1873_solutions

# Like this in the exercises on equivalance relations b

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online