[Math] Proof that every cauchy sequence converges in $\mathbb R^k$

real-analysis

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 $E$.

Best Answer

  1. Yes, every finite set is bounded, the same way any finite set of numbers has a maximum.

  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.

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