[proofplan] We prove completeness by reducing a Euclidean [Cauchy sequence](/page/Cauchy%20Sequence) to its finitely many coordinate sequences. The Euclidean distance controls each coordinate difference, so every coordinate sequence is Cauchy in $\mathbb{R}$ and hence converges by completeness of the [real numbers](/page/Real%20Numbers). The coordinate limits define a point $a \in \mathbb{R}^n$, and a finite-sum estimate shows that the original sequence converges to $a$ in the [Euclidean metric](/page/Euclidean%20Metric). [/proofplan] [step:Pass from a Euclidean Cauchy sequence to Cauchy coordinate sequences] Let $(x_m)_{m \in \mathbb{N}}$ be a $d_2$-Cauchy sequence in $\mathbb{R}^n$. For each $m \in \mathbb{N}$, write \begin{align*} x_m = (x_{m,1},\dots,x_{m,n}). \end{align*} Fix an index $i \in \{1,\dots,n\}$. Since $(x_m)_{m \in \mathbb{N}}$ is $d_2$-Cauchy, for every $\varepsilon>0$ there exists $M \in \mathbb{N}$ such that whenever $m,\ell \ge M$, \begin{align*} d_2(x_m,x_\ell)<\varepsilon. \end{align*} For such $m,\ell$, the definition of $d_2$ gives \begin{align*} |x_{m,i}-x_{\ell,i}|^2 \le \sum_{j=1}^n |x_{m,j}-x_{\ell,j}|^2. \end{align*} Taking square roots of both non-negative sides yields \begin{align*} |x_{m,i}-x_{\ell,i}| \le d_2(x_m,x_\ell)<\varepsilon. \end{align*} Thus the real sequence $(x_{m,i})_{m \in \mathbb{N}}$ is Cauchy in $\mathbb{R}$. [guided] The goal is to convert the Cauchy condition in $\mathbb{R}^n$ into an ordinary Cauchy condition for each real coordinate. Fix an index $i \in \{1,\dots,n\}$. The $i$-th coordinate sequence is the map \begin{align*} \mathbb{N} &\to \mathbb{R} \end{align*} sending $m$ to $x_{m,i}$. Let $\varepsilon>0$ be given. Since $(x_m)_{m \in \mathbb{N}}$ is Cauchy in the metric $d_2$, there exists $M \in \mathbb{N}$ such that for all $m,\ell \ge M$, \begin{align*} d_2(x_m,x_\ell)<\varepsilon. \end{align*} Now we compare a single coordinate difference with the full Euclidean distance. By the definition of $d_2$, \begin{align*} d_2(x_m,x_\ell)^2 = \sum_{j=1}^n |x_{m,j}-x_{\ell,j}|^2. \end{align*} Every term in this finite sum is non-negative, so the $i$-th term is bounded above by the whole sum: \begin{align*} |x_{m,i}-x_{\ell,i}|^2 \le \sum_{j=1}^n |x_{m,j}-x_{\ell,j}|^2. \end{align*} Taking square roots is valid because both sides are non-negative. Hence, for all $m,\ell \ge M$, \begin{align*} |x_{m,i}-x_{\ell,i}| \le d_2(x_m,x_\ell)<\varepsilon. \end{align*} This is exactly the Cauchy condition for the real sequence $(x_{m,i})_{m \in \mathbb{N}}$. Since the index $i$ was arbitrary, every coordinate sequence is Cauchy in $\mathbb{R}$. [/guided] [/step] [step:Use completeness of $\mathbb{R}$ to define the candidate limit] For each $i \in \{1,\dots,n\}$, the coordinate sequence $(x_{m,i})_{m \in \mathbb{N}}$ is Cauchy in $\mathbb{R}$. By the completeness of the real numbers, there exists $a_i \in \mathbb{R}$ such that \begin{align*} \lim_{m \to \infty} x_{m,i} = a_i. \end{align*} Define $a \in \mathbb{R}^n$ by \begin{align*} a = (a_1,\dots,a_n). \end{align*} [/step] [step:Reassemble the coordinate limits into Euclidean convergence] We show that $x_m \to a$ with respect to $d_2$. Let $\varepsilon>0$. Since $n \in \mathbb{N}$, we have $n \ge 1$, so $\sqrt{n}>0$. For each $i \in \{1,\dots,n\}$, the convergence $x_{m,i} \to a_i$ gives an integer $M_i \in \mathbb{N}$ such that for all $m \ge M_i$, \begin{align*} |x_{m,i}-a_i|<\frac{\varepsilon}{\sqrt{n}}. \end{align*} Define \begin{align*} M = \max\{M_1,\dots,M_n\}. \end{align*} Then for every $m \ge M$ and every $i \in \{1,\dots,n\}$, \begin{align*} |x_{m,i}-a_i|<\frac{\varepsilon}{\sqrt{n}}. \end{align*} Therefore \begin{align*} d_2(x_m,a)^2 = \sum_{i=1}^n |x_{m,i}-a_i|^2 < \sum_{i=1}^n \frac{\varepsilon^2}{n} = \varepsilon^2. \end{align*} Taking square roots gives \begin{align*} d_2(x_m,a)<\varepsilon. \end{align*} Thus $x_m \to a$ in the metric $d_2$. [guided] We now have coordinate limits $a_i \in \mathbb{R}$ for $i \in \{1,\dots,n\}$, and we have defined \begin{align*} a = (a_1,\dots,a_n) \in \mathbb{R}^n. \end{align*} To prove convergence in the Euclidean metric, we must prove that for every $\varepsilon>0$ there exists $M \in \mathbb{N}$ such that $m \ge M$ implies \begin{align*} d_2(x_m,a)<\varepsilon. \end{align*} Let $\varepsilon>0$. Since $n \in \mathbb{N}$ and the convention is that $\mathbb{N}$ starts at $1$, the number $\sqrt{n}$ is positive. For each coordinate $i \in \{1,\dots,n\}$, the real convergence $x_{m,i} \to a_i$ gives an integer $M_i \in \mathbb{N}$ such that for all $m \ge M_i$, \begin{align*} |x_{m,i}-a_i|<\frac{\varepsilon}{\sqrt{n}}. \end{align*} Because there are only finitely many coordinates, we can choose one index threshold that works for all of them: \begin{align*} M = \max\{M_1,\dots,M_n\}. \end{align*} Then $m \ge M$ implies $m \ge M_i$ for every $i \in \{1,\dots,n\}$, so every coordinate error satisfies \begin{align*} |x_{m,i}-a_i|<\frac{\varepsilon}{\sqrt{n}}. \end{align*} Substituting these coordinate bounds into the definition of the Euclidean metric gives \begin{align*} d_2(x_m,a)^2 = \sum_{i=1}^n |x_{m,i}-a_i|^2 < \sum_{i=1}^n \frac{\varepsilon^2}{n} = \varepsilon^2. \end{align*} The quantities $d_2(x_m,a)$ and $\varepsilon$ are non-negative, so taking square roots yields \begin{align*} d_2(x_m,a)<\varepsilon. \end{align*} This proves $x_m \to a$ in the metric $d_2$. [/guided] [/step] [step:Conclude that every Euclidean Cauchy sequence converges in $\mathbb{R}^n$] Starting from an arbitrary $d_2$-Cauchy sequence $(x_m)_{m \in \mathbb{N}}$ in $\mathbb{R}^n$, we constructed a point $a \in \mathbb{R}^n$ and proved that $d_2(x_m,a) \to 0$. Hence every $d_2$-Cauchy sequence in $\mathbb{R}^n$ converges to a point of $\mathbb{R}^n$. Therefore $(\mathbb{R}^n,d_2)$ is complete. [/step]