Given two balls and a point show there radii $c,d$ such that $B_c(x) \subseteq B_r(a) \cap B_s(b) $ Show that given two balls $B_r(a)$ and $B_s(b)$, and a point $x \in B_r(a) \cap B_s(b)$, there are radii $c$ and $d$ such  that
$B_c(x) \subseteq B_r(a) \cap B_s(b) $ and $B_d(x) \supseteq B_r(a) \cup B_s(b)$
attempt: Given given two balls $B_r(a)$ and $B_s(b)$, and a point $x \in B_r(a) \cap B_s(b)$ then $x \in B_r(a)$ and $x \in B_s(b)$. Let $c< b$ and $c<a$. Then $B_c(x)$ with center at $x$ and radius $c$ will lie inside both balls, or $B_c(x) \subseteq B_r(a) \cap B_s(b) $.
And let $d > r$ and $d>s$. Then $B_d(x)$ will have some region outside $B_r(a)$ and also $B_s(b)$. So $B_d(x)$ will habe $B_r(a)$ and $B_s(b)$ also some other region extra as radius $d >a,b$ . Thus,  $B_d(x) \supseteq B_r(a) \cup B_s(b)$.
Can anyone please verify this? Any feedback/help would really help. Thank you
 A: Your approach is not quite correct. To see this, take the metric space $R^1$, $x=\frac{1}{2}$. Notice that $x\in B_1(0)\cap B_1(1)=(0,1)$, but $(-\frac{1}{4},\frac{5}{4})=B_{\frac{3}{4}}(x)\nsubseteq B_1(0)\cap B_1(1)=(0,1)$ (draw the intervals). For the other case, take the same $x$ and the same balls, and note that $B_{\frac{5}{4}}(x)\nsupseteq B_1(0)\cup B_1(1)$. Thus, we need a slight modification of the argument. The key is that $B_c(x)$ must be fully contained in the intersection, and $B_d(x)$ must fully contain the union. For the first assertion, note that $y \in B_r(a) \cap B_s(b)$ must satisfy $d(y,a)<r$ and $d(y,b)<s$. Thus, this must be true $\forall y \in B_c(x)$. Notice that in essence we need to make sure that the radius of the ball is less than the distance from $x$ to the closest boundary of a ball. To satisfy this, it suffices to have $c<\min(r-d(x,a), s-d(x,b))$. Then, by the triangle inequality, $\forall y \in B_c(x)$, we have $d(y,a)\leq d(y,x)+d(x,a)<r-d(x,a)+d(x,a)=r$, and $d(y,b)\leq d(y,x)+d(x,b)<s-d(x,b)+d(x,b)=s$, so $y \in B_r(a) \cap B_s(b)$. I'll leave the other assertion to you.
