Let $X$ be a set (finite or infinite) and consider the Boolean algebra $P(X)$ and $A\subseteq X$. Denote the set of ultrafilters containing $A$ by $u(A)$.
Is the following true?
$$ u(A)=\bigcup_{a\in A} u(\{a\}) $$
That $\bigcup_{a\in A} u(\{a\})\subseteq u(A)$ is clear to me since if $\mathcal{F}\in u(\{a\})$, then since $\{a\}\subseteq A$, then $A\in\mathcal{F}$.
I attempted to show that equality does not hold by showing that there exists an ultrafilter without any $\{a\}$ for $a\in A$, but contains $A$. If we consider the set $E=\{A\}\cup\{\overline{\{a\}}\}_{a\in A}$, then this appears to have the FIP if $A$ is infinite so $E$ is contained in some ultrafilter and equality does not hold (if $A$ is infinite).
If $A$ is cofinite, then there is an ultrafilter $U$ of $P(X)$ such that $A \in U$ but no finite subset of $X$ is in $U$.
Just take $U$ to be an ultrafilter containing all cofinite subsets of $X$.
It follows that $U$ contains no finite subset of $X$ (because it contains their complements and an ultrafilter is a proper filter), whence $\{a\} \notin U$, for every $a \in A$, and therefore $U \notin u(\{a\})$.
Now $A \in U$, and so $U \in u(A)$, and so $u(A) \nsubseteq \bigcup_{a\in A}u(\{a\})$.