Example of infinite partition of a set $A$, with no elements in ultrafilter on $A$ Let $\mathcal{F}$ be an ultrafilter on a set $A$. It is easy to show that for every finite partition $\{X_n : n<m\}$ of $A$, there exists some $n<m$ such that $X_n\in\mathcal{F}$.
Similarly if the partition has cardinality less than $\kappa$, and the filter is $\kappa$-complete, the results holds too.
I'm trying to build an counter-example to see that the requirements are tight. Is there a simple example with an infinite partition of a set $A$, where all elements are not in some ultrafilter $\cal F$ on $A$? 
 A: Following the suggestion in comment we will build a counter-example with an infinite partition. Suppose that $\cal F$ is an ultrafilter on some $\kappa$, not $\kappa$-complete. Therefore there exist an intersection of less than $\kappa$ elements of $\cal F$, not in $\cal F$ : 
$$ Y = \bigcap_{<\kappa} F_i \notin \cal F \text{ with } F_i \in \cal F$$
For all $i$, we define $X_i$ as being the complementary of $F_i$. They cannot be in $\cal F$ by definition of filters.
$$ X_i = \kappa \setminus F_i \notin \cal F$$
Note that $Y$ is the complement of their union : 
$$ Y = \bigcap_{<\kappa} F_i = \bigcap_{<\kappa} (\kappa \setminus X_i) = \kappa \setminus\left(\bigcup_{<\kappa} X_i\right) $$
Now let $Y_i$ be the restriction of $X_i$ :
$$ Y_i = X_i\setminus \left(\bigcup_{j\neq i} X_j\right)$$
The $Y_i$'s are all disjoint and $\bigcup_{<\kappa} X_i=\bigcup_{<\kappa} Y_i$, so that the $Y_i$'s together with $Y$ forms a partitions of $\kappa$. By construction $Y \notin \cal F$, and for all $i$, $Y_i\subseteq X_i$ with $X_i \notin \cal F$ implies that $Y_i \notin \cal F$.
Therefore we have built an infinite partition of $\kappa$ with no elements in $\cal F$.
