Let $N$ be a countable set and let $Z$ be an arbitrary set with arbitrary elements $i\in N$ and $z\in Z$. Then, for each $i\in N$, define some set $M_i$. Further, for each $R\subseteq N$, let $M_R=\prod_{i\in R}M_i$ be the Cartesian product of all sets $M_i$ satisfying $i\in R$.
Given any $R\subseteq N$, let $m_R$ be any arbitrary element of $M_R$, and let $R^c=N\backslash R$ be its complement. Further, let $f:M_N\to Z$ be a function, and let $o$ be some well-defined object. Last, fix some $z^*\in Z$.
Then, object $o$ satisfies Property P if for each set $R\subseteq N$ and every vector $m_R\in M_R$, there exists some vector $m_{R^c}\in M_{R^c}$ for which $f(m_R,m_{R^c})=z^*$. Formally, \begin{gather} (\forall R\subseteq N)(\forall m_R\in M_R)(\exists m_{R^c}\in M_{R^c})(f(m_R,m_{R^c})=z^*) \end{gather}
I do not know whether Property P is vacuously satisfied when $R=N$, or whether Property P is never satisfied when $R=N$.
Let me try to explain my reasoning:
If $R=N$, then $R^c=N\backslash R=N\backslash N=N^c=\emptyset$, thus implying that $M_{R^c}=\emptyset$. Therefore, there exists no $m_{N^c}\in M_{N^c}$ satisfying $f(m_N,m_{N^c})=z^*$. By the same token, though, $f(m_N,m_{N^c})$ satisfies any condition I want to impose on it for all $m_{N^c}\in M_{N^c}$.
Can anybody help me see why I'm reaching two contradictory conclusions?
You state that $R = N$ implies $M_{R^C} = M_{\emptyset} = \emptyset$. However the empty cartesian product is not empty. The usual definition of the cartesian product in set theory gives $M_{\emptyset} = \{ \emptyset \}$. So the condition is neither vacuously satisfied nor false in general. The condition just states that $f(m_N,\emptyset)=z^{*}$.