Logic with Elements of Set Theory Relations & Cardinality January 2018

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 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}.

6. X = Z, An = {n, n + 1}, for n ∈ Z, A = {An : n ∈ Z}.

7. X = Z, An = {2n, 2n + 1}, for n ∈ Z, A = {An : n ∈ Z}.

8. X = Z, An = {3n, 3n + 1}, for n ∈ Z, A = {An : n ∈ Z}. 9. X = Z, A = 3Z, B = 2Z, C = 6Z + 3, A = {A, B, C}. 10. X ={permutations of all finite sets},A={even permutations},B ={odd permutations}, A = {A, B}. 4. If A = {1, 2, 3, 4}, give and example of a relation on A that is 1. reflexive and symmetric, but not transitive. 2. reflexive and transitive, but not symmetric. 3. symmetric and transitive, but not reflexive. 5. Prove or disprove that the following pairs of sets are equinumerous:

1. {−5, −4,..., 5} and {1, 2,..., 10}, 8. N − 3 and N + 4,

2. {−8, −6,..., 6, 8} and {1, 2,..., 8}, 9. 2N and 2N + 1,

3. {−10, −8, −6,..., 8, 10} and {0, 1,..., 10}, 10. N and 2N + 1,

4. 2{0,1,...,n−1} and {1, 2, 3,..., 2n}, 11. 3N + 1 and 5N + 3, n 5. N and {−5, −4,...} = N − 5, 12. {1, 2, 4,..., 2 ,...} and N, 1 2 3 1 6. N and N ∪ {∞}, 13. N and 0, 5 , 5 , 5 ,... = 5 N, 1 4 6 9 11 7. N and N ∪ {−∞, +∞}, 14. N and 5 ,..., 5 , 5 ,..., 5 , 5 ,... ,

15. N and Z, 21. N and Q, 27. (0, 1] and [0, 1], 33. R and R \{0},

16. Z and 3Z + 1, 22. N and N3, 28. (−5, −2) and [0, 1], 34. 2N and [0, 1],

1 1 1 π π N 17. Z and {1, 2 , 3 , 4 ,...}, 23. N and [0, 1], 29. R and − 2 , 2 , 35. 2 and R, 18. N and N2, 24. Q and [0, 1], 30. R and [0, 1], 36. R and R ∪ {−∞, +∞},

19. N2 and Z2, 25. [0, 1] and [2, 10], 31. R and (0, ∞), 37. R \ N and R, 20. N2 and Q, 26. [0, 1] and [e, π], 32. R and Q, 38. R \ Q and R. 6. Simplify:

1. 2 + 0, 6. 6ω − 3ω, 11. ω0, 16. 3ω5ω, 21. c · ωω, 2. 1 + 1, 7. 0 · ω, 12. ω1, 17. 2ω − 4ω2, 22. ωc,

c 3. ω + 5, 8. ω5, 13. ω2ω, 18. 3ω − 3ω3, 23. ωω ,

2 c 4. 3ω + 100, 9. 3 · (5 − 2), 14. 2ω , 19. cc, 24. cω , 5. ω − 24, 10. ω6 − 2ω, 15. ωω, 20. cω, 25. (cc − ωω)ω.

Further reading: http://www.logicmatters.net/resources//pdfs/godelexercises/Equinum_solutions.pdf http://www.math.ksu.edu/~nagy/real-an/ap-b-card.pdf http://www.ii.uni.wroc.pl/~wch/Dydaktyka/Logic08//07cardinalities.pdf http://www.sci.brooklyn.cuny.edu/~mate/misc/cardinalities.pdf

