Sachchidanand Prasad

30-01-2025

Problem: Let $(X,d)$ be a metric space. Given a point $x \in X$ and a real number $r > 0$, show that the set \[ U = \left\{ y \in X : d(x,y) > r \right\} \] is open in $X$.
topology metric-space
Solution: Let $y \in U$. We need to show that there exists $\epsilon > 0$ such that $B(y, \epsilon ) \subseteq U$. Look at the figure below and choose $\epsilon $ as shown in the picture, \[ \epsilon = d(x,y) - r. \]
Choice of epsilon
We claim that $B(y, \epsilon) \subseteq U $. Take $z \in B(y, \epsilon )$. Then, we have \begin{align*} d(y,z) < \epsilon & \implies d(y,z) < d(x,y) - r \\ & \implies d(y,z) + r < d(x,y) \\ & \kern 2.5cm \leq d(x,z) + d(y,z)\\ & \implies d(x,z) > r. \end{align*} Thus, $z \in U$.