# continuity and convergent nets

###### 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 $x$. ∎

Title continuity and convergent nets ContinuityAndConvergentNets 2013-03-22 18:37:53 2013-03-22 18:37:53 azdbacks4234 (14155) azdbacks4234 (14155) 5 azdbacks4234 (14155) Theorem msc 54A20 Net Continuous