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Let $X \subseteq \mathbb{R}^d$. We want to show that $X$ is closed iff every Cauchy sequence in $X$ converges in $X$.

MY attempt:

Let $X$ be a closed set, and pick a convergent sequence $(x_n)$ in $X$. Since $X$ is closed, then we know that $x_n \to x \in X$. Since convergence implies Cauchy, then we have found that every Cauchy sequence in $X$ is convergent, as desired.

Conversely, suppose every Cauchy sequence converges in $X$, we want to show that $X$ is closed. To this end, pick a sequence $(x_n) \subseteq X$ such that $x_n \to L$. We want to show that $L \in X$. Since $(x_n)$ converges, then it must be Cauchy. And by hypothesis, then we must havethat $L \in X$.

Is this correct? thanks for any feedback in advanced.

ILoveMath
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    To prove that from closedness you have completness you don't pick convergent sequence, but you need to pick a Cauchy sequence and prove that converges. – Alem Feb 09 '14 at 21:06

1 Answers1

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Assuming you know the fact that $\mathbb R^d$ is complete, let $X\subset\mathbb R^d$ closed. and $\{x_n\}_{n\in\mathbb N}\subset X$ be a Cauchy sequence. Then it is convergent in $\mathbb R^d$, i.e., $x_n\to x\in \mathbb R^d$. But as $X$ is closed and $\{x_n\}_{n\in\mathbb N}$, then its limit belongs to $X$. Thus $X$ is complete.

Conversely, let $X\subset\mathbb R^d$ complete and $\{x_n\}_{n\in\mathbb N}\subset X$ a convergent sequence, in $\mathbb R^d$, i.e., $x_n\to x\in \mathbb R^d$. Then $\{x_n\}_{n\in\mathbb N}$ is a Cauchy sequence, and as $X$ is complete, then it converges in $X$, and since the limit is unique, the $x\in X$. Thus $X$ is closed.

Yiorgos S. Smyrlis
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