[b]Proposition 2.12. [/b]A complex function [math]f:S\longrightarrow\mathbb{C}[/math] is continuous if and only if, for every[br]open set [math]U[/math] in [math]\mathbb{C}[/math], the set [math]f^{-1}(U)[/math] is open in [math]S[/math].[br][i]Proof[/i]. Suppose that [math]f[/math] is continuous and [math]U[/math] is open. Let [math]z_0\in f^{-1}\left(U\right)[/math]). Then [math]f(z_0)\in U[/math] so[br]there exists [math]\epsilon>0[/math] such that [math]N_{\epsilon}\left(f\left(z_0\right)\right)\subseteq U[/math]. By continuity of [math]f[/math] there exists [math]\delta>0[/math] such[br]that[br][center][math]f\left(N_{\delta}\left(z_0\right)\cap S\right)\subseteq N_{\epsilon}\left(f\left(z_0\right)\right)\subseteq U\mathbb{C}[/math][/center]Hence[center][math]N_{\delta}\left(z_0\right)\cap S\subseteq f^{-1}\left(U\right)[/math][br][/center]and [math]f^{-1}(U)[/math] is open.[br]Conversely, suppose that [math]f^{-1}\left(U\right)[/math] is open in [math]S[/math] for every open set [math]U[/math]. Given [math]z_0\in S[/math] and[br][math]\epsilon>0[/math], the set [math]N_{\epsilon}(f(z_0))[/math] is open, so [math]f^{-1}(N_{\epsilon}(f(z_0)))[/math] is open in [math]S[/math] and there exists [math]\delta>0[/math][br]such that[center][math]N_{\delta}\left(z_0\right)\cap S\subseteq f^{-1}\left(N_{\epsilon}\left(f\left(z_0\right)\right)\right)[/math][/center][br]Hence[center][math]f\left(N_{\delta}\left(z_0\right)\cap S\right)\subseteq N_{\epsilon}\left(f\left(z_0\right)\right)[/math][/center]so [math]f[/math] is continuous. [b]EOP[/b].