If $\mathcal{F}$ is a filter on $X$ show that $\mathcal{F}\subseteq \mathcal{U}$ Suppose $\mathcal{F}$ is a filter on $X$ but that $A\subset X$ is not in $\mathcal{F}$. I want to show that there is an ultrafilter $\mathcal{U}$ containing $\mathcal{F}$ such that $X\setminus A \in \mathcal{U}$.
By Zorn's lemma every filter is contained in an ultrafilter. So $\mathcal{F}$ is contained in some ultrafilter, $\mathcal{U}$. If $\mathcal{U}$ contains $X\setminus A$ then we are done.  Otherwise it must contain $A$. Can I somehow 'rearrange' the ultrafilter to contain $X\setminus A$?
 A: See my answer here : A filter is the intersection of the ultra filter refining it
A filter $\mathcal{F}$ is the intersection of the ultrafilters refining it, so if all ultrafilters $\mathcal{U}$ containing $\mathcal{F}$ contained $A$, then $\mathcal{F}$ would as well, contrary to hypothesis.
The more direct answer, and the actual idea, is that if $A\notin\mathcal{F}$ then for all $B\in\mathcal{F}$, $B\cap (X\setminus A) \neq \emptyset$, otherwise there would be a $B\subset A$ with $B\in\mathcal{F}$, which would imply $A\in \mathcal{F}$.
Therefore, $\{B\cap (X\setminus A) \mid B\in \mathcal{F}\}$ is a filter base. Pick any ultrafilter containing it, it must also contain $\mathcal{F}$, and we are done.
A: $X$ \ $A$ has non-empty intersection with every $B\in  F.$ Because if $B\in F$ and $B\cap (X$ \ $A)=\phi$ then $A\supset B\in F,$ implying $A\in F,$ contrary to the hypothesis. 
Therefore $F'=\{(B\cap (X$ \ $A))\cup C : B\in F\land C\subset X\}$ is a filter with $X$ \ $A\in  F'\supset F,$ and we may extend $F'$ to an ultrafilter.
