# proof of alternative characterization of filter

First, suppose that $\mathbf{F}$ is a filter. We shall show that, for any two elements $A$ and $B$ of $\mathbf{F}$, it is the case that $A\cap B\in \mathbf{F}$ if and only if $A\in \mathbf{F}$ and $B\in \mathbf{F}$.

By the definition of filter, if $A\in \mathbf{F}$ and $B\in \mathbf{F}$ then $A\cap B\in \mathbf{F}$. Since $A\supseteq A\cap B$ and $\mathbf{F}$ is a filter, $A\cap B\in \mathbf{F}$ implies $A\in \mathbf{F}$. Likewise, $A\cap B\in \mathbf{F}$ implies $B\in \mathbf{F}$.

Next, we shall show that any proper subset^{} $\mathbf{F}$ of the power
set^{} of $X$ such that $A\cap B\in \mathbf{F}$ if and only if $A\in \mathbf{F}$ and $B\in \mathbf{F}$ is a filter.

If the empty set^{} were to belong to $\mathbf{F}$ then for any $A\subset X$, we would have $A\cap \mathrm{\varnothing}=\mathrm{\varnothing}\in \mathbf{F}$. This would imply that every subset of $X$ belongs to
$\mathbf{F}$, contrary to our hypothesis^{} that $\mathbf{F}$ is a proper
subset of the power set of $X$.

If $A\subseteq B\subseteq X$ and $A\in \mathbf{F}$, then $A\cap B=A\in \mathbf{F}$. By our hypothesis, $B\in \mathbf{F}$.

The third defining property of a filter — If $A\in \mathbf{F}$ and $B\in \mathbf{F}$ then $A\cap B\in \mathbf{F}$ — is part of our hypothesis.

Title | proof of alternative characterization of filter |
---|---|

Canonical name | ProofOfAlternativeCharacterizationOfFilter |

Date of creation | 2013-03-22 14:43:05 |

Last modified on | 2013-03-22 14:43:05 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 5 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 03E99 |

Classification | msc 54A99 |