Theorem. Let \(f: X \to Y\). Then \(f\) is continuous if and only if for every convergent net \((x_\alpha)\) in \(X\), converging to \(x\), say, the net \((f(x_\alpha))\) converges to \(f(x)\).
Solution
If \(f\) is continuous and \(x_\alpha\to x\), then for every neighborhood \(V\) of \(f(x)\), \(f^{-1}(V)\) is a neighborhood of \(x\), so eventually \(x_\alpha\in f^{-1}(V)\), hence eventually \(f(x_\alpha)\in V\). Thus \(f(x_\alpha)\to f(x)\).
Conversely, suppose \(f\) preserves limits of all convergent nets. If \(f\) were not continuous, there would be a closed set \(C\subset Y\) such that \(f^{-1}(C)\) is not closed in \(X\). Choose \[ x\in\overline{f^{-1}(C)}\setminus f^{-1}(C). \] By the closure-net theorem, there is a net \(x_\alpha\in f^{-1}(C)\) converging to \(x\). By hypothesis, \(f(x_\alpha)\to f(x)\). Since all \(f(x_\alpha)\in C\) and \(C\) is closed, the limit \(f(x)\in C\), contradicting \(x\notin f^{-1}(C)\). Hence \(f\) is continuous.