when ad = bc. 5. X = Z: (a) x ∼ y, when x − y is odd, (d) x ∼ y, when x|y or y|x, (b) x ∼ y, when x − y is even, (e) x ∼ y, when x|y, 2 2 (c) x ∼ y, when |x − y| 6 1, (f) x ∼ y, when x + y is even, (a) x ∼ y, when x ≡ y mod p, for a fixed p ∈ N \{0}. 7. X = C: (a) x ∼ y, when |x| = |y|, (b) x ∼ y, when |x| 6 |y|. 8. X = R2: (a) x ∼ y, when (0, 0), x and y are collinear, (b) x ∼ y, when there exist lines passing through x and (0, 0) and through y and (0, 0) forming a right angle, (c) x ∼ y, when the triangle 4((0, 0), x, y) is equilateral, (d) x ∼ y, when lie on the same vertical line, (e) x ∼ y, when lie on the same vertical or horizontal line. 9. X = {lines in R3}: (a) x ∼ y, when x ⊥ y, (d) x ∼ y, when x 6 k y, (b) x ∼ y, when x k y, (e) x ∼ y, when x and y intersect, (c) x ∼ y, when x ⊥ y or x k y, (f) x ∼ y, when x and y do not intersect. 10. X = RR (f : R → R): (a) f ∼ g, when f + g is increasing, (b) f ∼ g, when f − g is constant, (c) f ∼ g, when f = k · g, for some k ∈ R, (d) f ∼ g, when f(A) = g(A) for some fixed A 6= ?, (e) f ∼ g, when f ←(B) = g←(B), for a fixed B 6= ?, (f) f ∼ g, when f(x) = g(x), for all x, except for some sequence {xn}n∈N of arguments. 11. X = 2Y : (a) A ∼ B, when A ∩ B = ?, (c) A ∼ B, when A ⊂ B, (b) A ∼ B, when A ∩ B 6= ?, (d) A ∼ B, when A ⊂ B or B ⊂ A, (e) A ∼ B, when A ⊂ B or B ⊂ A or A ∩ B = ?. 12. X = {propositional forms with n variables}: p ∼ q, when they have the same truth tables. 2. Describe all equivalences and partial orders on {0, 1} and {0, 1, 2}. 3. Are the following collections partitions of the given sets? What might be their equivalence relations? 1. X = {polynomials with real coefficients}, An := {polynomials of degree n}, A = {An : n ∈ N}. 3 2. X = R, A = (−∞, 0), B = (0, ∞), A = {A, B}. 3. X = R, A = (−∞, 0], B = [0, ∞), A = {A, B}. 4. X = R, An = [n, n + 1), n ∈ Z, A = {An : n ∈ Z}. 5. X = Z, A = 2Z + 1, B = 2Z, A = {A, B}.