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Let $X$ be a metric space. Suppose there exists $r >0$ such that $\overline{B(x,r)}$ is compact for every $x \in X$. Show that $X$ is complete.

Let $x_n$ be a Cauchy sequence in $X$. We want to show that $x_n \to a \in X$.

Since $x_n$ is Cauchy for all $\varepsilon >0$ there exists $k \in \Bbb N$ such that $n,m>k \implies d(x_n,x_m)< \varepsilon$.

Thus for $r$ there exists $n_0$ such that $n,m > n_0 \implies d(x_n,x_m)< r$.

I also have that for all $n$ there exists $r>0$ such that $\overline{B(x_n,r)}$ is compcat, but I cannot draw the conclusion from these facts that $x_n$ would converge?

Walker
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2 Answers2

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Let $\{x_n\}\subset X$ be a Cauchy-sequence. Then, there exists an $n_0\in\mathbb N$, such that $$ m,n\ge n_0\qquad\Longrightarrow\qquad d(x_m,x_n)<r. $$ In particular, $$ x_n\in B_r(x_{n_0}), \quad \text{for all $n\ge n_0$.} $$ Set $S=\{x_n: n\ge n_0\}$. Then $S\subset B_r(x_{n_0})$, and hence $\overline S\subset \overline{B_r}(x_{n_0})$, and since $\overline S$ is a closed subset of the compact $\overline{B_r}(x_{n_0})$, then $\overline S$ is also compact. Thus the sequence $\{x_n\}_{n\ge n_0}\subset \overline S$ possesses a convergent subsequence $x_{k_n}\to x\in\overline S$.

But if a Cauchy sequence possesses a convergent subsequence, it is convergent itself.

Yiorgos S. Smyrlis
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Let $(x_n) \subset X$ be a cauchy sequence.

Then $(x_n) $ is bounded (see here)i.e $\exists x_0\in X, r>0$ such that $(x_n) \subset\overline{B(x_0,r)}$.

Now by compactness of $\overline{B(x_0,r)}$ , $\exists (x_{n_k}) \subset (x_n) $ such that $(x_{n_k}) \to x_1 \in \overline{B(x_0,r)}$.

If a cauchy sequence have a convergent subsequence then the sequence converge (see here).

Done!

SoG
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