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
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 \}$ ).
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?