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Unformatted text preview: Homework 2 H´ ector Guillermo Cu´ ellar R´ ıos February 2, 2006 12.12 Let D be a nonempty set and suppose that f : D → R and g : D → R . Define the function f + g : D → R by ( f + g )( x ) = f ( x ) + g ( x ). (a) If f ( D ) and g ( D ) are bounded above, then prove that ( f + g )( D ) is bounded above and sup[( f + g )( D )] ≤ sup f ( D ) + sup g ( D ). Proof. Let sup f ( D ) = m and sup g ( D ) = n . We know that ∀ x ∈ D, f ( x ) ≤ m and g ( x ) ≤ n , then f ( x ) + g ( x ) ≤ m + n and since f ( x ) + g ( x ) = ( f + g )( x ) ( f + g )( x ) ≤ m + n thus ( f + g )( x ) is bounded above by m+n and by the completeness axiom ( f + g )( x ) has a least upper bound, let’s call it d , since d is the smallest of the upper bounds, d ≤ m + n . (b) Find an example to show that a strict inequality in part (a) may occur. Let D = { x : 0 ≤ x ≤ 10 } , f ( x ) = x and g ( x ) = x . sup f ( x ) = 10 and sup g ( x ) = 0 and sup [( f + g )( x )] = 0 and 0 ≤ 10 + 0....
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 Spring '08
 Akhmedov,A
 Empty set, Supremum, Order theory, upper bound, Closed set, 13.6 second

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