royden7 - Royden, Real Analysis 3rd ed. Chapter 7 Problem...

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Royden, Real Analysis 3rd ed. Chapter 7 Problem 7-3 b. Let X x = f y 2 X : ½ ( y;x ) < 1g : Then (i) X x is open. For each y 2 X x ; we have ½ ( y;x ) < 1 : By using the triangle inequality it can be shown easily that B ( y; r ) μ X x for any r > 0 : (ii) X c x is open. Similarly, it can be shown that B ( y; r ) μ X c x if y 2 X c x i.e. ½ ( y;x ) = 1 : Hence X c x is open which is the same to say X x is closed. Problem 7-7 L 1 ( ¡ 1 ; 1 ) and L 1 [0 ; 1] are separable. Take L 1 [0 ; 1] for instance. Any integrable function can be approximated by continuous func- tions, and then by polynomials (Stone-Weierstrass Theorem). Finally we consider the class of all polynomials with rational coe¢ cients, which is count- able. L 1 (0 ; 1) is not separable. Consider f c ( x ) = Â (0 ;c ) ( x ) for c 2 (0 ; 1) : Then k f c ¡ f d k 1 = 1 for c 6 = d: Problem 7-11 a. If ½ < "; then ¾ = ½ 1 + ½ < ½ < ": If ½ < " 2 < 1 2 ; then ½ = ¾ 1 ¡ ¾ < 2 ¾ < ": Hence ¾ is equivalent to ½: Problem 7-17 d. Let * x n = ( x n 1 ;x n 2 ; :: : ) be a Cauchy sequence in ( X ¤ ; ½ ¤ ) : By taking a proper subsequence we may assume ½ ¤ ( * x n ; * x n +1 ) < 2 ¡ n : Since f x nk : k ¸ 1 g is a Cauchy sequence, we can …nd N n such that ½ ( x nm ;x nl ) < 2 ¡ n and ½ ( x n;m ;x n +1 ;m ) < 2 ¡ n for m and l ¸ N n : (1) WLOG we may assume N 1 < N 2 < N 3 < ¢¢¢ : De…ne * x = ( x 1 N 1 ;x 2 N 2 ;x 3 N 3 ; : : : ) : We claim ½ ¤ ( * x n ; * x ) ! 0 : First we need to show f x n;N n g is a Cauchy sequence. For k ¸ 1 we have from (1)
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royden7 - Royden, Real Analysis 3rd ed. Chapter 7 Problem...

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