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I am working on a question and I am looking for some clarification. I can't seem to use what I know to complete the proof.

Let $\mathbf{x}_{0} \in \mathbb{R}^n$ and $R>0$. Prove that $U=\left \{ \mathbf{x} \in \mathbb{R}^n : \left \| \mathbf{x} - \mathbf{x_{0}} \right \| \leq R \right \}$ is complete.

What I know:

  1. $U$ is complete if every Cauchy sequence of points in $U$ converges to a point in $U$.

  2. If $\mathbf{x}_k$ is a Cauchy sequence then there is an integer $N$ such that $$||\mathbf{x}_k - \mathbf{x}_l|| < \epsilon$$ for all $k,l \geq N$

  3. $U$ is the set of points $\mathbf{x} \in \mathbb{R}^n$ such that the distance between $\mathbf{x}$ and $\mathbf{x}_0$ is less than or equal to $R$.

A Cauchy sequence in $U$ will be made up of points $\mathbf{x}_{k,1},\mathbf{x}_{k,1},...,\mathbf{x}_{k,n}$ such that $|| \mathbf{x}_{k,1} - \mathbf{x}_{0}||$, $|| \mathbf{x}_{k,2} - \mathbf{x}_{0}||,..., || \mathbf{x}_{k,n} - \mathbf{x}_{0}||$ $\leq R$. It seems that if the sequence is made up of points in $U$ then it must converge to a point in $U$. Any help and clarification would be appreciated.

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If you already know that $\mathbb{R}^n$ is complete, as one should, then just use that your $U$ is closed, and a closed subset of a complete space is complete: if $(x_n)$ is Cauchy in $U$, then it is Cauchy in $\mathbb{R}^n$ (being Cauchy only depends on the distances between the points, not the set in which they lie) and thus converges in $\mathbb{R}^n$ to some $x$. But as $U$ is closed and all $x_n$ are in $U$, so is $U$, so $x_n \rightarrow x$ in $U$ as well.

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Use the facts that a closed subset of a complete space is complete, $\mathbb R^n$ is complete, and the closed ball is closed.

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Can you use the fact that every compact metric space is complete? It makes the question very simple, since compact (in $\mathbb{R}^n$) means closed+bounded.

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I don't know how extensive your knowledge on compactness is, but assuming that you know some basic results, you may consider the following methodology. It is clear that $U$ is a bounded set. If you know how to prove that it is a closed set, too, then you can appeal to the Heine-Borel theorem to conclude that $U$ is a compact set. Then, using the fact that compact spaces are complete, you conclude the proof.

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