Diskrete Mathematik Solution 6

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1 ETH Zürich, D-INFK HS 2018, 30. October 2018 Prof. Ueli Maurer Marta Mularczyk Diskrete Mathematik Solution Partial Order Relations a) i) 11 and 12 are incomparable, since and ii) 4 and 6 are incomparable, since 4 6 and 6 4. iii) 5 and 15 are comparable, since iv) 42 and 42 are comparable, since b) The elements (a, b) A, such that (a, b) lex (2, 5) are: (2, 1), (2, 5) and (1, n) for all n N \ {0}. Justification: Let (a, b) A. We distinguish the following cases: Case a = 1: Since 1 2, we have (a, b) lex (2, 5) for any b. Case a = 2: Since 1 and 5 are the only natural numbers which divide 5, we have (a, b) lex (2, 5) only for b {1, 5}. Case a > 2: Since a 2, (a, b) lex (2, 5) cannot hold for any b. c) ({1, 3, 6, 9, 12}, ) is not a lattice, since 9 and 12 do not have a common upper bound. d) (A; 1 ) is a poset. To prove this, we show that 1 is a partial order on A. Reflexivity: For any a A, by the reflexivity of, we have a a, hence, a 1 a. Antisymmetry: Let a, b A be such that a 1 b and b 1 a. This means that b a and a b By the antisymmetry of, it follows that a = b. Transitivity: Let a, b, c A be such that a 1 b and b 1 c. This means that b a and c b. By the transitivity of, we have c a. Hence, a 1 c. 6.2 Hasse Diagrams a) The Hasse diagrams of the posets ({1, 2, 3}; ) and ({1, 2, 3, 5, 6, 9}; ) are as follows: In both cases, 1 is the least and the only minimal element. In the poset ({1, 2, 3}; ), the greatest and the only maximal element is 3. In the poset ({1, 2, 3, 5, 6, 9}; ) there is no greatest element. The maximal elements in this poset are 5, 6 and 9.

2 6.3 The Strictly Less Relation Let a, b, c A be such that a b and b c. By the definition of, this means that (1) a b, (2) a b, (3) b c, and (4) b c. From (1) and (3), and the transitivity of, it follows that a c. It remains to show that a c. Assume towards a contradiction that a = c. This, together with (1), implies that c b. But this, together with (3) and the antisymmetry of, implies that b = c, which is a contradiction to (4). Overall, we showed that for any a, b, c A, such that a b and b c, we have a c and a c. This, by the definition of, means that a c. 6.4 The Lexicographic Order For posets (A; ) and (B; ) the lexicographic order lex on A B is defined by (a 1, b 1 ) lex (a 2, b 2 ) : a 1 a 2 (a 1 = a 2 b 1 b 2 ) We show that lex fulfills all the properties of a partial order relation. Reflexivity: Take any (a 1, b 1 ) A B. Since is reflexive, we have b 1 b 1. Hence, it is true that (a 1 = a 1 b 1 b 1 ) and, thus, (a 1, b 1 ) lex (a 1, b 1 ). Antisymmetry: Take any (a 1, b 1 ) and (a 2, b 2 ) in A B such that (a 1, b 1 ) lex (a 2, b 2 ) and (a 2, b 2 ) lex (a 1, b 1 ). This means that a 1 a 2 (1) (a 1 = a 2 b 1 b 2 ) (2) and a 2 a 1 (3) (a 2 = a 1 b 2 b 1 ). (4) We have to show that (a 1, b 1 ) = (a 2, b 2 ). The proof proceeds by case distinction. (1) and (3): We have a 1 a 2 a 1 a 2 and a 2 a 1 a 2 a 1. But since is antisymmetric, it follows that a 1 = a 2, which is a contradiction with a 1 a 2. Therefore, this case cannot occur. (1) and (4): We have a 1 a 2 a 1 a 2 and a 2 = a 1 b 2 b 1, which is a contradiction. Therefore, this case also cannot occur. (2) and (3): We have a 1 = a 2 b 1 b 2 and a 2 a 1 a 2 a 1, which is a contradiction. Therefore, this case cannot occur as well. (2) and (4): We have a 1 = a 2 b 1 b 2 and a 2 = a 1 b 2 b 1. Since is antisymmetric, it follows that b 1 = b 2. But we also have a 1 = a 2 and, thus, (a 1, b 1 ) = (a 2, b 2 ). Transitivity: Take any (a 1, b 1 ), (a 2, b 2 ), (a 3, b 3 ) in A B such that (a 1, b 1 ) lex (a 2, b 2 ) and (a 2, b 2 ) lex (a 3, b 3 ). This means that a 1 a 2 (1) (a 1 = a 2 b 1 b 2 ) (2) and a 2 a 3 (3) (a 2 = a 3 b 2 b 3 ). (4) We have to show that (a 1, b 1 ) lex (a 3, b 3 ). The proof proceeds by case distinction.

