Unformatted text preview: 2 . Let N = max f N 1 ;N 2 g : Then n & N guarantees that j a n ² A j < "= 2 and j b n ² B j < "= 2 . Thus n & N ) j ( a n ² b n ) ² ( A ² B ) j = j ( a n ² A ) + ( B ² b n ) j ± j a n ² A j + j B ² b n j < " 2 + " 2 = ": 4. Suppose A and B are nonempty sets of real numbers with A ³ B: Then inf A & inf B: Proof. Assume that A and B are bounded below. Then both sets have &nite in&ma. Let & = inf A and ± = inf B: By de&nition, ± is a lower bound for B: Take any a 2 A: Then a 2 B by the containment A ³ B; so ± ± a: This shows that ± is a lower bound for A: Since & is the greatest lower bound of A we can conclude & & ±: If B is not bounded below, then ± = ²1 so the inequality & & ± is automatic. 1...
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 Spring '09
 Math, Real Numbers, Natural number, Archimedean

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