$x^{(k)}$ converges to $x$ according to $d_2$ iff for every $1\leq j\leq n$, the sequence $x^{(k)}_j$ converges to $x_j$.

Question

Let $\textbf{R}^{n}$ be a Euclidean space, and let $(x^{k})_{k=m}^{\infty}$ be a sequence of points in $\textbf{R}^{n}$. We write $x^{(k)} = (x^{(k)}_{1},x^{(k)}_{2},\ldots,x^{(k)}_{n})$, i.e., for $j = 1,2,\ldots,n$, $x^{(k)}_{j}\in\textbf{R}$ is the $j$-th coordinate of $x^{(k)}\in\textbf{R}$. Let $x = (x_{1},x_{2},\ldots,x_{n})$ be a point in $\textbf{R}^{n}$. Then the following two statements are equivalent.

(a) $(x^{(k)})_{k=m}^{\infty}$ converges to $x$ with respect to the Euclidean metric $d_{2}$.

(b) For every $1\leq j\leq n$, the sequence $(x^{(k)}_{j})_{k=m}^{\infty}$ converges to $x_{j}$.

MY ATTEMPT

Let us prove $(a)\Rightarrow(b)$ first.

According to the definition of limit, for every $\varepsilon > 0$, there is a $N\geq m$ such that \begin{align*} k\geq N & \Rightarrow |x^{(k)}_{j} - x_{j}| \leq (|x^{(k)}_{1} - x_{1}|^{2} + |x^{(k)}_{2} - x_{2}|^{2} + \ldots + |x^{(k)}_{n} - x_{n}|^{2})^{1/2}\\\\ & = \|x^{(k)} - x\| \leq \varepsilon \Rightarrow |x^{(k)}_{j} - x_{j}| \leq \varepsilon \end{align*} and the result holds.

Let us now prove $(b)\Rightarrow(a)$.

Once again, according to the definition of limit, for every $\varepsilon/\sqrt{n} > 0$, there is $N_{j} \geq m$ such that \begin{align*} k\geq N_{j} \Rightarrow |x^{(k)}_{j} - x_{j}| \leq \frac{\varepsilon}{\sqrt{n}} \Rightarrow |x^{(k)}_{j} - x_{j}|^{2} \leq \frac{\varepsilon^{2}}{n} \end{align*}

If we take $N = \max\{N_{1},N_{2},\ldots,N_{n}\}$, for every $\varepsilon > 0$, there is $N\geq m$ such that \begin{align*} k\geq N \Rightarrow \|x^{(k)} - x\| = (|x^{(k)}_{1} - x_{1}|^{2} + |x^{(k)}_{2} - x_{2}|^{2} + \ldots + |x^{(k)}_{n} - x_{n}|^{2})^{1/2} \leq \left(n\times\frac{\varepsilon^{2}}{n}\right)^{1/2} = \varepsilon \end{align*} and we are done.

Could someone please verify if I am reasoning correctly?

Answer 1

Your proof is correct.

Let me demonstrate how I would prove $(a) \implies (b)$ very rigorously, so you can compare.

Fix $\epsilon > 0$. Since $(x^{(k)})_{k=1}^\infty$ converges to $x$ for $d_2$, there exists $K \geq 1$ such that

$$k \geq K \implies d_2(x^{(k)},x)< \epsilon$$

Thus, whenever $k \geq N$ and $1 \leq j \leq n$, we get

$$|x_j^{(k)}-x_j| \leq \left(\sum_{i=1}^n |x_i^{(k)}-x_i|^2\right)^{1/2}= d_2(x^{(k)},x) < \epsilon$$

We thus have established:

$$\forall j \in \{1,2, \dots, n\}:\forall \epsilon > 0: \exists K\geq1: \forall k \geq K: |x_j^{(k)}-x_j|< \epsilon$$

which means that $(x_j^{(k)})_{k=1}^\infty$ converges to $x_j$ for every $j=1, 2, \dots, n$.