You can explicitly define such a filter as F' = {a ∧ c : a ∈ F, b ≤ c}. This is clearly a filter over B, with F ⊂ F'. If ¬b ∈ F', we have some particular a, c satisfying the above conditions. Then ¬b ≤ c, so that ¬c ≤ b, making ¬c ≤ c, implying that c ∨ c = 1, which means that c = 1, making ¬b = a ∧ 1 = a ∈ F, a contradiction. Therefore F' is proper.
The explicit proof that F' is proper is a bit long-winded, but I think it's intuitively pretty obvious. Your mileage may vary, of course.
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timeasmymeasure
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Date: 2010-04-11 01:56 pm (UTC)The explicit proof that F' is proper is a bit long-winded, but I think it's intuitively pretty obvious. Your mileage may vary, of course.
*wishes for LaTeX capacity for LJ*