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mth536_problem_7_1a

mth536_problem_7_1a - MTH536 Chapter 7 Problem 1a Gene...

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Unformatted text preview: MTH536 Chapter 7 Problem 1a Gene Quinn – p. 1/ Chapter 7 Problem 1a Show that the set of all -tuples of real numbers becomes a metric space under each of the following metrics: ¢ ¤ ¥£ ¨ ¤ §   ¤ ¦ §  §  #  ¦ ¦" " " ¦       ¡ § ©   ¤£  ¨ ¤ ! §  ©   ¤ ¡ §    – p. 2/ Proof of Proposition Any metric must satisfy the following four conditions: ¨ ¡ ¤ ¥£ $ ¨ ¦ § % ¨ & § ( ¤ & © ¦ ¨ £ ) § ¡ ¦ ¨ © ¤£ ¨ ¨ ¤ ¥£ ¦ §¤ © ¨ ¡ ¤ ¥£ ¨ £ 0 1 ¡ § ¡ 2  ¡ ¦ ¦ ¦2 § § ¡ ¨ ' ¡ ¤ ¥£ ¨ – p. 3/ Proof that Since 4 and  satisfy condition 1 ¨ ¤ §   ¤ §      § ©  3 ¢ ¤£ ¡ 3 is a sum of nonnegative reals for all ¦ ,   is always .  6 ¢ ¤ § 5 ¦ ¡ % & – p. 4/ Proof that Since 4 and  satisfy condition 1 ¨ ¤ §   ¤ §      § ©  3 ¢ ¤£ ¡ 3 is a sum of nonnegative reals for all ¤£  ¨ ¤ ! ¦ ,   is always  # .  6 ¢ ¤ § 5  ¦ ¤ ¡ §   § ¡ ©   ¦ ¦" " " is the maximum of a finite set of nonnegative real numbers, so for all , is always .  6  § 5 ¦ ¡ % & ¤ ¦   § % & – p. 4/ Proof that 4 and § ©  satisfy condition 2 , and $ 1 § 9 1 7© $ 1 9 ¦ 1 87¤ 3 Suppose . Then by definition, ¤7   § 7 © ¤ 3 ¦& – p. 5/ Proof that and ¤ © § ¡ ¤ ¥£ ¦ § ¨ ¦ § ¨ 7  ¤7  @  7 §  ©   ¦" & " " ¦ & © & © & 1 @  & 7 ©  1 9 1 $ ©  ¤7  © 7 §  1 9 7 §  ¦& $ ¦  © 7 ¡ ¢  ¤£ ¤7  Suppose It follows that . Then by definition, 87¤ 7© § 3 4 and 3  satisfy condition 2 , and – p. 5/ Proof that 4 and § ©  satisfy condition 2 , and $ 1 § 9 1 7© $ 1 9 ¦ 1   & § © 7 7  ¦" & @  © ¦ 7 & § ©   7 " " ¡ ¢ ¦ © & & 87¤ 3 Suppose . Then by definition, ¤7   § 7 © ¤ 3 It follows that ¤£ ¨  © ¤7  § © & 7 © ¤7    @  ¡ ¢ and ¤ ¥£  ¡ ¦ § ¨ ¤£ Now suppose . Since is a sum of nonnegative terms, it can only be zero if each term is zero, in which case for all and by definition . ¡ ¢ ¦ § ¤ © § 7© § ¤7 $ ¨ ¦& – p. 5/ Proof that 4 and § ©  satisfy condition 2 , and $ 1 § 9 1 7© $ 1 9 ¦ 1   & § © 7 7  ¦" & @  © ¦ 7 & § ©   7 " " ¡ ¢ ¦ © & & 87¤ 3 Suppose . Then by definition, ¤7   § 7 © ¤ 3 It follows that ¤£ ¨  © ¤7  § © & 7 © ¤7    @  ¡ ¢ and ¤ ¥£  ¡ ¦ § ¨ ¤£ Now suppose . Since is a sum of nonnegative terms, it can only be zero if each term is zero, in which case for all and by definition . ¡ ¢ ¦ § ¤ © § ¨ ¦& By a similar argument, is the maximum of a set of nonnegative terms and can only be zero if each term is zero, which again implies that . ¡ ¤ © § © &  7© § ¤7 $ – p. 5/ Proof that 4 and  satisfy condition 3 3 By the properties of absolute value, ¤7    ¤7 £ ¨ © 7 §  $ 1 9 1  ¤7  ¦ § § 7 © 7  so interchanging 3 and leaves both and unchanged. ¤ § ¡ ¢ ¡ – p. 6/ Proof that Let ¤ ¦ ¦§ 2 5 3 4 ¡ ¦ § 7 © @  ¢ ¤ ¥£ ¨  6 . Then  ¤7  7 §  7 © @   ¤7  7 2 7 2 7 §  satisfies condition 4 – p. 7/ Proof that Let ¤ ¦ ¦§ 2 5 3 4 ¡ ¦ § 7 7 1 @   7 ¤ £ 7 2   7 2 @  7 §  ¨ @  © ¢ ¤ ¥£ ¨  6 . Then  ¤7  7 §  7 ©  ¤7  7 2 7 2 7 §  satisfies condition 4 – p. 7/ Proof that Let ¤ ¦ ¦§ 2 5 3 4 7 © @  ¤7  7 2   7 2 7 §  © ¡ ¦ 2 ¨ ¡ ¦2 § ¨  ¢ £ ¢ ¤£ 7 §  ¨  7 2   7 2 @  @  7 1  7 ¤ £ @  7 @  7  ¡ ¦ § © ¢ ¤ ¥£ ¨  6 . Then  ¤7  7 §  7 ©  ¤7  7 2 7 2 7 §  satisfies condition 4 – p. 7/ Proof that Let ¤ ¦§ ¦  satisfies condition 4 . Then for some , A ¤£  ¨ © 8B¤  § § B ¦ 3 ,   B 2 B §   6 $ 1 A 1 ¤B  1 B 2 5 ¡  2 – p. 8/ Let But Proof that and ¤ ¦ ¦ 2 5 ¡ ¤£ ¦ § CB¤  B 2 B §  7   7 2 7 §  7 2  1 7   ¤7  B 2   B 2 B §  1 B 2  1 ¨ © 8B¤  B §  ¤B  1   6 3  satisfies condition 4 . Then for some , A $ 1 A , – p. 8/ ¤B  so ¤ ¦  ¦§ 2 5 ¡  ¤£  ¦ § CB¤  B 2 B §  7 1   7  B 2  1 ¨ © 8B¤  B §  ¤B  1  6 Let But and Proof that B 2 B 2 B § 1 7   ¤7  7 2  7   7 2 7 §  © ¡ ¤£ ¦ 2 ¨ ¡ ¦2 § ¨ – p. 8/  3  satisfies condition 4 . Then for some , A  7 2 7 §  7 2   ¤7  B 2   B 2  B §  1 $ 1 A ,  £  ...
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