Centers of nested balls in a normed space make a Cauchy sequence.

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Exercise 1.21 of An Introduction to Banach Space Theory by Megginson:

Let $X$ be normed space. If $(B_n)$ is a nested sequence of closed
balls in $X$, then the centers of the balls form a Cauchy sequence.

How to prove it?

It also asked for a counterexample for metric space, which is easy: in an ultrametric space, like the p-adic number field, any point in a ball is a center of the ball.

analysis vector-spaces normed-spaces

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asked Jul 31 at 8:49

MaudPieTheRocktorate

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up vote
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Exercise 1.21 of An Introduction to Banach Space Theory by Megginson:

Let $X$ be normed space. If $(B_n)$ is a nested sequence of closed
balls in $X$, then the centers of the balls form a Cauchy sequence.

How to prove it?

It also asked for a counterexample for metric space, which is easy: in an ultrametric space, like the p-adic number field, any point in a ball is a center of the ball.

analysis vector-spaces normed-spaces

share|cite|improve this question

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Exercise 1.21 of An Introduction to Banach Space Theory by Megginson:

Let $X$ be normed space. If $(B_n)$ is a nested sequence of closed
balls in $X$, then the centers of the balls form a Cauchy sequence.

How to prove it?

It also asked for a counterexample for metric space, which is easy: in an ultrametric space, like the p-adic number field, any point in a ball is a center of the ball.

analysis vector-spaces normed-spaces

share|cite|improve this question

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

Exercise 1.21 of An Introduction to Banach Space Theory by Megginson:

Let $X$ be normed space. If $(B_n)$ is a nested sequence of closed
balls in $X$, then the centers of the balls form a Cauchy sequence.

How to prove it?

It also asked for a counterexample for metric space, which is easy: in an ultrametric space, like the p-adic number field, any point in a ball is a center of the ball.

analysis vector-spaces normed-spaces

share|cite|improve this question

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

asked Jul 31 at 8:49

MaudPieTheRocktorate

1,431421

1,431421

add a comment | 

add a comment | 

2 Answers
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Let $barB(x_n,r_n)$
be a decreasing sequence of closed balls. We
have $x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)in barB(x_n+1,r_n+1)subseteq barB(x_n,r_n)$
so $leftVert x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)-x_nrightVert leq r_n$ which says $leftVert
x_n+1-x_nrightVert +r_n+1leq r_n$ i.e. $leftVert
x_n+1-x_nrightVert leq r_n-r_n+1$. This implies that $r_n$
is decreasing ( which is also obvious from the fact that the diameters of $%
barB(x_n,r_n)$ are decreasing). Let $r_ndownarrow r$. Iteration of
$leftVert x_n+1-x_nrightVert leq r_n-r_n+1$ yields $leftVert
x_n+m-x_nrightVert leq r_n-r_n+mrightarrow 0$ as $n,mrightarrow
infty $

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
  – MaudPieTheRocktorate
  Jul 31 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
  – Kavi Rama Murthy
  Jul 31 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, was thinking the wrong way.
  – nicomezi
  Jul 31 at 9:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

If, in a normed space, you have two balls, $B_r(a)$ and $B_r'(a')$ and if $B_r(a)supset B_r'(a')$, then $|a-a'|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+frac ra-a'(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$|p-a|=|p-a'|+|a'-a|.$$Therefore $r=|p-a'|+|a-a'|geqslant r'+|a-a'|$ and so $|a-a'|leqslant r-r'$.

