Prove that for functions \(f: \mathbb{R} \to \mathbb{R}\), the \(\epsilon\)-\(\delta\) definition of continuity implies the open set definition.
Solution
Assume the epsilon-delta definition. Let \(V\subseteq\mathbb R\) be open. We show \(f^{-1}(V)\) is open.
Take \(x\in f^{-1}(V)\). Since \(f(x)\in V\) and \(V\) is open, choose \(\varepsilon>0\) with \[ (f(x)-\varepsilon,f(x)+\varepsilon)\subseteq V. \] By epsilon-delta continuity at \(x\), there is \(\delta>0\) such that \[ |t-x|<\delta\implies |f(t)-f(x)|<\varepsilon. \] Thus \[ (x-\delta,x+\delta)\subseteq f^{-1}(V). \] Every point of \(f^{-1}(V)\) therefore has an open interval contained in \(f^{-1}(V)\), so \(f^{-1}(V)\) is open. Hence \(f\) is continuous in the open-set sense.