Foundations of Mathematics (HS 2016) Solution to Exercises 4.6 and 4.8 Mathilde Bouvel Exercise 4.6 Prove (in natural mathematical language, not formally) the following remark from the lecture: a. if R is a weak partial (resp. total) order on A, then R n f(x; x): x 2 Ag is a strict partial (resp. total) order on A; b. if R is a strict partial (resp. total) order on A, then R [ f(x; x): x 2 Ag is a weak partial (resp. total) order on A. Recall that the set f(x; x): x 2 Ag is called the diagonal. Solution for a. in the case of R a weak partial order About R, we know that it is reflexive, antisymmetric, and transitive. We want to prove that R0 = R n f(x; x): x 2 Ag is a strict partial order on A, that is to say that R0 is transitive and irreflexive on A. Proof of transitivity of R0: Consider x, y and z in A such that xR0y and yR0z. That xR0y means that x 6= y and xRy. Similarly, we have y 6= z and yRz. It then follows that xRz by transitivity of R. If x 6= z, then we also have xR0z (since again, (x; z) does not belong to the diagonal). And we will now see that the case x = z is impossible. Indeed, if x = z, yRz would mean yRx. So we would have xRy and yRx, and the antisymmetry of R implies that x = y, which we know is not the case. Therefore, we have shown that xR0z, so R0 is transitive. Proof of irreflexivity of R0: By definition of R0, for any x 2 A,(x; x) 2= R0 (because the diagonal has been removed). This means exactly that R0 is irreflexive. Solution for a. in the case of R a weak total order About R, we know that it is a weak total order on A. That is to say, R is a weak partial order on A which in addition satisfies totality. We want to prove that R0 = R n f(x; x): x 2 Ag is a strict total order on A. By the above proof, we already know that R0 is a strict partial order on A. So, it is only left to show that R0 satisfies trichotomy. Proof of trichotomy of R0: Consider x and y in A. We want to prove that one of the following holds: xR0y, or yR0x, or x = y. This is clearly true if x = y. So, let us assume that x 6= y. It 1 implies that (x; y) and (y; x) do not belong to the diagonal. Because R satisfies totality, we know that xRy or yRx holds. And because (x; y) and (y; x) do not belong to the diagonal, this implies that xR0y or yR0x holds. This concludes the proof that R0 satisfies trichotomy. Solution for b. in the case of R a strict partial order About R, we now know that it is transitive and irreflexive. We want to prove that R0 defined as the union of R and the diagonal is a weak partial order, that it to say that R0 is reflexive, transitive and antisymmetric. Proof of reflexivity of R0: Because R0 contains the diagonal, R0 is by definition reflexive (xRx for all x 2 A). Proof of transitivity of R0: Consider x, y and z in A such that xR0y and yR0z. We distinguish several cases. • If x 6= y and y 6= z, then it follows from xR0y and yR0z that both (x; y) and (y; z) belong to R. By transitivity of R, we obtain that xRz, and in particular xR0z (since R ⊆ R0). • If x = y and y 6= z, then we have yRz as before. Now, yRz means xRz, which implies as above that xR0z. • The case x 6= y and y = z is solved similarly. • If x = y = z, then xR0z indeed holds because (x; z) = (x; x) belongs to the diagonal, which is included in R0. Proof of antisymmetry of R0: Consider x and y in A such that xR0y and yR0x. Assume that x 6= y. Then (x; y) and (y; x) do not belong to the diagonal, so xRy and yRx. By transitivity of R, we get that xRx. But this contradicts that R is irreflexive. So, it holds that x = y, which concludes the proof that R0 is antisymmetric. Solution for b. in the case of R a strict total order About R, we know that it is a strict total order on A. That is to say, R is a strict partial order on A which in addition satisfies trichotomy. We want to prove that R0 = R [ f(x; x): x 2 Ag is a weak total order on A. By the above proof, we already know that R0 is a weak partial order on A. So, it is only left to show that R0 satisfies totality. Proof of totality of R0: Consider x and y in A. Because R satisfies trichotomy, it holds that xRy or yRx or x = y. • If x = y, then (x; y) belongs to the diagonal, so xR0y. • If xRy, since R ⊆ R0, then xR0y. • If yRx, since R ⊆ R0, then yR0x. In all cases, we obtain that xR0y _ yR0x holds, proving that R0 is total. 