Now, if $r_n$ is the radius of $B_n$, then $(r_n)_ninmathbb N$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $varepsilon>0$, $|r_m-r_n|<varepsilon$ if $m,ngg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $varepsilon$.

share|cite|improve this answer

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

add a comment | 

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2 Answers
2

active

oldest

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2 Answers
2

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote

Let $barB(x_n,r_n)$
be a decreasing sequence of closed balls. We
have $x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)in barB(x_n+1,r_n+1)subseteq barB(x_n,r_n)$
so $leftVert x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)-x_nrightVert leq r_n$ which says $leftVert
x_n+1-x_nrightVert +r_n+1leq r_n$ i.e. $leftVert
x_n+1-x_nrightVert leq r_n-r_n+1$. This implies that $r_n$
is decreasing ( which is also obvious from the fact that the diameters of $%
barB(x_n,r_n)$ are decreasing). Let $r_ndownarrow r$. Iteration of
$leftVert x_n+1-x_nrightVert leq r_n-r_n+1$ yields $leftVert
x_n+m-x_nrightVert leq r_n-r_n+mrightarrow 0$ as $n,mrightarrow
infty $

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
  – MaudPieTheRocktorate
  Jul 31 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
  – Kavi Rama Murthy
  Jul 31 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, was thinking the wrong way.
  – nicomezi
  Jul 31 at 9:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $barB(x_n,r_n)$
be a decreasing sequence of closed balls. We
have $x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)in barB(x_n+1,r_n+1)subseteq barB(x_n,r_n)$
so $leftVert x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)-x_nrightVert leq r_n$ which says $leftVert
x_n+1-x_nrightVert +r_n+1leq r_n$ i.e. $leftVert
x_n+1-x_nrightVert leq r_n-r_n+1$. This implies that $r_n$
is decreasing ( which is also obvious from the fact that the diameters of $%
barB(x_n,r_n)$ are decreasing). Let $r_ndownarrow r$. Iteration of
$leftVert x_n+1-x_nrightVert leq r_n-r_n+1$ yields $leftVert
x_n+m-x_nrightVert leq r_n-r_n+mrightarrow 0$ as $n,mrightarrow
infty $

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
  – MaudPieTheRocktorate
  Jul 31 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
  – Kavi Rama Murthy
  Jul 31 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, was thinking the wrong way.
  – nicomezi
  Jul 31 at 9:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $barB(x_n,r_n)$
be a decreasing sequence of closed balls. We
have $x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)in barB(x_n+1,r_n+1)subseteq barB(x_n,r_n)$
so $leftVert x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)-x_nrightVert leq r_n$ which says $leftVert
x_n+1-x_nrightVert +r_n+1leq r_n$ i.e. $leftVert
x_n+1-x_nrightVert leq r_n-r_n+1$. This implies that $r_n$
is decreasing ( which is also obvious from the fact that the diameters of $%
barB(x_n,r_n)$ are decreasing). Let $r_ndownarrow r$. Iteration of
$leftVert x_n+1-x_nrightVert leq r_n-r_n+1$ yields $leftVert
x_n+m-x_nrightVert leq r_n-r_n+mrightarrow 0$ as $n,mrightarrow
infty $

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

Let $barB(x_n,r_n)$
be a decreasing sequence of closed balls. We
have $x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)in barB(x_n+1,r_n+1)subseteq barB(x_n,r_n)$
so $leftVert x_n+1+fracr_n+1leftVert x_n+1-x_nrightVert %
(x_n+1-x_n)-x_nrightVert leq r_n$ which says $leftVert
x_n+1-x_nrightVert +r_n+1leq r_n$ i.e. $leftVert
x_n+1-x_nrightVert leq r_n-r_n+1$. This implies that $r_n$
is decreasing ( which is also obvious from the fact that the diameters of $%
barB(x_n,r_n)$ are decreasing). Let $r_ndownarrow r$. Iteration of
$leftVert x_n+1-x_nrightVert leq r_n-r_n+1$ yields $leftVert
x_n+m-x_nrightVert leq r_n-r_n+mrightarrow 0$ as $n,mrightarrow
infty $

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

answered Jul 31 at 9:01

Kavi Rama Murthy

19.5k2829

19.5k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
  – MaudPieTheRocktorate
  Jul 31 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
  – Kavi Rama Murthy
  Jul 31 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, was thinking the wrong way.
  – nicomezi
  Jul 31 at 9:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
  – MaudPieTheRocktorate
  Jul 31 at 9:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
  – Kavi Rama Murthy
  Jul 31 at 9:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, was thinking the wrong way.
  – nicomezi
  Jul 31 at 9:30
  

