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royden8 - ROYDEN REAL ANALYSIS 3RD ED CHAPTER 8 Problem...

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ROYDEN, REAL ANALYSIS 3RD ED. CHAPTER 8 Problem 8-10 a. Let us assume A 1 \ A 2 = Á . For x 2 A 1 [ A 2 ;x 2 A 1 or x 2 A 2 . Enough to consider the case x 2 A 1 . Then x = 2 A 2 . Or x 2 A n A 2 which is open in A . Hence there exists an open set O , such that (1) x 2 O 1 ½ A n A 2 ; which means O 1 \ A 2 = Á: Since f j A 1 is continuous, there exists, for any open set U of f ( x ), an open set O 2 in A such that (2) O 2 \ A 1 μ f ¡ 1 ( U ) : Then x 2 O 1 \ O 2 and by (1) ( O 1 \ O 2 ) \ A = ( O 1 \ O 2 ) \ A 1 μ f ¡ 1 ( U ) : That is f j A is continuous at x . b. A 1 = ( ¡ 1 ; 0] and A 2 = (0 ; 1) : f 1 ( x ) ´ 0 on A 1 and f 2 ( x ) = 1 =x on A 2 . c. For x 2 A 1 n A 2 we can ¯nd an open set O 1 such that x 2 O 1 and O 1 \ ( A 2 ¡ A 1 ) = Á: Together with (2), we have x 2 O 1 \ O 2 and O 1 \ O 2 \ ( A 1 [ A 2 ) = O 1 \ O 2 \ A 1 μ f ¡ 1 ( U ) : Hence f A 1 [ A 2 is continuous at x . Similar arguments work for x 2 A 2 n A 1 . Typesetby A M S -T E X 1
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2 Since F μ U r ½ O for any r = p ¢ 2 ¡ m , we immediately get f j F = 0 and f j e O = 1 : Because U 1 ´ X; 0 · f · 1. It remains to show the continuity of f . Suppose f ( x ) = ® . By de¯nition x 2 U ® + " and x = 2 U ® ¡ " μ ¹ U ® ¡ " for any " > 0 with ® § " = p ¢ 2 ¡ n . Since U ® + " is open, there is an open set N 1 ( x ) μ U ® + " which implies (1) f ( y ) · ® + " for y 2 N 1 ( x ) : As x 2 ( ¹ U ® ¡ " ) c which is open, there is an open set N 2 ( x ) such that N 2 ( x ) ½ ¹ U ® ¡ " ) c or equivalently N 2 ( x ) \ ¹ U ® ¡ " = Á . Since U ® μ U s for ® < s , we have N 2 ( x ) ½ U ® = ; and thus (2) f ( y ) ¸ ® ¡ " for y 2 N 2 ( x ) : Let N ( x ) = N 1 ( x ) \ N 2 ( x ) : Combining (1) and (2) j f ( y ) ¡ ® j · " for y 2 N ( x ) : d. ( ) ) Apply c. with O c = B .
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