Completeness of R^k

Theorem (Completeness of $\mathbb{R}^k$). For every integer $k\ge 1$, the Euclidean space $\mathbb{R}^k$ equipped with its usual distance is a complete metric space .

Context. This is the finite-dimensional completeness statement used throughout analysis, optimization, and convex geometry.

Proof sketch. Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}^k$, with $x_n=(x_n^{(1)},\dots,x_n^{(k)})$. The Cauchy property implies each coordinate sequence $(x_n^{(j)})$ is Cauchy in $\mathbb{R}$. Since $\mathbb{R}$ is complete, each coordinate converges to some $a^{(j)}\in\mathbb{R}$. Let $a=(a^{(1)},\dots,a^{(k)})$. One checks that $x_n\to a$ by estimating the distance between $x_n$ and $a$ in terms of coordinate differences, hence $(x_n)$ converges in $\mathbb{R}^k$.

Example. The sequence $x_n=(1/n,1/n^2,\dots,1/n^k)$ is Cauchy and converges to $0\in\mathbb{R}^k$.