# Question on order ideals

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). –  Asaf Karagila Jan 17 '12 at 9:08

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.
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. –  kjo Jan 17 '12 at 16:49