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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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 Fall '08
 STAFF
 Math, Calculus

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