1873_solutions

Like this in the exercises on equivalance relations b

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Unformatted text preview: uch j we have f x j  j f x j f Aj and the fact that j f Aj j Bj guarantees that  f x . Therefore  cannot belong to the range of f and we have shown that the function f is not onto the product Ü i I B i . b. Explain why part a is a stronger form of this earlier exercise. Suppose that S i is strictly subequivalent to U for each i I. König’s inequality tells us that  i I S i is strictly subequivalent to the product Ü i I U. Now all we have to observe is that Ü U  UI. iI c. Suppose that S is a given set and that to each member x of the set S are associated the sets A x  x and B x  0, 1 . Explain how König’s inequality can be applied to this association to show that S is strictly subequivalent to the set 0, 1 S and deduce that Cantor’s inequality is a special case of König’s inequality. Since  x S xS and since x is strictly subequivalent to 0, 1 whenever x S we deduce from König’s inequality that S is strictly subequivalent to 0, 1 S and we already know that 0, 1 S ß p S . Exercises on Well Orders 1. Prove that the lexicographic order  in 1, 2 Z  that was defined earlier is a well order of 1, 2 The desired result follows at once from Exercise 3. 2. Prove that if  is lexicographic order from the left in the set 1, 2, 3 1, 2, 3 Z. The desired result follows at once from Exercise 3. Z. Z  then  is a well order of 3. Prove that if is a well order in a set S and  is lexicographic order from the left in the set S well order of S. S, then  is a Solution: We leave it as an exercise to show that the relation  is a total order in S S. We shall now show that every nonempty subset of S S has a member that is least with respect to the order . Suppose that E is a nonempty subset of S S. We define E. P  x S There is at least one member y of S such that x, y Since the set P is nonempty, it must have a least member that we shall call u. We now define v to be the least member of the set yS u, y E. Now write a simple but careful explanation of why the pair u, v must be the member of S S that is least with respect to the order  . 68 4. Given that a and b belong to a well ordered set S and that a  b, prove that a   b  . The desired result follows at once from the fact that a  b and b  b  . 5. Given that a and b belong to a well ordered set S and that a   b  , prove that a  b. From Exercise 4 we know that neither of the conditions a  b or b  a can hold. 6. Prove that if a belongs to a well ordered set then a  a  . We know that a  a  and that no member of the set can lie between a and a  . Therefore a is the predecessor of a  . 7. Prove that if a member a of a well ordered set has a predecessor then a  a  . Suppose that a has a predecessor. We know that a  a and that no member of the set can lie between a and a. Therefore a is the successor of a . 8. Prove that if a belongs to a well ordered set S then a fails to be a successor if and only if x S x  a  x  a . Suppose that a belongs to a well ordered set S. Suppose that a is no...
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