{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

royden7

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

This preview shows pages 1–2. Sign up to view the full content.

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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online