# 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 $x$.
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Title | continuity and convergent nets |
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Canonical name | ContinuityAndConvergentNets |

Date of creation | 2013-03-22 18:37:53 |

Last modified on | 2013-03-22 18:37:53 |

Owner | azdbacks4234 (14155) |

Last modified by | azdbacks4234 (14155) |

Numerical id | 5 |

Author | azdbacks4234 (14155) |

Entry type | Theorem |

Classification | msc 54A20 |

Related topic | Net |

Related topic | Continuous |