Build an ultrafilter finer than the Frechet filter. I need to build an ultrafilter finer than the Fréchet (Filter finite complements).
In an infinite set $X=\mathbb{R}$, $\mathcal{F}_{c}=\{A\subseteq X\mid A^{c}\ \ \ \text{is a finite set}\}$ is the filter of Finite complements .
I tried whit this.
$\mathcal{H}=\{U\cup\mathbb{Q}: U\in \mathcal{F}\}\cup\mathcal{F}$ and we have
1) $\emptyset\notin\mathcal{H}$
2) Let $A,B\in \mathcal{H}$ then we have that $A=U\cup\mathbb{Q}$ and $B= V\cup\mathbb{Q}$;
$$A\cap B=(U\cup \mathbb{Q})\cap (V\cup \mathbb{Q})=(U\cap V)\cup \mathbb{Q}$$ 
also $U\cap V\in\mathcal{F}$ then $A\cap B\in\mathcal{H}$.
3) Let $A\in\mathcal{H}$  then $A=U\cap\mathbb{Q}$. The set is $A\uparrow$ defined by $A\uparrow=\{C:A\subseteq C\}$ Then $A\uparrow =U\uparrow\cup\mathbb{Q}\uparrow$.
This is the idea that occurred to me but I'm not sure.
 A: If $U\in\mathcal F$ then $U\cup\mathbb Q\in\mathcal F$, because $U\subseteq U\cup\mathbb Q$ and $\mathcal F$ is a filter. Hence $\{U\cup\mathbb Q:U\in\mathcal F\}\subseteq\mathcal F$ and so $\mathcal H=\mathcal F$.
By the way, note that $(U\cap V)\uparrow=U\uparrow\cup V\uparrow$ holds only when $U\subseteq V$ or $V\subseteq U$; that's because $U\cap V\in(U\cap V)\uparrow$ while $U\cap V\notin U\uparrow\cup V\uparrow$ unless $U\subseteq V$ or $V\subseteq U$.
I guess what you were trying to do was construct the smallest filter $\mathcal H$ which extends $\mathcal F$ and contains $\mathbb Q$ as an element. A correct definition of that filter would be$$\mathcal H=\{U\subseteq X:F\cap\mathbb Q\subseteq U\text{ for some }F\in\mathcal F\}=\{U\subseteq X:\mathbb Q\setminus U\text{ is finite }\}.$$But that's not an ultrafilter.
Anyway, you will not be able to construct a free ultrafilter explicitly. You will have to use the Ultrafilter Theorem, which says than any proper filter (in particular the Fréchet filter) can be extended to an ultrafilter; or the (equivalent) Prime Ideal Theorem; or use some general maximal principal such as Zorn's Lemma to extend the Fréchet filter to a maximal filter, and then show that a maximal filter is an ultrafilter.
