$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
FullPage
result
Concepts
If
Only If
proof

Component of Cauchy Sequences

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. The sequence $(\vec{x}_k)$ is cauchy if and only if $(x_{k, i})_{k=1}^{\infty}$ if cauchy for each $1\leq i\leq n$

Concepts

This relates Cauchy sequences in vector spaces and cauchy sequences of components of the vector.
It would not make sense if one component of the vector is a cauchy sequence (or not at all a cauchy sequence) while the vector itself is. Where is that one component going. It's not getting any closer to the point we want.

If

We have that $(x_{k, i})_{k=1}^{\infty}$ for each $1 \leq i\leq n$ converces to $a_i$. By definition, this means for each $1\leq i\leq n$, for any $\epsilon>0$, there exists some $N_i\in\mathbb{N}$ such that $|x_{k, i}-a_i|<\epsilon$ for all $k\geq N_i$.
If we take $N=\text{max}\{N_i:1\leq i\leq n\}$, then $|x_{k, i}-a_i|<\epsilon$ for all $1\leq i\leq n$ and $k\geq N$.

Only If

We have that $(\vec{x}_k)_{k=1}^\infty$ converges to $\vec{a}$. That means for all $\epsilon>0$, there exists some $N\in\mathbb{N}$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$.

Proof

Coming soon

Component of Cauchy Sequences

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. The sequence $(\vec{x}_k)$ is cauchy if and only if $(x_{k, i})_{k=1}^{\infty}$ if cauchy for each $1\leq i\leq n$

Concepts

This relates Cauchy sequences in vector spaces and cauchy sequences of components of the vector.
It would not make sense if one component of the vector is a cauchy sequence (or not at all a cauchy sequence) while the vector itself is. Where is that one component going. It's not getting any closer to the point we want.

If

We have that $(x_{k, i})_{k=1}^{\infty}$ for each $1 \leq i\leq n$ converces to $a_i$. By definition, this means for each $1\leq i\leq n$, for any $\epsilon>0$, there exists some $N_i\in\mathbb{N}$ such that $|x_{k, i}-a_i|<\epsilon$ for all $k\geq N_i$.
If we take $N=\text{max}\{N_i:1\leq i\leq n\}$, then $|x_{k, i}-a_i|<\epsilon$ for all $1\leq i\leq n$ and $k\geq N$.

Only If

We have that $(\vec{x}_k)_{k=1}^\infty$ converges to $\vec{a}$. That means for all $\epsilon>0$, there exists some $N\in\mathbb{N}$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$.

Proof

Coming soon
FullPage
result
concepts
If
Only If
proof
FullPage
result
Concepts
If
Only If
proof

Component of Cauchy Sequences

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. The sequence $(\vec{x}_k)$ is cauchy if and only if $(x_{k, i})_{k=1}^{\infty}$ if cauchy for each $1\leq i\leq n$

Concepts

This relates Cauchy sequences in vector spaces and cauchy sequences of components of the vector.
It would not make sense if one component of the vector is a cauchy sequence (or not at all a cauchy sequence) while the vector itself is. Where is that one component going. It's not getting any closer to the point we want.

If

We have that $(x_{k, i})_{k=1}^{\infty}$ for each $1 \leq i\leq n$ converces to $a_i$. By definition, this means for each $1\leq i\leq n$, for any $\epsilon>0$, there exists some $N_i\in\mathbb{N}$ such that $|x_{k, i}-a_i|<\epsilon$ for all $k\geq N_i$.
If we take $N=\text{max}\{N_i:1\leq i\leq n\}$, then $|x_{k, i}-a_i|<\epsilon$ for all $1\leq i\leq n$ and $k\geq N$.

Only If

We have that $(\vec{x}_k)_{k=1}^\infty$ converges to $\vec{a}$. That means for all $\epsilon>0$, there exists some $N\in\mathbb{N}$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$.

Proof

Coming soon

Component of Cauchy Sequences

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. The sequence $(\vec{x}_k)$ is cauchy if and only if $(x_{k, i})_{k=1}^{\infty}$ if cauchy for each $1\leq i\leq n$

Concepts

This relates Cauchy sequences in vector spaces and cauchy sequences of components of the vector.
It would not make sense if one component of the vector is a cauchy sequence (or not at all a cauchy sequence) while the vector itself is. Where is that one component going. It's not getting any closer to the point we want.

If

We have that $(x_{k, i})_{k=1}^{\infty}$ for each $1 \leq i\leq n$ converces to $a_i$. By definition, this means for each $1\leq i\leq n$, for any $\epsilon>0$, there exists some $N_i\in\mathbb{N}$ such that $|x_{k, i}-a_i|<\epsilon$ for all $k\geq N_i$.
If we take $N=\text{max}\{N_i:1\leq i\leq n\}$, then $|x_{k, i}-a_i|<\epsilon$ for all $1\leq i\leq n$ and $k\geq N$.

Only If

We have that $(\vec{x}_k)_{k=1}^\infty$ converges to $\vec{a}$. That means for all $\epsilon>0$, there exists some $N\in\mathbb{N}$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$.

Proof

Coming soon
FullPage
result
concepts
If
Only If
proof