Suppose $(X,d)$ is a metric space where $A\subseteq X$ is totally covered i.e. $\forall \epsilon >0$ $\exists \{a_1,a_2,\ldots,a_m\}\in A $ s.t $A\subseteq \bigcup_{1,\ldots,m}B(a_i,\epsilon)$. Given that $B\subseteq A$ show that $B$ is totally covered.
I can't seem to show – with the given definition – that the centers of the balls would now come from $B.$
Best Answer
So $B\subseteq\bigcup_{i=1}^{m}B_{\epsilon/2}(a_{i})$ by choosing $\epsilon/2$-cover for $A$, and assume without loss of generality that $B_{\epsilon/2}(a_{i})\cap B\ne\emptyset$ for all $i=1,...,m$. Pick a $b_{i}\in B_{\epsilon/2}(a_{i})\cap B$. We claim that $\{B_{\epsilon}(b_{i})\}_{i=1}^{m}$ cover $B$:
For $b\in B$, pick an $i$ such that $b\in B_{\epsilon/2}(a_{i})$. Then $d(b,b_{i})\leq d(b,a_{i})+d(a_{i},b_{i})<\epsilon$, so $b\in B_{\epsilon}(b_{i})$, as expected.