$ \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}} $
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Component-wise Convergence (Sequences)

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. Then the sequence $(\vec{x}_k)$ converges to a point $\vec{a}=(a_1, \dots, a_n)$ if and only if $\lim_{k\to\infty}x_{k, i}=a_i$ for $1\leq i\leq n$.

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Proof

Suppose that $(\vec{x}_k)$ converges to $\vec{a}$
Let $i\in\{1, 2, \dots, n \}$ and $\epsilon>0$. Convergence means that there exists some $N$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$. By the definition of the Euclidean norm, $$\begin{align*} ||\vec{x}_k-\vec{a}|| &=(\sum_{j=1}^n (x_{j, k}-a_j)^2)^{1/2} \\ & \geq ||x_{k, i}-a_i||, \end{align*}$$ (as in the distance between two components can't be greater than the norm). This means that for all $k\geq N_i=N$, we have $|x_{k, i}-a_i|\leq ||\vec{x}_k-\vec{a}||<\epsilon$, which means that each component converges, as desired.

Now assume that $x_k, j$ converges to $a_j$.
Let $\epsilon>0$, $\bar{\epsilon} =\epsilon/\sqrt{n}$. Because we have component-wise convergence, then for each $i\in\{1, \dots, n\}$, there exists some $N_i$ such that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N_i$.
Now let $N=\max\{N_i\}$, which implies that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N$. Then that means $$\begin{align*} ||\vec{x}_k-\vec{a}||&=(\sum_{i=1}^n (x_{k, i}-a_i)^2)^{1/2} \\ &\leq (\sum_{i=1}^n \bar{\epsilon}^2)^{1/2} \\ &=(n(\epsilon/\sqrt{n})^2)^{1/2}\\ &=\epsilon, \end{align*}$$ and hence $||\vec{x}_k-\vec{a}||<\epsilon$ for $k\geq N$.

Component-wise Convergence (Sequences)

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. Then the sequence $(\vec{x}_k)$ converges to a point $\vec{a}=(a_1, \dots, a_n)$ if and only if $\lim_{k\to\infty}x_{k, i}=a_i$ for $1\leq i\leq n$.

Concepts

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If

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Only If

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Proof

Suppose that $(\vec{x}_k)$ converges to $\vec{a}$
Let $i\in\{1, 2, \dots, n \}$ and $\epsilon>0$. Convergence means that there exists some $N$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$. By the definition of the Euclidean norm, $$\begin{align*} ||\vec{x}_k-\vec{a}|| &=(\sum_{j=1}^n (x_{j, k}-a_j)^2)^{1/2} \\ & \geq ||x_{k, i}-a_i||, \end{align*}$$ (as in the distance between two components can't be greater than the norm). This means that for all $k\geq N_i=N$, we have $|x_{k, i}-a_i|\leq ||\vec{x}_k-\vec{a}||<\epsilon$, which means that each component converges, as desired.

Now assume that $x_k, j$ converges to $a_j$.
Let $\epsilon>0$, $\bar{\epsilon} =\epsilon/\sqrt{n}$. Because we have component-wise convergence, then for each $i\in\{1, \dots, n\}$, there exists some $N_i$ such that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N_i$.
Now let $N=\max\{N_i\}$, which implies that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N$. Then that means $$\begin{align*} ||\vec{x}_k-\vec{a}||&=(\sum_{i=1}^n (x_{k, i}-a_i)^2)^{1/2} \\ &\leq (\sum_{i=1}^n \bar{\epsilon}^2)^{1/2} \\ &=(n(\epsilon/\sqrt{n})^2)^{1/2}\\ &=\epsilon, \end{align*}$$ and hence $||\vec{x}_k-\vec{a}||<\epsilon$ for $k\geq N$.
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proof

Component-wise Convergence (Sequences)

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. Then the sequence $(\vec{x}_k)$ converges to a point $\vec{a}=(a_1, \dots, a_n)$ if and only if $\lim_{k\to\infty}x_{k, i}=a_i$ for $1\leq i\leq n$.

