Clarification on Rudin's proof for theorem 3.11 c.): "In $\mathbb{R}^k$, every Cauchy sequence converges."

Question

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).

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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?

Answer 1

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.

Answer 2

For a sequence $(x_n)$ in a metric space, the closure of the set $E = \{x_n:n\in \mathbb N\}$ is itself, along with all the accumulation points sequence $(x_n)$.

If $(x_n)$ is Cauchy, it has at most one accumulation point, which is the limit of $(x_n)$ if it exists. Since $E$ is bounded, and therefore has compact closure (in $\mathbb R^n$), the sequence $(x_n)$ has a limit point in $\overline E$, by the Bolzano-Weierstrass theorem, and therefore converges.