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Prove that $\mathbb{R}^m$ with Euclidean metric is complete.


Every Cauchy sequence in $\mathbb{R}$ is bounded and has monotonic subsequence thus has convergent subsequence. Since Cauchy sequence which has convergent subsequence is convergent itself we get that $\mathbb{R}$ is complete.

Now let $(a_n^1,\ldots, a_n^m)$ be a Cauchy sequence in $\mathbb{R}^m$. Obviously this implies that each $(a_n^i)$, $i=1,\ldots,m$ is a Cauchy sequence. Since $\mathbb{R}$ is complete we know that each $(a_n^i)$ converges to $a^i\in\mathbb{R}$. So for every $\epsilon>0$ there is $N^i\in\mathbb{N}$ such that $|a_n^i-a^i|<\frac{\epsilon}{\sqrt{m}}$. Let $N=\max_{i=1,\ldots, m} {N^i}$. Then if $n>N$ we have

$$\sqrt{(a_n^1-a^1)^2+\ldots+(a_n^m-a^m)^2}<\epsilon$$

Is this correct? Thanks.

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    $\begingroup$ The proof is basically correct, it remains to show why the last inequality holds. $\endgroup$ – Crostul Jan 7 '16 at 20:17
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In general $\mathbb{C}^n$ and $\mathbb{R}^n$ are complete metric spaces. Since $\mathbb{C}^n=\mathbb{R}^{2n}$ it's sufficient to prove the result for $\mathbb{R}^n$. Let $\lbrace x_n \rbrace \subset \mathbb{R}^n$ Cauchy sequence, then $\exists N \in \mathbb{N}$ : $|x_n - x_N| < \epsilon=1$ $\forall n \geq N$ where $|\cdot |$ is the Euclidean metric , place $R:=\max \lbrace |x_1|, |x_2|,...,|x_N| \rbrace$, with an application of the triangle inequality we have that $\lbrace x_n \rbrace \subset K:=\lbrace x \in \mathbb{R}^n : |x| \leq R+1 \rbrace$, where $K$ is a compact subspace of $\mathbb{R}^n$. Now you just know that a topological space $(X, \mathcal{T})$ that admits countable bases, then it is compact if and only if it is sequentially compact. Obviously $K$ satisfies these properties, therefore $\exists x_{n_k} \rightarrow x$, and also $x_n \rightarrow x$.

If you know that $\mathbb{R}$ is complete, also your proof is correct.

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