Concepts

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If

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Only If

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Proof

Suppose that $(\vec{x}_k)$ converges to $\vec{a}$
Let $i\in\{1, 2, \dots, n \}$ and $\epsilon>0$. Convergence means that there exists some $N$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$. By the definition of the Euclidean norm, $$\begin{align*} ||\vec{x}_k-\vec{a}|| &=(\sum_{j=1}^n (x_{j, k}-a_j)^2)^{1/2} \\ & \geq ||x_{k, i}-a_i||, \end{align*}$$ (as in the distance between two components can't be greater than the norm). This means that for all $k\geq N_i=N$, we have $|x_{k, i}-a_i|\leq ||\vec{x}_k-\vec{a}||<\epsilon$, which means that each component converges, as desired.

Now assume that $x_k, j$ converges to $a_j$.
Let $\epsilon>0$, $\bar{\epsilon} =\epsilon/\sqrt{n}$. Because we have component-wise convergence, then for each $i\in\{1, \dots, n\}$, there exists some $N_i$ such that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N_i$.
Now let $N=\max\{N_i\}$, which implies that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N$. Then that means $$\begin{align*} ||\vec{x}_k-\vec{a}||&=(\sum_{i=1}^n (x_{k, i}-a_i)^2)^{1/2} \\ &\leq (\sum_{i=1}^n \bar{\epsilon}^2)^{1/2} \\ &=(n(\epsilon/\sqrt{n})^2)^{1/2}\\ &=\epsilon, \end{align*}$$ and hence $||\vec{x}_k-\vec{a}||<\epsilon$ for $k\geq N$.

Component-wise Convergence (Sequences)

Let $(\vec{x}_k)_{k=1}^{\infty}$ be a sequence of points in $\mathbb{R}^n$ where each point is of the form $\vec{x}_k=(x_{k, 1}, x_{k, 2}, \dots, x_{k, n})$. Then the sequence $(\vec{x}_k)$ converges to a point $\vec{a}=(a_1, \dots, a_n)$ if and only if $\lim_{k\to\infty}x_{k, i}=a_i$ for $1\leq i\leq n$.

Concepts

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If

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Only If

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Proof

Suppose that $(\vec{x}_k)$ converges to $\vec{a}$
Let $i\in\{1, 2, \dots, n \}$ and $\epsilon>0$. Convergence means that there exists some $N$ such that $||\vec{x}_k-\vec{a}||<\epsilon$ for all $k\geq N$. By the definition of the Euclidean norm, $$\begin{align*} ||\vec{x}_k-\vec{a}|| &=(\sum_{j=1}^n (x_{j, k}-a_j)^2)^{1/2} \\ & \geq ||x_{k, i}-a_i||, \end{align*}$$ (as in the distance between two components can't be greater than the norm). This means that for all $k\geq N_i=N$, we have $|x_{k, i}-a_i|\leq ||\vec{x}_k-\vec{a}||<\epsilon$, which means that each component converges, as desired.

Now assume that $x_k, j$ converges to $a_j$.
Let $\epsilon>0$, $\bar{\epsilon} =\epsilon/\sqrt{n}$. Because we have component-wise convergence, then for each $i\in\{1, \dots, n\}$, there exists some $N_i$ such that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N_i$.
Now let $N=\max\{N_i\}$, which implies that $|x_{k, i}-a_i|<\bar{\epsilon}$ for all $k\geq N$. Then that means $$\begin{align*} ||\vec{x}_k-\vec{a}||&=(\sum_{i=1}^n (x_{k, i}-a_i)^2)^{1/2} \\ &\leq (\sum_{i=1}^n \bar{\epsilon}^2)^{1/2} \\ &=(n(\epsilon/\sqrt{n})^2)^{1/2}\\ &=\epsilon, \end{align*}$$ and hence $||\vec{x}_k-\vec{a}||<\epsilon$ for $k\geq N$.
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