Why is the cardinality both supermodular and submodular(or modular)? It's easy to understand that cardinality is submodular, but how can it be supermodular at the same time? The supermodular function of the cardinality is the opposite of its submodular function, then I thought it would never be submodular again. It's very incomprehensible for me that a function can be both submodular and supermodular, or called modular, so could anyone please instantiate and exemplify it. Thanks. 
 A: Note that a function $f \colon \def\P{\mathfrak P}\P(\Omega) \to \mathbf R$ is submodular iff 
$$ f(S) + f(T) \ge f(S\cap T) + f(S \cup T), \qquad S,T \subseteq \Omega\tag 1 $$
and supermodular iff the reverse inequality 
$$ f(S) + f(T) \le f(S\cap T) + f(S \cup T), \qquad S,T \subseteq \Omega \tag 2$$
holds. 
Now consider cardinality, denoted by $f(S) = |S|$. It is well-known, that 
$$ |S| + |T| = |S \cap T| + |S\cup T| \tag 3$$
holds (we will prove it later). Hence, $f=|\cdot|$ fulfills both inequalities (1) and (2), as it has equality in both cases. 
To prove (3) note that 
$$ S \cup T = (S \cap T) \cup (S\setminus T) \cup (T \setminus S) $$
and the three sets are disjoint. Hence, 
$$ |S\cup T| = |S\cap T| + |S \setminus T| + |T \setminus S| \tag 4$$
just by counting elements. As 
$$ S = (S\cap T) \cup (S \setminus T) $$
is also a disjoint union
$$ |S| = |S \cap T| + |S \setminus T| \tag{5a} $$
Along the same line of thought
$$ |T| = |S \cap T| + |T \setminus S|\tag{5b} $$
Now use eqns. (5) in (4), we get 
$$ |S\cup T| = |S\cap T| + |S| - |S\cap T| + |T| - |S \cap T| \iff |S\cup T| + |S\cap T| = |S| + |T| $$
which proves (3).