3 (1) and (3): We have a 1 a 2 and a 2 a 3. Since by Task 6.3 is transitive, a 1 a 3. Hence, (a 1, b 1 ) lex (a 3, b 3 ). (1) and (4): We have a 1 a 2 and a 2 = a 3 b 2 b 3. Hence, a 1 a 3 and, therefore, (a 1, b 1 ) lex (a 3, b 3 ). (2) and (3): We have a 1 = a 2 b 1 b 2 and a 2 a 3. Hence, a 1 a 3 and, therefore, (a 1, b 1 ) lex (a 3, b 3 ). (2) and (4): We have a 1 = a 2 b 1 b 2 and a 2 = a 3 b 2 b 3. It follows that a 1 = a 3. Since is transitive, we also have b 1 b 3. Therefore, (a 1, b 1 ) lex (a 3, b 3 ). 6.5 Countability i) The set of all Java programs is countable. Every Java program can be seen as a finite binary sequence. That is, there is an injection from the set of all Java programs to the set {0, 1} of finite binary sequences. By Theorem 3.16, this set is countable. ii) This set is uncountable. To prove this, we notice that {0, 1} A, which implies that {0, 1} A (Lemma 3.13). Since {0, 1} is uncountable, A must be uncountable as well (if A was countable, the transitivity of would imply that {0, 1} is countable, which is a contradiction). An alternative proof. We can also apply directly the diagonalization argument. Assume towards a contradiction that there is a bijection f : N A. Let β i,j denote the j-th number in the i-th sequence. We define a new sequence as follows: α def = R 10(β 0,0 + 1), R 10(β 1,1 + 1), R 10(β 2,2 + 1),..., where R 10(a) denotes the remainder when a is divided by 10. Of course, α A. Moreover, there is no n N such that α = f(n), since α disagrees with a sequence f(n) on position n. iii) This set is uncountable. We can define an injective function f : [0, 1] C by f(x) = (x, 1 x 2 ). Hence, we have [0, 1] C. Since [0, 1] is uncountable, C must be uncountable as well (if C was countable, the transitivity of would imply that [0, 1] is countable as well, which is a contradiction). Note: The fact that the interval [0, 1] is uncountable follows from Theorem 3.21 and the fact that any element of {0, 1} can be interpreted as the binary expansion of a number in the interval [0, 1], and vice versa. iv) This set is uncountable. Let S denote the set of all equivalence relations on N. We give an injection f : P(N \ {0}) S. The claim follows, since P(N \ {0}) is uncountable (the elements of P(N \ {0}) correspond to semi-infinite binary sequences, which are uncountable by Theorem 3.21). To define the injection f : P(N \ {0}) S, consider an A P(N \ {0}). We partition N into A {0} and N \ (A {0}) and define the equivalence relation f(a) such that two numbers are f(a)-related if they are in the same set of the partition. Clearly, f is injective, since for two different sets A and A, the equivalence classes of 0 are different for the relations f(a) and f(a ) and hence f(a) f(a ).

4 6.6 Nonincreasing Functions To prove that A is countable, we will define an injection φ : A N. Since the set N is countable (by Theorem 3.20.) and the dominates relation ( in the lecture notes) is transitive (by Lemma 3.13.), it will then follow that A is countable as well. To define φ, take any f A. Since f is nonincreasing, the image Im(f) is a subset of {0,..., f(0)}. Since this is a finite set and f is nonincreasing, there must exist an N N, such that for all n N, f is constant (that is, f(n) = f(n) for all n N). Let N 0 be the smallest such N. We set φ(f) = (f(0), f(1),..., f(n 0 )) N. Note that φ is an injection, because a sequence (f 0, f 1,..., f N0 ) N clearly defines at most one nonincreasing function. The function f uniquely defines a finite sequence (f(0), f(1),..., f(n 0 )) N. In this way we defined an injection φ : A N. By Theorem 3.20., the set N is countable. By the transitivity of the dominates relation (Lemma 3.13.), it follows that A is countable as well. 6.7 The Diagonalization Argument a) Let β i,j be the j-th bit in the i-th sequence. In the lecture, the sequence α that is not in the list was constructed as α def = β 0,0, β 1,1, β 2,2,.... A second such sequence α can be defined as α def = α 0, β 0,1, β 1,2, β 2,3,..., that is, α disagrees with the i-th sequence on the bit i+1, and the first bit α 0 can be arbitrary. Then, for α to be different than α, we can set α 0 = β 0,0. Overall, the constructed sequence α is β 0,0, β 0,1, β 1,2,.... b) It is not possible to list all these sequences. Justification: Take any list L of semi-infinite binary sequences and let M denote the set of sequences that are not in L. Assume towards a contradiction that the elements of M can be listed, that is, assume that M is countable. Then, the union L M of two countable sets is countable. However, L M = {0, 1}, which is a contradiction to Theorem (Optional) The Hunt for the Red October At any time t we can fire a torpedo to position s = x t+y for some x and y. The submarine sinks if its speed and the starting position happened to be x and y. Thus, at any time t we can make a guess about x and y and sink the submarine based on that guess. We now have to systematically check all the pairs (x, y) Z Z. Hence, we need a surjective function f : N Z Z that will assign to a time t a pair (x, y). (Surjectivity guarantees that every (x, y) will be tested at some time t.) Since Z Z is countable (by Example 3.65 and Corollary 3.18), there exists an injective function g : Z Z N. We can now define f as { (a, b) if (a, b) g((a, b)) = n f(n) := (0, 0) otherwise

5 By the injectivity of g, we have {(a, b)} = g 1 ({g((a, b))}) for all (a, b) Z Z. Also, for any (a, b) there exists an n N such that g((a, b)) = n and, therefore, there exists an n N such that f(n) = (a, b). Hence, f is surjective and we will eventually sink the submarine.