Unformatted text preview: Modern Analysis 1
Homework 01
1. Let F be an ordered ﬁeld. Let A and B be subsets of F for which sup A
and sup B exist (in F ); deﬁne subsets A + B and AB of F by
A + B = {a + b : a ∈ A, b ∈ B },
AB = {ab : a ∈ A, b ∈ B }.
(i) Does sup(A + B ) exist in F ? If so identify it; if not, show why.
(ii) Does sup(AB ) exist in F ? If so identify it; if not, show why.
Solution
For convenience, write α = sup A and β = sup B .
(i) If a ∈ B and b ∈ B then a α and b β (because α and β are upper
bounds) whence (as noted in class) a + b α + β ; this shows that α + β is
an upper bound for A + B . Now let u ∈ F be any upper bound for A + B .
If a ∈ A and b ∈ B are arbitrary, then u a + b so that u − b a; as a here
is arbitrary, we deduce that u − b is an upper bound for A so that u − b α
(because α is least). Rearrange to obtain u − α
b and deduce similarly
that u − α β . Finally, the conclusion u α + β shows that α + β is the
least upper bound for A + B .
(ii) In general, the set AB need not be bounded above: one of A or B may
contain a negative element while the other contains ‘arbitrarily large’ negative
elements. If A and B are also bounded below, then AB will be bounded above
(but its supremum, if it exists, need not be αβ ; why?). However, if A and B
contain only positive (or rather, nonnegative) elements, then sup(AB ) exists
and equals αβ ; simply replace diﬀerences by quotients in the argument for
part (i). 1 ...
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 Fall '09
 Robinson
 Supremum, Order theory, upper bound, α, sup

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