\nA Dedekind cut is a pair (A, B) where A, B ? Q such that:
\n(a) A 6= ?, B 6= ?;
\n(b) If a ? A and c < a, then c ? A;
\n(c) If b ? B and c > b, then c ? B;
\n(d) If b \/? B and a < b, then a ? A;
\n(e) If a \/? A and b > a, then b ? B;
\n(f) For each a ? A there is some b > a so that b ? A;
\n(g) For each b ? B there is some a < b so that a ? B;
\n1. Let A = {x ? Q : x < 0 or x
\n2 < 2} and B = {x ? Q : x > 0 and x
\n2 > 2}. Prove that
\n(A, B) is a Dedekind cut.
\n2. Let (A1, B1) and (A2, B2) be Dedekind cuts and define A3 := {a1 + a2 : ai ? Ai
\n, i ?
\n{1, 2}}, B3 := {b1 + b2 : bi ? Bi
\n, i ? {1, 2}}. Prove that (A3, B3) is a Dedekind cut.
\n3. Let X, Y be non-empty sets and f : X ? Y ? R have bounded range. Define
\nf1(x) := sup{f(x, y) : y ? Y }, f2(y) := sup{f(x, y) : x ? X}.
\nShow that
\nsup{f(x, y) : x ? X, y ? Y } = sup{f1(x) : x ? X} = sup{f2(y) : y ? Y }.
\n4. Let f and f1 be as in Problem 3 and define g2(y) := inf{f(x, y) : x ? X}. Prove that
\nsup{g2(y) : y ? Y } = inf{f1(x) : x ? X} and show via an example that strict inequality is
\npossible. (The shorthand for this inequality is: supy
\ninfx f(x, y) = infx supy f(x, y).)
\n5. Let X be a non-empty set and suppose f : X ? R has bounded range. If a ? R, show
\nthat
\nsup{a + f(x) : x ? X} = a + sup{f(x) : x ? X}
\nand
\ninf{a + f(x) : x ? X} = a + inf{f(x) : x ? X}.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"