# testing continuity via filters

###### Proposition 1.

Let $X\mathrm{,}Y$ be topological spaces^{}. Then a function $f\mathrm{:}X\mathrm{\to}Y$ is continuous^{} iff it sends converging filters to converging filters.

###### Proof.

Suppose first $f$ is continuous. Let $\mathbb{F}$ be a filter in $X$ converging to $x$. We want to show that $f(\mathbb{F}):=\{f(F)\mid F\in \mathbb{F}\}$ converges^{} to $f(x)$. Let $N$ be a neighborhood^{} of $f(x)$. So there is an open set $U$ such that $f(x)\in U\subseteq N$. So ${f}^{-1}(U)$ is open and contains $x$, which means that ${f}^{-1}(U)\in \mathbb{F}$ by assumption^{}. This means that $f{f}^{-1}(U)\in f(\mathbb{F})$. Since $f{f}^{-1}(U)\subseteq U\subseteq N$, we see that $N\in f(\mathbb{F})$ as well.

Conversely, suppose $f$ preserves converging filters. Let $V$ be an open set in $Y$ containing $f(x)$. We want to find an open set $U$ in $X$ containing $x$, such that $f(U)\subseteq V$. Let $\mathbb{F}$ be the neighborhood filter of $x$. So $\mathbb{F}\to x$. By assumption, $f(\mathbb{F})\to f(x)$. Since $V$ is an open neighborhood of $f(x)$, we have $V\in f(\mathbb{F})$, or $f(F)\subseteq V$ for some $F\in \mathbb{F}$. Since $F$ is a neighborhood of $x$, it contains an open neighborhood $U$ of $x$. Furthermore, $f(U)\subseteq f(F)\subseteq V$. Since $x$ is arbitrary, $f$ is continuous. ∎

Title | testing continuity via filters |
---|---|

Canonical name | TestingContinuityViaFilters |

Date of creation | 2013-03-22 19:09:31 |

Last modified on | 2013-03-22 19:09:31 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 4 |

Author | CWoo (3771) |

Entry type | Result |

Classification | msc 26A15 |

Classification | msc 54C05 |

Related topic | Filter |