# The Metric Space R^n and Sequences ... Remark by Carothers, page 47 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading N. L. Carothers' book: "Real Analysis". ... ...

I am focused on Chapter 3: Metrics and Norms ... ...

I need help with a remark by Carothers concerning convergent sequences in \mathbb{R}^n ...

Now ... on page 47 Carothers writes the following: In the above text from Carothers we read the following:

" ... ... it follows that a sequence of vectors $$\displaystyle x^{ (k) } = ( x_1^k, \ ... \ ... \ , x_n^k)$$ in $$\displaystyle \mathbb{R}^n$$ converges (is Cauchy) if and only if each of the coordinate sequences $$\displaystyle ( x_j^k )$$ converges in $$\displaystyle \mathbb{R}$$ ... ... "

My question is as follows:

Why exactly does it follow that a sequence of vectors $$\displaystyle x^{ (k) } = ( x_1^k, \ ... \ ... \ , x_n^k)$$ in $$\displaystyle \mathbb{R}^n$$ converges (is Cauchy) if and only if each of the coordinate sequences $$\displaystyle ( x_j^k )$$ converges in $$\displaystyle \mathbb{R}$$ ... ... ?

Help will be appreciated ...

Peter

#### steep

##### Member
it may be convenient to switch norms slightly here...
in particular with

$\mathbf z:= \mathbf x - \mathbf y$

and $\mathbf z \in \mathbb R^n$

convince yourself that
$\big \Vert \mathbf z \big \Vert_2 \leq \big \Vert \mathbf z \big \Vert_1 \leq \sqrt{n}\cdot \big \Vert \mathbf z \big \Vert_2$

where the first inequality is triangle inequality and 2nd one is cauchy-schwarz (with 1's trick).

to a large extent the 1 norm allows you linearize the distance computed on each component.... can you prove the Carothers comment of convergence iff each $x_k$ converges in $\mathbb R$ using the 1 norm? The first leg should be easy -- select $\frac{\epsilon}{n}$ for each component and import favorite single variable real analysis results. The second leg is similar...
- - - -
Then using the above chain of inequalities this givess you the result with the 2 norm / standard metric on x,y.

#### Peter

##### Well-known member
MHB Site Helper
it may be convenient to switch norms slightly here...
in particular with

$\mathbf z:= \mathbf x - \mathbf y$

and $\mathbf z \in \mathbb R^n$

convince yourself that
$\big \Vert \mathbf z \big \Vert_2 \leq \big \Vert \mathbf z \big \Vert_1 \leq \sqrt{n}\cdot \big \Vert \mathbf z \big \Vert_2$

where the first inequality is triangle inequality and 2nd one is cauchy-schwarz (with 1's trick).

to a large extent the 1 norm allows you linearize the distance computed on each component.... can you prove the Carothers comment of convergence iff each $x_k$ converges in $\mathbb R$ using the 1 norm? The first leg should be easy -- select $\frac{\epsilon}{n}$ for each component and import favorite single variable real analysis results. The second leg is similar...
- - - -
Then using the above chain of inequalities this givess you the result with the 2 norm / standard metric on x,y.

Thanks for the help steep ...

... a sequence of vectors $$\displaystyle x^{ (k) } = ( x_1^k, \ ... \ ... \ , x_n^k)$$ in $$\displaystyle \mathbb{R}^n$$ converges if and only if each of the coordinate sequences $$\displaystyle ( x_j^k )$$ converges in $$\displaystyle \mathbb{R}$$ ... ...

I think we may proceed as follows where $$\displaystyle z = x - y$$ ...

$$\displaystyle \| z \mid \mid_2 \ = \| \sum_{ j = 1}^n z_j e_j \| \leq \sum_{ j = 1}^n \| z_j e_j \| = \sum_{ j = 1}^n \mid z_j \mid \| e_j \| = \sum_{ j = 1}^n \mid z_j \mid$$

Thus $$\displaystyle \| x - y \| = \left( \sum_{ j = 1}^n \mid x_j - y_j \mid^2 \right)^{ \frac{1}{2} } \leq \sum_{ j = 1}^n \mid x_j - y_j \mid$$ ... ... ... (1)

Now ... given (1) above ...

... if $$\displaystyle ( x_j^k)_{ k = 1}^{ \infty }$$ converges to a limit $$\displaystyle y \in \mathbb{R}^n$$ ...

... then for every $$\displaystyle \frac{ \epsilon }{ n } \gt 0 \ \exists \ N(j ; \epsilon )$$ such that for $$\displaystyle k \geq N(j ; \epsilon )$$ ...

... we have $$\displaystyle \mid x_j^k - y \mid \lt \frac{ \epsilon }{ n }$$ ...

But then we have $$\displaystyle x^{ (k) }$$ converges to $$\displaystyle y$$ since ...

