real_analysisii_homework2 - Homework 2 Hctor Guillermo Cullar R e e os February 2 2006 12.12 Let D be a nonempty set and suppose that f D R and g D R

# real_analysisii_homework2 - Homework 2 Hctor Guillermo...

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Homework 2 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. (c) State and prove the analog of part (a) for infima. If f ( D ) and g ( D ) are bounded below, then prove that ( f + g )( D ) is bounded below and inf[( f + g )( D )] inf f ( D ) + inf g ( D ).

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• Fall '08
• Akhmedov,A
• Empty set, Supremum, Order theory, upper bound, Closed set, 13.6 second

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