Let $(X,d)$ be a metric space such that, for every $x \in X$ and $r>0$, the closed ball $$\overline{B}(x,r)=\{y \in X:d(x,y)\leq r\}$$ is compact. Prove that $X$ is complete.

My attempt: Let $\{x_n \}$ be a Cauchy sequence in $X$. Then, given $\epsilon>0$ there exist a positive integer $N$ such that, $n \geq N, m \geq N$ implies $d(x_n,x_m)<\epsilon$. In particular, let $m=N$ and $d(x_n, x_N)<\epsilon$ for all $n\geq N$. Thus $\{x_n\}_{n=N}^{\infty} \subseteq \overline{B}(x_N,\epsilon)$. But, since $\overline{B}(x_N,\epsilon)$ is compact, it is also complete, so $\{x_n\}_{n=N}^{\infty}$ is Cauchy, then converges in $\overline{B}(x_N,\epsilon)$, and then converges in $X$.

Is this correct?

Now, prove that $\overline{B}(0,1)$ is not compact on $(S,d)$ with $S$ the set of all bounded sequence of real numbers and $d(s_n,t_n)=\sup \{|s_n - t_n|\}$

  • $\begingroup$ Yes, well done! $\endgroup$
    – sranthrop
    Mar 5 '15 at 22:05

Yes, the proof is correct. Now let's show that the closed ball on $S$ is not compact. Suppose it is. For each $n \in \mathbb N$, let $x_n$ be the sequence such that $x_n(n)=1$ and $x_n(m)=0$ if $n \neq m$, e.g., $x_0=\langle1, 0, 0, \cdots\rangle$, $x_2=\langle 0, 1, 0, cdots \rangle$, etc. For each $n$, $d(x_n, 0)=1$, so $x_n \in \bar B(0, 1)$. If $n \neq m$, then $d(x_n, x_m)=1$. Let $K=\{x_n: n \in \omega\}$.

$K$ is closed: If $x \notin K$, then $B(x, \frac{1}{2})\cap K$ has at most one element (suppose there are two and use the triangular inequality), so there exists $r>0$ with $B(x, r)\cap K = \emptyset$.

Since $K$ is a closed set contained in a compact space ($\bar B(0, 1)$), $K$ is compact.

For each $n$, let $U_n=B(x_n, \frac{1}{2})$. Then $\{U_n: n \in \omega\}$ is a open cover of $K$, but since $U_n \cap K=\{x_n\}$, there is no finite subcover.


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