If \((X,d)\) is a metric space then \(\{x\in X:d(x,x_0)<\epsilon\}\) is said to be the open ball of radius \(\epsilon\). Prove that an open ball is an open set.
Solution
Let \[ B(x_0,\epsilon)=\{x\in X:d(x,x_0)<\epsilon\}, \qquad \epsilon>0. \]
1. Every point \(y\in B(x_0,\epsilon)\) is contained in an open ball centered at \(y\) that lies inside \(B(x_0,\epsilon)\).
Fix \[ y\in B(x_0,\epsilon). \] Then \[ d(y,x_0)<\epsilon, \] so \[ r=\epsilon-d(y,x_0)>0. \] Let \(z\in B(y,r)\). By the triangle inequality, \[ d(z,x_0) \le d(z,y)+d(y,x_0) <r+d(y,x_0) =\epsilon. \] Hence \[ z\in B(x_0,\epsilon). \] Thus \[ B(y,r)\subseteq B(x_0,\epsilon). \]
2. The ball \(B(x_0,\epsilon)\) is open.
By 1, for every point \(y\in B(x_0,\epsilon)\) there is a radius \(r>0\) such that \[ y\in B(y,r)\subseteq B(x_0,\epsilon). \] This is exactly the metric-topology criterion for \(B(x_0,\epsilon)\) to be open.