Given a function $r\colon X\times Y \longrightarrow [0,\infty]$, and $\alpha\in[0,\infty]$, we define the following relation: $$ (x,y)\in r_{\alpha}\Longleftrightarrow r(x,y)\leq\alpha.$$
Let $\mathscr{U}$ be an ultrafilter on $X$ and $\mathscr{V}$ an ultrafilter on $Y$. Then I'm trying to prove the following equality: $$\sup_{U\in\mathscr{U},V\in\mathscr{V}}\ \inf_{x\in U, y\in V} r(x,y) = \inf\big\{\alpha\in[0,\infty]\ \big| r_{\alpha}\mathscr{U}\subseteq\mathscr{V}\big\}. $$
Any help will be appreciated!
I think I found a solution to this question.
First, assume $r_{\gamma}\mathscr{U}\subseteq\mathscr{V}$. Take an arbitrary $A\in\mathscr{U}$ and $B\in\mathscr{V}$. Then one has that $r_{\gamma}A\in\mathscr{V}$ and hence there is some $y\in r_{\gamma}A\cap B$. Choose $x\in A$ such that $r(x,y)\leq\gamma$. In particular, one has $$\inf_{x\in A,y\in B}r(x,y)\leq\gamma,$$ so that $\sup_{A\in\mathscr{U},B\in\mathscr{V}}\inf_{x\in A,y\in B}r(x,y)$ is indeed a lower bound for $\{\gamma\in[0,\infty] | r_{\gamma}\mathscr{U}\subseteq\mathscr{V}\}$.
To prove the other inequality, assume $\gamma<\inf\{\alpha\in[O,\infty] | r_{\gamma}\mathscr{U}\subseteq\mathscr{V}\}$. Hence, $r_{\gamma}\mathscr{U}\not\subseteq\mathscr{V}$ and thus there is an $A\in\mathscr{U}$ so that $r_{\gamma}A\notin\mathscr{V}$. This means there must be a $B\in\mathscr{V}$ for which one has $r_{\gamma}A\cap B=\emptyset$. Take now an arbitrary $x\in A$ and $y\in B$. Since $y\notin r_{\gamma}A$, one has $$ r(x,y)\geq\gamma.$$ This shows that $\gamma\leq\sup_{A\in\mathscr{U},B\in\mathscr{V}}\inf_{x\in A,y\in B}r(x,y)$, which in turn proves (by the randomness of $\gamma$) the remaining inequality.
Is this a correct proof?