4 Some solutions Ex. 1 (b) ∼ = ∅. Reflexivity NO. We have x 6∼ y for all x, y ∈ X, so in particular x 6∼ x. Irreflexivity YES. See above. Symmetry YES. The key here and subsequently is the “(false ⇒ anything) is true” rule: x is never related to x, making the assumption false. Hence the whole sentence is true never mind the conclusion. Asymmetry YES. The same rule applies: x is never related to y, so the assumption is false. Antisymmetry YES. The same rule applies: x is never related to y nor y to x, so the assumption is false. Completeness NO, for x is never related to y nor y to x. Transitivity YES. This property is again in a form of implication with a false assumption. Ex. 1 (a) ∼ = X2. Reflexivity YES. We have x ∼ y for all x, y ∈ X, so in particular x ∼ x. Irreflexivity NO. See above. Symmetry YES, for x is always related to y and y to x. Asymmetry NO. See above. Antisymmetry NO, if X has more than one element. For, if x 6= y ∈ X, then x ∼ y and y ∼ x. Completeness YES. It’s trivial. Transitivity YES. Trivial again. Hence, this is an equivalence relation. Ex. 1 (c) ∼ = idX . Reflexivity YES. We have x ∼ x, for x ∈ X. Irreflexivity NO. See above. Symmetry YES. Only x is related to x. Asymmetry NO, as x ∼ y means y = x, so y ∼ x. Antisymmetry YES, for if x ∼ y then y = x. Completeness NO, if X has moe than one element. For then, if x 6= y ∈ X, then neither x ∼ y nor y ∼ x. Transitivity YES. Trivial: x = y and y = z means that x = z. Therefore, that one is an equivalence and a partial order. Ex. 1 (d) x ∼f y iff f(x) = f(y). Reflexivity YES. We have f(x) = f(x), for x ∈ X. Irreflexivity NO. See above. Symmetry YES. If f(x) = f(y) then f(y) = f(x). Asymmetry NO. See above. Antisymmetry YES, for injections. NO for other functions. Completeness YES for constant functions, NO for others. Transitivity YES. Trivial: f(x) = f(y) and f(y) = f(z) means that f(x) = f(z). We conclude that this relation is always an equivalence. For injections it is a partial order. Since constant injections can be defined only on singletons, the relation is a total order only when X has one element.

5 Ex. 2 (b) x ∼ y iff xy = 0. Reflexivity NO. For x = 1 we have 1 · 1 = 1 6= 0, so 1 6∼ 1. Irreflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 ∼ 0. Symmetry YES. If xy = 0 then yx = 0. Asymmetry NO. See above. Antisymmetry NO. For x = 0 and y = 1 we have 0 · 1 = 0 but 0 6= 1. Completeness NO. 1 6∼ 2 nor 2 ∼ 1. Transitivity NO. 1 · 0 = 0 and 0 · 2 = 0, but 1 · 2 = 2 6= 0. Ex. 2 (c) x ∼ y iff xy 6= 0. Reflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 6∼ 0. Irreflexivity NO. For x = 1 we have 1 · 1 = 1 6= 0, so 1 ∼ 1. Symmetry YES. If xy 6= 0 then yx 6= 0. Asymmetry NO. See above. Antisymmetry NO. 2 · 1 6= 0 but 2 6= 1. Completeness NO. 1 6∼ 0 nor 0 ∼ 1. Transitivity YES. For x ∼ y and y ∼ z mean that neither of x, y, z is zero. Thus, xz 6= 0. Ex. 2 (d) x ∼ y iff xy > 0. Reflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 6∼ 0. Irreflexivity NO. For x = 1 we have 1 · 1 = 1 > 0, so 1 ∼ 1. Symmetry YES. If xy > 0 then yx > 0. Asymmetry NO. See above. Antisymmetry NO. 2 · 1 = 1 · 2 > 0 but 2 6= 1. Completeness NO. 1 6∼ 0 nor 0 ∼ 1. Transitivity YES. For x ∼ y and y ∼ z mean that the signs of x, y and z equal. Thus, xz > 0. Ex. 2 (e) x ∼ y iff xy < 0. Reflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 6∼ 0.

Irreflexivity YES, x · x > 0. Symmetry YES. If xy < 0 then yx < 0. Asymmetry NO. See above. Antisymmetry NO. We have −2 · 1 = 1 · (−2) < 0 but −2 6= 1. Completeness NO. 1 6∼ 0 nor 0 ∼ 1. Transitivity NO. For x ∼ y and y ∼ z mean that the signs of x, y are opposite and so are the signs of y, z. Thus, xz > 0. Ex. 2 (f) x ∼ y iff xy = 1. Reflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 6∼ 0. Irreflexivity NO. For x = 1 we have 1 · 1 = 1, so 1 ∼ 1.

6 Symmetry YES. If xy = 1 then yx = 1 as well.

Asymmetry NO. See above.

1 1 1 Antisymmetry NO. We have 2 · 2 = 2 · 2 = 1 but 2 6= 2 . Completeness NO. 1 6∼ 0 nor 0 ∼ 1.

1 Transitivity NO. Take x = z = 2 and y = 2. Obviously x ∼ y and y ∼ z, but x · z = 4 6= 1, so x 6∼ z. Ex. 2 (g) x ∼ y iff xy < 1. Reflexivity NO. For x = 1 we have 1 · 1 = 1, so 1 6∼ 1. Irreflexivity NO. For x = 0 we have 0 · 0 = 0, so 0 ∼ 0.

Symmetry YES. If xy < 1 then yx < 1 as well. Asymmetry NO. See above.

1 1 1 1 1 1 1 Antisymmetry NO. We have 3 · 2 = 2 · 3 = 6 < 1 but 3 6= 2 . Completeness NO. 1 6∼ 2 nor 2 ∼ 1.

