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?