# Proof that every cauchy sequence converges in $\mathbb R^k$

I'm having a hard time understanding this proof (the portion in bold).

1. I know $E_N$ is bounded but how is the finite set $\{x_1, \ldots, x_{n-1}\}\,$ bounded? (Is it because every finite set in $\mathbb R^k$ is bounded?)

2. I didn't get the last sentence at all ("Since every bounded …, (c) follows from (b)). Can you please explain this?

Theorem

(a) In any metric space $X$, every convergent sequence is a Cauchy sequence.

(b) If $X$ is a compact metric space and if $\{p_n\}$ is a Cauchy sequence in $X$, then $\{p_n\}$ converges to some point of $X$.

(c) In $\Bbb R^k$, every Cauchy sequence converges.

Proof

Let $\{\mathbf x_k\}$ be a Cauchy sequence in $\Bbb R^k$. Define $E_n$ as in $(b)$, with $\Bbb x_i$ in place of $p_i$. For some $N$, $\operatorname{diam}E_n<1$. The range of $\{\mathbf x_k\}$ is the union of $E_n$ and the finite set $\{\mathbf x_1, \dots, \mathbf x_{N-1}\}$. Hence $\{\mathbf x_k\}$ is bounded. Since every bounded subset of $\Bbb R^k$ has compact closure in $\Bbb R^k$ (Theorem 2.41), (c) follows from (b).

Theorem 2.41

If a set $E$ in $\Bbb R^k$ has one of the following three properties, then it has the other two:

(a) $E$ is closed and bounded.

(b) $E$ is compact.

(c) Every infinite subset of $E$ has a limit point in $E$.

• which book is this
– Manx
Oct 8, 2019 at 15:07
• Principal of Mathematical Analysis by Rudin Oct 30, 2019 at 14:53

2. Since $\{x_n\}$ is bounded, its closure is closed and bounded and hence compact from Theorem 2.41. In a compact set, every sequence has a convergent subsequence, so in particular $\{x_n\}$ has a convergent subsequence. However, since $\{x_n\}$ is Cauchy, convergence of a subsequence implies convergence of the full sequence.
• In fact they're both standard results. The first is Theorem 2.42 in Rudin and the second is seen directly from the definition of Cauchy sequences. Let $\{x_n\}$ be a Cauchy sequence with subsequence $\{x_{n_k}\}$ converging to $x.$ Let $\varepsilon>0$ and let $N>0$ be such that $n_1>n_2>N$ implies $|x_{n_1}-x_{n_2}|<\frac{1}{2}\varepsilon$ and also $n_k>N$ implies $|x_{n_k}-x|<\frac{1}{2}\varepsilon.$ Combining these by the triangle inequality, $n>N$ implies $|x_n-x|<\varepsilon.$ So $x_n$ converges to $x.$ Jul 22, 2013 at 14:57
• Take a sequence $\{x_n\}$ in a compact set in $\mathbb{R}^k.$ Since a compact set in $\mathbb{R}^k$ is in particular bounded, $\{x_n\}$ is a bounded infinite subset of $\mathbb{R}^k$, and so by the theorem has a limit point $x.$ For each $n\in\mathbb{N}$ let $y_n$ be an element of the sequence $\{x_n\}$ within $1/n$ of $x$, with $y_{n-1}$ and all elements of the sequence $\{x_n\}$ before it removed from consideration. Then $\{y_n\}$ is a subsequence of $\{x_n\}$ which converges to $x.$ Jul 23, 2013 at 0:24