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A sequence of vectors $\{a_i\}$ in $\mathbb{R}^n$ is a Cauchy sequence if for each $\epsilon> 0$ there exists $I \in \mathbb{N}$ such that $\|a_i -a_j\| < \epsilon$ for all $i, j \geq I$. Prove that Cauchy sequences are convergent.

I was told I have to show that if the vectors are Cauchy that their elements must also be Cauchy, but I'm not sure how to go about doing this. The rest of the proof should fall into place accordingly.

Could someone show me how this particular part is done?

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Hint:

  • Show that every Cauchy sequence is bounded;
  • Then there will be a convergent subsequence (Bolzano-Weierstrass), call it $(x_r)_{r \in \mathbb N'}$. Let $\displaystyle a = \lim_{r \in \mathbb N'} x_r$, then $$ \lim_{r \in \mathbb N'} |x_r - a| = 0 \,\,\,\text{and}\,\,\, \lim_{k \in \mathbb N, r \in \mathbb N'} |x_k - x_r| = 0$$

Now from $|x_k - a| \leq |x_k - x_r| + |x_r- a|$, you may get your result.

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Let $a_i=(a_i^1,a_i^2,\ldots, a_i^n)$. Then for each $k\in[n]$ one has $$|a_i^k-a_j^k|\leq\|a_i-a_j\|\ ,$$ hence $\bigl( a_i^k)_{i\geq1}$ is a Cauchy sequence in ${\mathbb R}$ and converges to some $\alpha^k\in{\mathbb R}$. Put $(\alpha^1,\ldots,\alpha^n)=:\alpha\in{\mathbb R}^n$. The triangle inequality in ${\mathbb R}^n$ then implies $$\|a_i-\alpha\|\leq\sum_{k=1}^n|a_i^k-\alpha^k|\to0\qquad(i\to\infty)\ ,$$ and this guarantees $\lim_{i\to\infty}a_i=\alpha$.

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