... for every $$\displaystyle \epsilon \gt 0 \ \exists \ N( \epsilon )$$ such that for $$\displaystyle k \geq N( \epsilon )$$ we have ...

... $$\displaystyle \| x^{ (k) } - y \| \leq \sum_{ j = 1}^n \mid x_j^k - y \mid \lt \frac{ \epsilon }{ n } + \ ... \ ... \ \frac{ \epsilon }{ n } = \epsilon$$

Is that correct?

Proof is similar for Cauchy sequences in \mathbb{R}^n ...

Peter

Last edited:

#### steep

##### Member
Thanks for the help steep ...

... a sequence of vectors $$\displaystyle x^{ (k) } = ( x_1^k, \ ... , x_n^k)$$ in $$\displaystyle \mathbb{R}^n$$ converges if and only if each of the coordinate sequences $$\displaystyle ( x_j^k )$$ converges in $$\displaystyle \mathbb{R}$$ ... ...

I think we may proceed as follows where $$\displaystyle z = x - y$$ ...

$$\displaystyle \| z \mid \mid_2 \ = \| \sum_{ j = 1}^n z_j e_j \| \leq \sum_{ j = 1}^n \| z_j e_j \| = \sum_{ j = 1}^n \mid z_j \mid \| e_j \| = \sum_{ j = 1}^n \mid z_j \mid$$

Thus $$\displaystyle \| x - y \| = \left( \sum_{ j = 1}^n \mid x_j - y_j \mid^2 \right)^{ \frac{1}{2} } \leq \sum_{ j = 1}^n \mid x_j - y_j \mid$$ ... ... ... (1)

Now ... given (1) above ...

... if $$\displaystyle ( x_j^k)_{ k = 1}^{ \infty }$$ converges to a limit $$\displaystyle y \in \mathbb{R}^n$$ ...

... then for every $$\displaystyle \frac{ \epsilon }{ n } \gt 0 \ \exists \ N(j ; \epsilon )$$ such that for $$\displaystyle k \geq N(j ; \epsilon )$$ ...

... we have $$\displaystyle \mid x_j^k - y \mid \lt \frac{ \epsilon }{ n }$$ ...

But then we have $$\displaystyle x^{ (k) }$$ converges to $$\displaystyle y$$ since ...

... for every $$\displaystyle \epsilon \gt 0 \ \exists \ N( \epsilon )$$ such that for $$\displaystyle k \geq N( \epsilon )$$ we have ...

... $$\displaystyle \| x^{ (k) } - y \| \leq \sum_{ j = 1}^n \mid x_j^k - y \mid \lt \frac{ \epsilon }{ n } + \ ... \ ... \ \frac{ \epsilon }{ n } = \epsilon$$

Is that correct?

Proof is similar for Cauchy sequences in \mathbb{R}^n ...

Peter
I think this is basically right. You may be approaching it in a more succinct manner than I am... I have this in my head as 2 steps, first sufficiency, then necessity. The above clearly gives sufficiency. I'm not sure I saw the second leg, necessity, though.

Another way to get it is to use the infinity /max norm, so

$\big\Vert \mathbf z \big \Vert_\infty^2 = \max\big(z_1^2, z_1^2, ...., z_n^2\big) \leq \sum_{i=1}^n z_i^2 = \big \Vert \mathbf z\big \Vert_2^2$
hence taking square roots over non-negative numbers gives

$\big\Vert \mathbf z \big \Vert_\infty \leq \big \Vert \mathbf z\big \Vert_2$

so for the second leg, suppose that there is (at least one) coordinate j that doesn't converge -- i.e. where there is no N large enough such that $\vert z_j^{(n)} \vert \lt \epsilon_0$, for all $n\geq N$... then if the sequence still convergences you'd have

$\epsilon_0 \leq \vert z_j^{(n)} \vert \leq \big \Vert \mathbf z^{(n)}\big \Vert_\infty \leq \big \Vert \mathbf z^{(n)}\big \Vert_2 \lt \epsilon$
for some $n \geq N$ for any $N$. Selecting $\epsilon := \epsilon_0$ then contradicts the definition of convergence.

There's probably a slightly nicer way of showing it, but this is at the heart of the necessity of component-wise convergence.

#### Peter

##### Well-known member
MHB Site Helper
I think this is basically right. You may be approaching it in a more succinct manner than I am... I have this in my head as 2 steps, first sufficiency, then necessity. The above clearly gives sufficiency. I'm not sure I saw the second leg, necessity, though.

