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How prove that $[a,b]$ is complete in euclidean metric space? I know that $\mathbb{R}$ is complete space, so maybe I can find function $f: \mathbb{R} \rightarrow [a,b]$ which will be surjective function?

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Or you can think that $\Bbb R$ is complete and $[a,b]$ is a closed subset of a complete space – Haha Nov 9 '13 at 13:20
$\exp : \mathbb R \to (0, \infty)$ is surjective. Yet $(0, \infty)$ is not complete. You need more conditions if you want to go this route. Showing that $[a, b]$ is a closed subset of $\mathbb R$ is much easier. – Ayman Hourieh Nov 9 '13 at 13:23

You don’t need to find any function. If $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $[a,b]$, then it’s also a Cauchy sequence in $\Bbb R$, which is complete, so $\langle x_n:n\in\Bbb N\rangle$ converges to some $x\in\Bbb R$. Now use the fact that $[a,b]$ is a closed set in $\Bbb R$.

(In fact it’s a general theorem that if $\langle X,d\rangle$ is a complete metric, space, and $F$ is a closed subset of $X$, then $\langle F,d_F\rangle$ is complete, where $d_F=d\upharpoonright(F\times F)$.)

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Hint: $\sin\colon \mathbb R \to [-1,1]$ is continuous and onto.

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@Tomek Here is good enough,but in general even if two spaces are homeomorphic and the first one is complete it's doesn't mean that the second is ,for example $(\Bbb R,|.|),(\Bbb R,d)$ with $d=|arctanx-arctany|$. The first one is complte but the second is not – Haha Nov 9 '13 at 13:23

The segment $[a,b]$ in $\mathbb{R}^n$ is the image of the continuous function \begin{align} f\colon [0,1] &\to \mathbb{R}^n\\ t&\mapsto a+t(b-a) \end{align} (where $[0,1]\subseteq\mathbb{R}$). Since $[0,1]$ is compact, also its image is, so it's complete, for any compact subset in a metric space is complete.

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