[proofplan] Let $(x_k)_{k \in \mathbb{N}}$ be a Cauchy sequence in $\mathbb{R}$. The strategy is to construct an explicit candidate limit by selecting, for each $k$, a rational number $q_k \in \mathbb{Q}$ that approximates $x_k$ to within $1/k$ (such a $q_k$ exists because $\mathbb{Q}$ is dense in $\mathbb{R}$ — a fact that follows directly from the Cauchy-completion construction). The triangle inequality together with the Cauchy property of $(x_k)$ forces $(q_k)$ to be a rational Cauchy sequence, so its equivalence class $x := [(q_k)] \in \mathbb{R}$ is a well-defined real number. We then show $x_k \to x$ in $\mathbb{R}$ by combining the rational approximation $|x_k - q_k| < 1/k$ with the definition of $|x - q_k|$ as a limit of rational quantities. [/proofplan] [step:Approximate each $x_k$ by a rational using density of $\mathbb{Q}$ in $\mathbb{R}$] Let $(x_k)_{k \in \mathbb{N}}$ be a Cauchy sequence in $\mathbb{R}$, i.e. for every $\varepsilon > 0$ in $\mathbb{R}$ there exists $K \in \mathbb{N}$ with $|x_k - x_l| < \varepsilon$ for all $k, l \ge K$. By the [Density of $\mathbb{Q}$ in $\mathbb{R}$](/theorems/740), for each $k \in \mathbb{N}$ there exists $q_k \in \mathbb{Q}$ such that \begin{align*} |x_k - q_k| < \frac{1}{k}. \end{align*} (Here we identify $q_k$ with its image $[(q_k, q_k, q_k, \dots)] \in \mathbb{R}$ under the canonical embedding $\iota: \mathbb{Q} \hookrightarrow \mathbb{R}$, $q \mapsto [(q, q, q, \dots)]$.) Define the sequence $(q_k)_{k \in \mathbb{N}} \subset \mathbb{Q}$. [/step] [step:Show that $(q_k)$ is a rational Cauchy sequence] We verify $(q_k)$ is Cauchy in $\mathbb{Q}$. Fix $\varepsilon \in \mathbb{Q}_{>0}$. Choose $K_1 \in \mathbb{N}$ with $1/K_1 < \varepsilon/3$. Since $(x_k)$ is Cauchy in $\mathbb{R}$, choose $K_2 \in \mathbb{N}$ such that $|x_k - x_l| < \varepsilon/3$ for all $k, l \ge K_2$. Set $K := \max\{K_1, K_2\}$. For $k, l \ge K$, the triangle inequality gives \begin{align*} |q_k - q_l| \le |q_k - x_k| + |x_k - x_l| + |x_l - q_l| < \frac{1}{k} + \frac{\varepsilon}{3} + \frac{1}{l} < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon, \end{align*} where we used $1/k, 1/l \le 1/K \le 1/K_1 < \varepsilon/3$. Hence $(q_k)$ is Cauchy in $\mathbb{Q}$. [/step] [step:Define the candidate limit $x := [(q_k)] \in \mathbb{R}$] Since $(q_k)$ is a rational Cauchy sequence, its equivalence class \begin{align*} x := [(q_k)] \in \mathbb{R} \end{align*} is a well-defined real number, where $\mathbb{R}$ is the set of equivalence classes of rational Cauchy sequences under $(a_n) \sim (b_n) \iff |a_n - b_n| \to 0$. [/step] [step:Show $q_k \to x$ in $\mathbb{R}$] We claim that the embedded rationals $\iota(q_k) = [(q_k, q_k, q_k, \dots)]$ converge to $x$ in $\mathbb{R}$. By definition, $x - \iota(q_k) = [(q_n - q_k)_{n \in \mathbb{N}}]$ (where $k$ is fixed and $n$ varies). The absolute value on $\mathbb{R}$ is defined on equivalence classes by $|[(a_n)]| := [(|a_n|)]$, i.e. it extends the rational absolute value through the Cauchy completion; equivalently, identifying a real number $[(a_n)]$ with the limit of its rational representatives, we have $|[(a_n)]| = \lim_{n \to \infty} |a_n|$ (the limit existing in $\mathbb{R}$ because $(|a_n|)$ is itself rational Cauchy by the reverse triangle inequality). Applying this with $a_n = q_n - q_k$: \begin{align*} |x - \iota(q_k)| = \lim_{n \to \infty} |q_n - q_k|, \end{align*} the limit being taken in $\mathbb{R}$ (this limit exists because $(|q_n - q_k|)_{n \in \mathbb{N}}$ is itself a Cauchy sequence in $\mathbb{Q}$ for each fixed $k$, by Step 2 and the reverse triangle inequality). Fix $\varepsilon > 0$ in $\mathbb{R}$. By Step 2, $(q_n)$ is Cauchy: choose $K \in \mathbb{N}$ such that $|q_n - q_l| < \varepsilon/2$ for all $n, l \ge K$. For $k \ge K$, taking $l = k$ and letting $n \to \infty$, \begin{align*} |x - \iota(q_k)| = \lim_{n \to \infty} |q_n - q_k| \le \frac{\varepsilon}{2} < \varepsilon \quad \text{for all } k \ge K. \end{align*} Hence $\iota(q_k) \to x$ in $\mathbb{R}$ as $k \to \infty$. [/step] [step:Conclude $x_k \to x$ via the triangle inequality] We combine the approximation $|x_k - \iota(q_k)| < 1/k$ from Step 1 with the convergence $\iota(q_k) \to x$ from Step 4. Fix $\varepsilon > 0$ in $\mathbb{R}$. Choose $K_1 \in \mathbb{N}$ with $1/K_1 < \varepsilon/2$. By Step 4, choose $K_2 \in \mathbb{N}$ with $|\iota(q_k) - x| < \varepsilon/2$ for all $k \ge K_2$. Set $K := \max\{K_1, K_2\}$. For $k \ge K$, the triangle inequality on $\mathbb{R}$ gives \begin{align*} |x_k - x| \le |x_k - \iota(q_k)| + |\iota(q_k) - x| < \frac{1}{k} + \frac{\varepsilon}{2} \le \frac{1}{K_1} + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Therefore $x_k \to x$ in $\mathbb{R}$, proving that the Cauchy sequence $(x_k)$ converges to a limit in $\mathbb{R}$. Since $(x_k)$ was arbitrary, $\mathbb{R}$ is complete. [guided] We have three real numbers in play: the original term $x_k$, the embedded rational $\iota(q_k)$, and the candidate limit $x = [(q_n)]$. Two estimates are already in hand: \begin{align*} |x_k - \iota(q_k)| &< \frac{1}{k} \qquad \text{(Step 1, by choice of } q_k\text{)}, \\ |\iota(q_k) - x| &\to 0 \text{ as } k \to \infty \qquad \text{(Step 4)}. \end{align*} The strategy is the standard "insert and split": to bound $|x_k - x|$ we insert $\iota(q_k)$ and apply the triangle inequality on $\mathbb{R}$, which holds because the absolute value on $\mathbb{R}$ inherited from the Cauchy completion is a norm: \begin{align*} |x_k - x| = |(x_k - \iota(q_k)) + (\iota(q_k) - x)| \le |x_k - \iota(q_k)| + |\iota(q_k) - x|. \end{align*} We now make each summand smaller than $\varepsilon/2$. Fix $\varepsilon > 0$ in $\mathbb{R}$. For the first summand, we use $|x_k - \iota(q_k)| < 1/k$ and choose $K_1 \in \mathbb{N}$ large enough that $1/K_1 < \varepsilon/2$ — such $K_1$ exists by the Archimedean property of $\mathbb{R}$. Then for every $k \ge K_1$ we have $1/k \le 1/K_1 < \varepsilon/2$, so \begin{align*} |x_k - \iota(q_k)| < \frac{1}{k} \le \frac{1}{K_1} < \frac{\varepsilon}{2}. \end{align*} For the second summand, Step 4 supplies $K_2 \in \mathbb{N}$ such that $|\iota(q_k) - x| < \varepsilon/2$ whenever $k \ge K_2$. Set $K := \max\{K_1, K_2\}$ so that both bounds hold simultaneously. For $k \ge K$: \begin{align*} |x_k - x| \le |x_k - \iota(q_k)| + |\iota(q_k) - x| < \frac{1}{K_1} + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, this is precisely the statement $x_k \to x$ in $\mathbb{R}$. The Cauchy sequence $(x_k)$ thus converges to the real number $x \in \mathbb{R}$, and because $(x_k)$ was an arbitrary Cauchy sequence we conclude that $\mathbb{R}$ is complete. [/guided] [/step]