continuity and convergent nets

Theorem.

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

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 $\alpha \ge {\alpha }_0$. It follows that $f(x_{\alpha })\in V$ for $\alpha \ge {\alpha }_0$, hence that $f(x_{\alpha })\to f(x)$. 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 $\alpha \ge {\alpha }_0\in A$ such that $f(x_{\alpha })\notin V$, hence such that $x_{\alpha }\notin f^{-1}(V)$; as $x_{\alpha }\to x$ by hypothesis, this implies that $f^{-1}(V)$ cannot be a neighborhood of $x$, and thus that $f$ fails to be continuous at $x$.∎