If $f$ is continuous, $x_\alpha\to x$ and $V$ is a neighborhood of $f(x)$ , then $U=f^{-1}(V)$ is an open neighborhood of $x$ and for some $\beta$ , $\beta\preceq\alpha$ implies $x_\alpha\in U$ and $f(x_\alpha)\in V$ .The other direction. Let $V$ be open, $U=f^{-1}(V)$ , and $x\in U$ . We need to show that there is an open neighborhood of $x$ contained in $U$ . Suppose there is no such neighborhood. Then $x\in\overline{X-U}$ . Using the previous exercise, there is a net $(x_\alpha)$ of points of $X-U$ converging to $x$ . But in this case $V$ contains no points of $(f(x_\alpha))$ which contradicts the assumption that the image of any net converging to $x$ converges to $f(x)$ .