Theorem. In $\mathbb{R}^k$, every Cauchy sequence converges.
Proof: Let $\{x_n\}$ be a Cauchy sequence in $\mathbb{R}^k$. Define $E_N = \{x_k\mid k \geq N\}$. For some $N \in \mathbb{N}$, $\mathrm{diam}E_n < 1$. The range of $\{x_n\}$ is the union of $E_{N}$ and the finite set $\{x_1,\dots,x_{N-1}\}$. Hence $\{x_n\}$ is bounded. Since every bounded subset of $\mathbb{R}^k$ has compact closure in $\mathbb{R}^k$ (Theorem 2.41), (c) follows from (b).
While I think that I understand why the sequence is bounded (pick an $r = 1 + \max\{d(x_i, x_j)\mid 1\leq i < j \leq N\}$), I don't don't know how we can argue that the (range of the) sequence is closed. Specifically, the theorem 2.41 says that a subset of $\mathbb{R}^k$ which is closed and bounded is also compact. Is closedness somehow evident for sequences/Cauchy sequences?
Best Answer
Rudin says "every bounded set has compact closure." He is applying Theorem 2.41 to the closure of the set of points in the sequence, which is automatically closed; he is implicitly using that the closure of a bounded set is bounded.