1 Transitivity NO. Take x = z = 2 and y = 3 . Obviously x ∼ y and y ∼ z, but x · z = 4 > 1, so x 6∼ z. Ex. 5 (c) x ∼ y iﬀ x | y. Reﬂexivity YES. We always have x | x.

Irreﬂexivity NO. See above.

Symmetry NO. 1 | 2 but 2 - 1. Asymmetry NO. Take y = x.

Antisymmetry YES, since x | y and y | x imply that x = y. Completeness NO. 3 6∼ 2 nor 2 ∼ 3.

Transitivity YES. If y = kx for some k ∈ Z and z = my for some m ∈ Z, then z = kmx and km ∈ Z. This proves that the relation is a partial order which is not total. Ex. 5 (b) x ∼ y iff x | y or y | x. Reflexivity YES. We always have x | x. Irreflexivity NO. See above. Symmetry YES. It’s obvious.

Asymmetry NO. Take y = x. Antisymmetry NO. 1 | 2, so 1 ∼ 2 and 2 ∼ 1. Completeness NO. 3 6∼ 2 nor 2 ∼ 3.

Transitivity NO. Take x = 3, y = 12 and z = 2. Obviously x ∼ y (since 3 | 12) and y ∼ z (since 2 | 12), but 2 - 3, nor 3 | 2. Ex. 8 (a) x ∼ y iff x ⊥ y. Reflexivity NO. No line is at right angles with itself. Irreflexivity YES. See above.

Symmetry YES. It’s obvious. Asymmetry NO. Take y = x.

7 Antisymmetry NO. {(x, 0, 0) ∈ R3 : x ∈ R} ⊥ {(0, y, 0) ∈ R3 : y ∈ R}, but these lines are certainly distinct. Completeness NO. Parallel lines are unrelated.

Transitivity NO. Take x¯ = {(x, 0, 0) ∈ R3 : x ∈ R}, y¯ = {(0, y, 0) ∈ R3 : y ∈ R} and z¯ = {(x, 1, 0) ∈ R3 : x ∈ R}. Obviously x ∼ y and y ∼ z, but x¯ k z¯, so they are not perpendicular. Ex. 8 (b) x ∼ y iff x k y. Reflexivity YES. A line is parallel to itself. Irreflexivity NO. See above. Symmetry YES. It’s obvious. Asymmetry NO. See above.

Antisymmetry NO. {(x, 0, 0) ∈ R3 : x ∈ R} k {(x, 1, 0) ∈ R3 : x ∈ R}, but these lines are certainly distinct. Completeness NO. Perpendicular lines are unrelated. Transitivity YES. It’s obvious. So this one is an equivalence. Ex. 8 (e) x ∼ y iff x and y intersect. Reflexivity YES. A line intersects itself. Irreflexivity NO. See above. Symmetry YES. It’s obvious. Asymmetry NO. See above.

Antisymmetry NO. {(x, 0, 0) ∈ R3 : x ∈ R} intersects {(0, y, 0) ∈ R3 : y ∈ R}, but these lines are certainly distinct. Completeness NO. Distinct parallel lines are unrelated.

Transitivity NO. x¯ = {(x, 0, 0) ∈ R3 : x ∈ R} intersects y¯ = {(0, y, 0) ∈ R3 : y ∈ R}, y¯ intersects z¯ = {(x, 1, 0) ∈ R3 : x ∈ R} but x¯ k z¯ and these are distinct, so they do not cross. Ex. 11 (a) x ∼ y iff x ∩ y = ∅. Reflexivity NO if Y 6= ∅. Indeed, let x 6= ∅, x ⊂ Y . Then x ∩ x = x 6= ∅. Irreflexivity NO. Take x = ∅. Then x ∼ x. Symmetry YES. It’s obvious. Asymmetry NO. See above.

Antisymmetry NO if Y 6= ∅. Indeed, let x 6= ∅ = y and x ⊂ Y . Then x ∩ y = ∅, so x ∼ y, but x 6= y. Completeness NO if Y 6= ∅. See reflexivity. Transitivity NO if Y 6= ∅. Take x = z 6= ∅ = y. Then x ∼ y, y ∼ z but z 6∼ z. Ex. 11 (a) x ∼ y iff x ⊂ y. Reflexivity YES, as we always have x ⊂ x. Irreflexivity NO. See above.

Symmetry NO if Y 6= ∅. Let y 6= x = ∅. Then x ∼ y but y 6∼ x. Asymmetry NO. Take x = y. Antisymmetry YES. Obvious.

Completeness NO if Y 6= ∅. See symmetry. Transitivity YES. Obvious.