proof of alternative characterization of filter

First, suppose that 𝐅 is a filter. We shall show that, for any two elements A and B of 𝐅, it is the case that AB𝐅 if and only if A𝐅 and B𝐅.

By the definition of filter, if A𝐅 and B𝐅 then AB𝐅. Since AAB and 𝐅 is a filter, AB𝐅 implies A𝐅. Likewise, AB𝐅 implies B𝐅.

Next, we shall show that any proper subsetMathworldPlanetmathPlanetmath 𝐅 of the power setMathworldPlanetmath of X such that AB𝐅 if and only if A𝐅 and B𝐅 is a filter.

If the empty setMathworldPlanetmath were to belong to 𝐅 then for any AX, we would have A=𝐅. This would imply that every subset of X belongs to 𝐅, contrary to our hypothesisMathworldPlanetmathPlanetmath that 𝐅 is a proper subset of the power set of X.

If ABX and A𝐅, then AB=A𝐅. By our hypothesis, B𝐅.

The third defining property of a filter — If A𝐅 and B𝐅 then AB𝐅 — 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