  
  
  
  

Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
– MaudPieTheRocktorate
Jul 31 at 9:02

Thanks. I also found that the solution is a part of this solution to this problem: math.stackexchange.com/a/22484/256345
– MaudPieTheRocktorate
Jul 31 at 9:02

@nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
– Kavi Rama Murthy
Jul 31 at 9:27

@nicomezi Hint: $x_n+1-x_n$ is common to the two terms on the left.
– Kavi Rama Murthy
Jul 31 at 9:27

Thanks, was thinking the wrong way.
– nicomezi
Jul 31 at 9:30

Thanks, was thinking the wrong way.
– nicomezi
Jul 31 at 9:30

add a comment | 

up vote
1
down vote

If, in a normed space, you have two balls, $B_r(a)$ and $B_r'(a')$ and if $B_r(a)supset B_r'(a')$, then $|a-a'|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+frac ra-a'(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$|p-a|=|p-a'|+|a'-a|.$$Therefore $r=|p-a'|+|a-a'|geqslant r'+|a-a'|$ and so $|a-a'|leqslant r-r'$.

Now, if $r_n$ is the radius of $B_n$, then $(r_n)_ninmathbb N$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $varepsilon>0$, $|r_m-r_n|<varepsilon$ if $m,ngg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $varepsilon$.

share|cite|improve this answer

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

add a comment | 

up vote
1
down vote

If, in a normed space, you have two balls, $B_r(a)$ and $B_r'(a')$ and if $B_r(a)supset B_r'(a')$, then $|a-a'|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+frac ra-a'(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$|p-a|=|p-a'|+|a'-a|.$$Therefore $r=|p-a'|+|a-a'|geqslant r'+|a-a'|$ and so $|a-a'|leqslant r-r'$.

Now, if $r_n$ is the radius of $B_n$, then $(r_n)_ninmathbb N$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $varepsilon>0$, $|r_m-r_n|<varepsilon$ if $m,ngg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $varepsilon$.

share|cite|improve this answer

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

add a comment | 

up vote
1
down vote

up vote
1
down vote

If, in a normed space, you have two balls, $B_r(a)$ and $B_r'(a')$ and if $B_r(a)supset B_r'(a')$, then $|a-a'|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+frac ra-a'(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$|p-a|=|p-a'|+|a'-a|.$$Therefore $r=|p-a'|+|a-a'|geqslant r'+|a-a'|$ and so $|a-a'|leqslant r-r'$.

Now, if $r_n$ is the radius of $B_n$, then $(r_n)_ninmathbb N$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $varepsilon>0$, $|r_m-r_n|<varepsilon$ if $m,ngg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $varepsilon$.

share|cite|improve this answer

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

If, in a normed space, you have two balls, $B_r(a)$ and $B_r'(a')$ and if $B_r(a)supset B_r'(a')$, then $|a-a'|$ is $r-r'$, at most. You can prove it as follows (see the picture below): if $a=a'$, it's trivial. Otherwise, consider the ray whose origin is $a$ and passes through $a'$. Let $p$ be the point of the ray whose distance to $a$ is $r$ (that is, $p=a+frac ra-a'(a-a')$). Then $a$, $a'$ and $p$ are colinear, with $a'$ being between the other two. So$$|p-a|=|p-a'|+|a'-a|.$$Therefore $r=|p-a'|+|a-a'|geqslant r'+|a-a'|$ and so $|a-a'|leqslant r-r'$.

Now, if $r_n$ is the radius of $B_n$, then $(r_n)_ninmathbb N$ is a decreasing sequence of non-negative real numbers. Therefore, it converges. So, given $varepsilon>0$, $|r_m-r_n|<varepsilon$ if $m,ngg1$. But then the distance between the centers of $B_m$ and $B_n$ is smaller than $varepsilon$.

share|cite|improve this answer

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 10:58

edited Jul 31 at 10:58

edited Jul 31 at 10:58

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

answered Jul 31 at 9:06

José Carlos Santos

112k1696172

112k1696172

add a comment | 

add a comment | 

 

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