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I'm trying to prove the following result:

Let $r:X\longrightarrow Y$ be a relation and $\mathcal{U}$ and $\mathcal{V}$ be ultrafilters such that $r^{-1}V \in\mathcal{U}$ for all $V\in\mathcal{V}$. There exists an ultrafilter $\mathcal{X}$ on $\Gamma_r = \{(x,y)\in X\times Y | xry\}$ for which one has $\pi_X(\mathcal{X}) = \mathcal{U}$ and $\pi_Y(\mathcal{X}) = \mathcal{V}$.

My guess was to put $C_{U,V} = \pi_X^{-1}(U)\cap\pi_Y^{-1}(V) = U\times V \cap\Gamma_r$ for all $U\in\mathcal{U}$ and $V\in\mathcal{V}$. Because of the assumption on $r$, these sets are never empty and satisfy the finite intersection property. Hence, they form a basis for a filter $\mathcal{X}$ on $\Gamma_r$. Obviously one has $\pi_X(\mathcal{X})=\mathcal{U}$ and $\pi_Y(\mathcal{X})=\mathcal{V}$.

Yet, I'm stuck at proving that this filter $\mathcal{X}$ is an ultrafilter. First of all, is this indeed an ultrafilter, and if not, is there one that satisfies the conditions? Any help would be appreciated!

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Hint: A (proper) filter can be extended to an ultrafilter.

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  • $\begingroup$ Yes, I agree, but will any ultrafilter extending the basis consisting of the $C_{U,V}$ satisfy the conditions? $\endgroup$
    – user75548
    May 3 '13 at 10:51
  • $\begingroup$ @user75548: Yes. If $\mathscr{W}$ is an ultrafilter extending the filter of $C_{U,V}$’s, and $\pi_X[W]\notin\mathscr{U}$ for some $W\in\mathscr{W}$, then $U=X\setminus\pi_X[W]\in\mathscr{U}$, so $(U\times Y)\cap r\in\mathscr{W}$, which is impossible, since it’s disjoint from $W$. You can argue similarly for the other projection. $\endgroup$ May 4 '13 at 2:21

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