Another way to get it is to use the infinity /max norm, so

$\big\Vert \mathbf z \big \Vert_\infty^2 = \max\big(z_1^2, z_1^2, ...., z_n^2\big) \leq \sum_{i=1}^n z_i^2 = \big \Vert \mathbf z\big \Vert_2^2$
hence taking square roots over non-negative numbers gives

$\big\Vert \mathbf z \big \Vert_\infty \leq \big \Vert \mathbf z\big \Vert_2$

so for the second leg, suppose that there is (at least one) coordinate j that doesn't converge -- i.e. where there is no N large enough such that $\vert z_j^{(n)} \vert \lt \epsilon_0$, for all $n\geq N$... then if the sequence still convergences you'd have

$\epsilon_0 \leq \vert z_j^{(n)} \vert \leq \big \Vert \mathbf z^{(n)}\big \Vert_\infty \leq \big \Vert \mathbf z^{(n)}\big \Vert_2 \lt \epsilon$
for some $n \geq N$ for any $N$. Selecting $\epsilon := \epsilon_0$ then contradicts the definition of convergence.

There's probably a slightly nicer way of showing it, but this is at the heart of the necessity of component-wise convergence.

Hi steep ...

Thanks again for your considerable assistance ...

Thought I would try a direct approach to demonstrate that ...

... if a sequence of vectors $$\displaystyle x^{ (k) } = ( x_1^k, \ ... \ ... \ , x_n^k)$$ in $$\displaystyle \mathbb{R}^n$$ converges ...

... then ... each of the coordinate sequences $$\displaystyle ( x_j^k )$$ converges in $$\displaystyle \mathbb{R}$$ ... ...

Proceed as follows, where $$\displaystyle z = x - y$$ ...

$$\displaystyle \mid z_j \mid = \| z_j \|_2 = \sum_{ j = 1}^1 (z_j^2)^{ \frac{1}{2} } = \left( \sum_{ j = 1}^1 z_j^2) \right)^{ \frac{1}{2} } \leq \left( \sum_{ j = 1}^n z_j^2) \right)^{ \frac{1}{2} } = \| z \|_2$$ ... ... ... ... ... (2)

Given (2) above ... we have ...

... if $$\displaystyle x^{ (k) }$$ converges to $$\displaystyle y$$ in $$\displaystyle \mathbb{R}^n$$ ...

... then ... for every $$\displaystyle \epsilon \gt 0 \ \exists \ N( \epsilon )$$ such that for $$\displaystyle k \geq N( \epsilon )$$ we have ...

... $$\displaystyle \| x^{ (k) } - y \|_2 \leq \epsilon$$ ...

But then, for arbitrary j, we have $$\displaystyle ( x_j^k)_{ k = 1}^{ \infty }$$ converges to a limit $$\displaystyle y_j \in \mathbb{R}$$ ...

... since for every $$\displaystyle \epsilon \gt 0 \ \exists \ N( \epsilon )$$ such that for $$\displaystyle k \geq N( \epsilon )$$ we have ...

$$\displaystyle \mid x_j -y_j \mid = \| x_j -y_j \|_2 \leq \| x^{ (k) } - y \|_2 \leq \epsilon$$ ...

Is that correct?

Thanks once again for the help!

Peter

#### steep

##### Member
...
But then, for arbitrary j, we have $$\displaystyle ( x_j^k)_{ k = 1}^{ \infty }$$ converges to a limit $$\displaystyle y_j \in \mathbb{R}$$ ...

... since for every $$\displaystyle \epsilon \gt 0 \ \exists \ N( \epsilon )$$ such that for $$\displaystyle k \geq N( \epsilon )$$ we have ...

$$\displaystyle \mid x_j -y_j \mid = \| x_j -y_j \|_2 \leq \| x^{ (k) } - y \|_2 \leq \epsilon$$ ...
re-reading this thread with a fresh set of eyes I see that your post 3 really was

$\text{convergence in each component} \longrightarrow \text{convergence of vector in } \mathbb R^n$

and this current post 5 is the other leg
$\text{convergence of vector in } \mathbb R^n \longrightarrow \text{convergence in each component}$

and yes I think it works. The only nitpick I'll do is to make sure to use strictness in the inequality with the epsilon, i.e.
$\| x^{ (k) } - y \|_2 \lt \epsilon$

any other items are immaterial... and looking back through my posts it looks like I overloaded $n$ both for the dimension in reals as well as the number in the sequence, so no need to nitpick too much #### Peter

##### Well-known member
MHB Site Helper
re-reading this thread with a fresh set of eyes I see that your post 3 really was

$\text{convergence in each component} \longrightarrow \text{convergence of vector in } \mathbb R^n$

and this current post 5 is the other leg
$\text{convergence of vector in } \mathbb R^n \longrightarrow \text{convergence in each component}$

and yes I think it works. The only nitpick I'll do is to make sure to use strictness in the inequality with the epsilon, i.e.
$\| x^{ (k) } - y \|_2 \lt \epsilon$

any other items are immaterial... and looking back through my posts it looks like I overloaded $n$ both for the dimension in reals as well as the number in the sequence, so no need to nitpick too much Thanks for all your help, steep ...

I really appreciate it ...

Peter