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On p. 80 of his General topology, John Kelley gives the following theorem:

Let $X$ be a distributive lattice, and let $A \subseteq X$ be an ideal and $B\subseteq X$ a filter such that $A\cap B = \varnothing$. Then there exist ideal A'\supseteq A and filter B'\supseteq B such that A' \cap B' = \varnothing and A' \cup B' = X.

[NB: Both in the theorem statement and below I've departed somewhat from the wording and terminology of Kelley's original. In particular, Kelley uses the term dual ideal instead of filter.]

In the proof of this theorem, Kelley considers the family $\cal{A}$ of all ideals in $X$ that contain $A$ and are disjoint from $B$, and proposes a maximal member of $\cal{A}$ (ordered by set inclusion) as a candidate for the ideal A' claimed by the theorem. (Kelley bases the existence of such maximal ideal A' on the Hausdorff Maximal Principle, p. 32, or equivalently, the Axiom of Choice.)

Then, in the next step of the proof, Kelley asserts that the smallest ideal that contains A' and some arbitrary element $c \in X$ corresponds to the set

P = \{x:x\leq c \;\;\; \mathrm{or} \;\;\; x\leq c \vee y \;\; \mathrm{for} \; \mathrm{some} \; y \in A'\}

True or not, this assertion confuses me because

  1. If A' \neq \varnothing, I don't see how $P$ could contain any element that is not already contained in the set Q = \{x:x\leq c \vee y \;\; \mathrm{for} \; \mathrm{some} \; y \in A'\} (since $\forall u, v\in X\;[\;u \leq u \vee v\;]$, it follows from the transitivity of $\leq$ that \forall y \in A', $ \{ x:x\leq c\} \subseteq \{ x : x \leq c \vee y \}$ ).

  2. The theorem is trivially true when $A = \varnothing$ and $B = X$ (both necessary for $A' = \varnothing$), which makes me doubt the idea that covering this trivial case is the sole reason for including the otherwise obfuscating "$x \leq c$" clause.

Is the "$x \leq c$" clause less superfluous than it looks?

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    It should be mentioned that this claim is *equivalent* to the axiom of choice (where as the usual restriction from a lattice to a Boolean algebra is notably less).2012-01-17

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So far as I can see, the $x\le c$ clause is there precisely to cover the case $A=\varnothing,B=X$. Yes, that case is trivial, but it has to be dealt with, and with this definition of $P$ one avoids having to treat it as a separate case.

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    Thanks. This habit, so widespread in mathematical writing, of piling on the cute technical/notational tricks (thereby obscuring the basic ideas) all for the sake of a pathetically small gain in "generality" is one that I find most misguided. As if "having to treat it as a separate case", at least in situations like this one, meant a massive expenditure of effort on either the writer's or the reader's part. The adult way to handle a case like this goes along the lines of "the assertion is trivially true for the case $A = \varnothing, B = X$, therefore assume that $A \neq \varnothing$," etc.2012-01-17