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Foundations of Mathematics (HS 2016) Solution to Exercises 4.6 and 4.8

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Foundations of Mathematics (HS 2016) Solution to Exercises 4.6 and 4.8 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.
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