# hw18solutions - g a = 0 then g x = g x-a g a = 0 For the...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 22, 2010 Assignment 18 1. Let f be a continuous real function on R . Let Z ( f ) (the zero set of f ) be the set of all p R at which f ( p ) = 0. Prove that Z ( f ) is closed. The set E = { x R : x 6 = 0 } = ( -∞ , 0) (0 , ) is open. (This is most easily seen by noting that E c = { 0 } is trivially closed as it has no limit points.) Thus, since f is continuous, f - 1 ( E ) is open. But f - 1 ( E ) = { x R : f ( x ) 6 = 0 } . Thus, ( f - 1 ( E )) c = { x R : f ( x ) = 0 } = Z ( f ) but since f - 1 ( E ) is open, we have that Z ( f ) is closed. 2. Let g : R R satsify the relation g ( x + y ) = g ( x ) g ( y ) for all x,y R . Show that if g is continuous at x = 0, then g is continuous at every point of R . Also if we have g ( a ) = 0 for some a R , then g ( x ) = 0 for all x R . The latter property follows easily as if
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Unformatted text preview: g ( a ) = 0, then g ( x ) = g ( x-a ) g ( a ) = 0. For the former, assuming that g is continuous at x = 0, we need to show that g ( y ) → g ( x ) as y → x for every x ∈ R . But g ( y ) = g ( x ) g ( y-x ). As y → x , y-x → 0, and since g is continuous at 0, it follows that g ( y ) = g ( x ) g ( y-x ) → g ( x ) g (0). Moreover since 0 = 0 + 0, we have g (0) = g (0) g (0) and thus g (0) is either 0 and g ( x ) ≡ 0 and is trivially continuous or g (0) = 1. In this latter case, we have g ( y ) → g ( x ) as desired. 1...
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## This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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