6. Given a positive integer $n$. Find the largest real number $\mu$ such that: for any set $C$ of $4 n$ points in the "open" unit square $U$, there exists an "open" rectangle $V$ in $U$ satisfying the following properties:
(1) The sides of the open rectangle $V$ are parallel to the sides of $U$;
(2) The open rectangle $V$ contains exactly one point from the set $C$;
(3) The area of the open rectangle $V$ is at least $\mu$
6. $\mu_{\max }=\frac{1}{2 n+2}$.

Without loss of generality, let $U$ be the set of points in the Cartesian coordinate system $\left\{\begin{array}{l}01$, otherwise swap the horizontal and vertical coordinates.
First, we provide a lemma.
Lemma If segment $A_{0} A_{m}$ has $m(m \geqslant 2)$ points inside, denoted as $A_{1}, A_{2}, \cdots, A_{m}$, and
$$
\begin{array}{l}
\left|A_{0} A_{1}\right|1$,

it is known that there exists $i \in\{0,1,3, \cdots, m-2\}$, such that
$\left|A_{i} A_{i+2}\right| \geqslant \frac{1}{\frac{m+1}{2}}=\frac{2}{m+f(m)}$.
Returning to the original problem.
Let the points with horizontal coordinate $x_{j}$ be sorted by their vertical coordinates as $A_{1}, A_{2}, \cdots, A_{a_{i}}$, and let $A_{0}\left(x_{j}, 0\right)$, $A_{a_{j+1}}\left(x_{j}, 1\right)$.
By the lemma, there exists $i \in\left\{0,1, \cdots, a_{j-1}\right\}$, such that
$\left|A_{i} A_{i+2}\right| \geqslant \frac{2}{a_{j+1}+f\left(a_{j+1}\right)}$,
the rectangle $V:\left\{\begin{array}{l}x_{j-1}<x<x_{j+1}, \\ y_{i}<y<y_{i+2}\end{array}\right.$ contains the point set $U$ and only contains one point $A_{i+1}\left(y_{i} 、 y_{i+2}\right.$ are the vertical coordinates of $A_{i}$ and $A_{i+2}$, respectively).

Assume $\mu=\frac{1}{2 n+2}$ cannot be achieved. Then for any $j \in\{1,2, \cdots, k\}$, we have
$\left(x_{j+1}-x_{j-1}\right)\left(y_{i+2}-y_{i}\right)<\frac{1}{2 n+2}$,
$x_{j+1}-x_{j-1}<\frac{a_{j}+1+f\left(a_{j+1}\right)}{4 n+4}$.
Let $S$ be the set of all even numbers in $2,3, \cdots, k+1$, and $T$ be the set of all odd numbers in $2,3, \cdots, k-1$.

Since $y_{i+2}-y_{i}=\left|A_{i} A_{i+2}\right|$, and the largest elements in sets $S$ and $T$ are both at least $k-2$, and $x_{2}<x_{3}<\cdots<x_{k-1}$, we have
$$
\begin{array}{l}
\sum_{j \in S}\left(x_{j+1}-x_{j-1}\right) \geqslant x_{k-1}-x_{1}, \\
\sum_{j \in T}\left(x_{j+1}-x_{j-1}\right) \geqslant x_{k-1}-x_{2} .
\end{array}
$$
(Provided by Gu Bin)