How to prove a Cauchy sequence converges in ℝ?

This is the cornerstone theorem of real analysis. Here is a complete proof. **Theorem:** In $\mathbb{R}$, every Cauchy sequence converges. **Proof:** Let $(a_n)$ be a Cauchy sequence. **Step 1: Boundedness.** Take $\varepsilon = 1$. There exists $N$ such that $|a_n - a_m| < 1$ for all $m,n \geq N$. In particular, $|a_n - a_N| < 1$ for all $n \geq N$, so $|a_n| \leq |a_N| + 1$ for $n \geq N$. Hence $(a_n)$ is bounded. **Step 2: Construct subsequential limit via supremum.** Define $b_k = \sup\{a_n : n \geq k\}$. Since $(a_n)$ is bounded, each $b_k$ is finite. The sequence $(b_k)$ is decreasing and bounded below, so it converges. Let $L = \lim b_k$. This is the limit superior: $L = \limsup a_n$. **Step 3: Show $a_n \to L$.** Given $\varepsilon > 0$, since $(a_n)$ is Cauchy, choose $N$ such that $m,n \geq N \implies |a_m - a_n| < \varepsilon$. Since $b_N = \sup\{a_n : n \geq N\}$, there exists $n_1 \geq N$ such that $a_{n_1} > b_N - \varepsilon$. Also $a_{n_1} \leq b_N$. For any $n \geq N$: $$|a_n - L| \leq |a_n - a_{n_1}| + |a_{n_1} - b_N| + |b_N - L|$$ The first term is $< \varepsilon$ by Cauchy property. The second term is $< \varepsilon$ by choice of $n_1$. The third term is $< \varepsilon$ for sufficiently large $N$ since $b_k \to L$. Thus $|a_n - L| < 3\varepsilon$, proving convergence. **Why does this fail in $\mathbb{Q}$?** The Cauchy property only depends on distances, and $\mathbb{Q}$ embeds in $\mathbb{R}$ with the same metric. A Cauchy sequence in $\mathbb{Q}$ whose limit is $\sqrt{2}$ (e.g., decimal approximations) does not converge in $\mathbb{Q}$ because $\sqrt{2} \ otin \mathbb{Q}$. The limit exists in $\mathbb{R}$ but not in $\mathbb{Q}$, showing that $\mathbb{Q}$ is not complete. Completeness is the defining property that distinguishes $\mathbb{R}$ from $\mathbb{Q}$.