2 Exercise 4.8 Recall the definition of the lexicographic product of two relations (which are most of the time taken to be order relations in this context): Given two relations R1 and R2, the lexicographic product of R1 and R2 is the relation R with domain dom(R1) × dom(R2) defined by 0 0 0 0 0 (x; y)R(x ; y ) if and only if (xR1x ) _ (x = x ^ yR2y ): Prove the following property of strict total orders (in natural mathematical language, not formally): if < is a strict total order on A and ≺ is a strict total order on B, then the lexicographic product of < and ≺ is a strict total order on A × B. Does it also hold with weak (instead of strict) total orders? Notation: We denote by C the lexicographic product of < and ≺. Solution for strict total orders About <, we know that it is transitive, irreflexive, and satisfies trichotomy. The same holds for ≺. For C to be a strict total order, we need to prove that C is transitive, irreflexive, and satisfies trichotomy. We will denote A = dom(<) × dom(≺), which is the domain of C. Proof of irreflexivity: Consider (x; y) 2 A. Because < is irreflexive, x < x does not hold. Because ≺ is irreflexive, y ≺ y does not hold, so that the formula (x = x ^ y ≺ y) does not hold. This implies that (x < x) _ (x = x ^ y ≺ y) does not hold, that is to say that (x; y) C (x; y) does not hold. Therefore C is irreflexive. Proof of transitivity: Consider three elements (x; y), (x0; y0) and (x00; y00) in A, and assume that 0 0 0 0 00 00 00 00 (x; y) C (x ; y ) and (x ; y ) C (x ; y ). We want to prove that (x; y) C (x ; y ). 0 0 0 0 0 That (x; y) C (x ; y ) holds means that either (x < x ) or (x = x ^ y ≺ y ), and similarly for (x0; y0) and (x00; y00). (Note that both cannot hold at the same time, since < is irreflexive.) We distinguish four cases, according to which of these properties are satisfied. • If (x < x0) and (x0 < x00), then by transitivity of <, x < x00, from which 00 00 it follows that (x; y) C (x ; y ). • If (x < x0) and (x0 = x00 ^ y0 ≺ y00), then (x < x00), and we conclude as above. • If (x = x0 ^ y ≺ y0) and (x0 < x00), then again (x < x00), and we conclude as above. • If (x = x0 ^y ≺ y0) and (x0 = x00 ^y0 ≺ y00), then x = x00 and by transitivity 00 00 00 of ≺ we have y ≺ y . It follows that (x; y) C (x ; y ). Remark: Note that the proof of irreflexivity and transitivity above do not use the fact that < and ≺ satisfy trichotomy. This is enough to ensure that the statement is also true for strict partial orders (instead of strict total orders). 3 Proof of trichotomy: Consider two elements (x; y) and (x0; y0) in A. We want to prove that one of 0 0 0 0 0 0 the following holds: (x; y) C (x ; y ) or (x ; y ) C (x; y) or (x; y) = (x ; y ). Because < and ≺ satisfy trichotomy, we know that the two following formulas are true: x < x0 _ x0 < x _ x = x0 and y ≺ y0 _ y0 ≺ y _ y = y0. We distinguish several cases according to which formulas make these disjunctions true. 0 0 0 • If x < x , then it holds that (x; y) C (x ; y ). 0 0 0 • If x < x, then it holds that (x ; y ) C (x; y). 0 0 0 0 • If x = x and y ≺ y , then it holds that (x; y) C (x ; y ). 0 0 0 0 • If x = x and y ≺ y, then it holds that (x ; y ) C (x; y). • If x = x0 and y = y0 then it holds that (x; y) = (x0; y0). 0 0 0 0 0 0 In all cases, the formula (x; y) C (x ; y ) _ (x ; y ) C (x; y) _ (x; y) = (x ; y ) holds, and we conclude that C satisfies trichotomy. Solution for weak total orders The claim is not true for weak total orders.