Let $(E,d)$ a metric space. Supose that exists $r>0$ such that $\overline{B(x,r)}$ is compact for each $x\in E$. Let $A\subset E$ compact. Prove that the set $$ \lbrace x\in E: d(x,A)\leq s\rbrace$$
is compact for each $s<r$.
Anyone have an idea? I proved that $(E,d)$ is complete, but I don't know if that is really necessary. If anyone can help me I would really appreciate it.
The easiest way may be to use the equivalence of compactness and sequential compactness for metric spaces. So you need to show that for $0 < s < r$, the set
$$A(s) = \{ x\in E : d(x,A) \leqslant s\}$$
is sequentially compact. Hence, let $(x_n)_{n\in\mathbb{N}}$ a sequence in $A(s)$. For all $n$, choose an $y_n \in A$ with $d(x_n,y_n) < \frac{r+s}{2}$. Since by assumption $A$ is compact, $(y_n)_{n\in\mathbb{N}}$ has a convergent subsequence. Without loss of generality, suppose that $y_n \to y_\ast \in A$. Choose an $n_0 \in \mathbb{N}$ so that $d(y_n,y_\ast) < \frac{r-s}{2}$ for all $n \geqslant n_0$. Then we have $d(x_n,y_\ast) \leqslant d(x_n,y_n) + d(y_n,y_\ast) < \frac{r+s}{2} + \frac{r-s}{2} = r$ for $n\geqslant n_0$. Since $\overline{B(y_\ast,r)}$ is compact by assumption, we can conclude that $(x_n)_{n\in\mathbb{N}}$ has a convergent subsequence. Since $A(s)$ is closed, its limit lies in $A(s)$, hence we saw that $A(s)$ is sequentially compact.
