Theorem   Let $X$ and $Y$ be topological spaces. A function $f:X\rightarrow Y$ is continuous at a point $x\in X$ if and only if for each net $(x_\alpha)_{\alpha\in A}$ in $X$ converging to $x$ , the net $(f(x_\alpha))_{\alpha\in A}$ converges to $f(x)$ .

Proof. If $f$ is continuous, $(x_\alpha)_{\alpha\in A}$ converges to $x$ , and $V$ is an open neighborhood of $f(x)$ in $Y$ , then $f^{-1}(V)$ is an open neighborhood of $x$ in $X$ , so there exists $\alpha_0\in A$ such that $x_\alpha\in f^{-1}(V)$ for $\alpha\geq\alpha_0$ . It follows that $f(x_\alpha)\in V$ for $\alpha\geq\alpha_0$ , hence that $f(x_\alpha)\rightarrow f(x)$ . Conversely, suppose there exists a net $(x_\alpha)_{\alpha\in A}$ in $X$ converging to $x$ such that $(f(x_\alpha))_{\alpha\in A}$ does not converge to $f(x)$ , so that, for some open subset $V$ of $Y$ containing $f(x)$ and every $\alpha_0\in A$ , there exists $\alpha\geq\alpha_0\in A$ such that $f(x_{\alpha})\notin V$ , hence such that $x_\alpha\notin f^{-1}(V)$ ; as $x_\alpha\rightarrow x$ by hypothesis, this implies that $f^{-1}(V)$ cannot be a neighborhood of $x$ , and thus that $f$ fails to be continuous at $x$ .