1. **Problem Statement:** Prove that the interior of a ball in a metric space is always an open set. 2. **Definition of a Ball:** A ball centered at a point $x$ with radius $r > 0$ in a metric space $(X,d)$ is defined as $$ B(x,r) = \{y \in X : d(x,y) < r\} $$ 3. **Definition of an Open Set:** A set $U \subseteq X$ is open if for every point $u \in U$, there exists some radius $\epsilon > 0$ such that the ball $B(u,\epsilon) \subseteq U$. 4. **Goal:** Show that for every point $y \in B(x,r)$, there exists $\epsilon > 0$ such that $B(y,\epsilon) \subseteq B(x,r)$. 5. **Proof:** - Since $y \in B(x,r)$, by definition $d(x,y) < r$. - Let $\epsilon = r - d(x,y) > 0$. - For any $z \in B(y,\epsilon)$, we have $d(y,z) < \epsilon$. - By the triangle inequality, $$ d(x,z) \leq d(x,y) + d(y,z) < d(x,y) + \epsilon = d(x,y) + (r - d(x,y)) = r $$ - Therefore, $z \in B(x,r)$. 6. **Conclusion:** Since for every $y \in B(x,r)$ there exists $\epsilon > 0$ such that $B(y,\epsilon) \subseteq B(x,r)$, the ball $B(x,r)$ is open by definition. **Final answer:** The interior of a ball is always open.