It seems fairly trivial for filter F defined on a set S - If A and B are in F, then so is A ∧ B. Therefore if A and its complement are in F then the empty set is in F. And the empty set is a subset of all sets, which means all sets are in F, making F an improper filter.
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timeasmymeasure
no subject
Date: 2010-06-01 05:57 pm (UTC)