Let $n$ be a positive integer. Find the smallest positive integer $k$ such that for any set $S$ of $n$ points in the interior of the unit square, there exists a set of $k$ rectangles such that the following hold:  
- The sides of each rectangle are parallel to the sides of the unit square.  
- Each point in $S$ is not in the interior of any rectangle.  
- Each point in the interior of the unit square but not in $S$ is in the interior of at least one of the $k$ rectangles. (The interior of a polygon does not contain its boundary.)
We claim the answer is $k=2 n+2$. The lower bound is given by picking  
$$ S=\left\{\left(s_{1}, s_{1}\right),\left(s_{2}, s_{2}\right), \ldots,\left(s_{n}, s_{n}\right)\right\} $$  
for some real numbers $0<s_1 < s_2 < \ldots < s_n$. The four rectangles covering each of  
$$ \left(s_{1}-\varepsilon, s_{1}\right),\left(s_{1}, s_{1}-\varepsilon\right),\left(s_{n}+\varepsilon, s_{n}\right),\left(s_{n}, s_{n}+\varepsilon\right) $$  
cannot cover any other points in $S^{\prime}$; all other rectangles can only cover at most 2 points in $S^{\prime}$, giving a bound of  
$$ k \geq 4+\frac{\left|S^{\prime}\right|-4}{2}=2 n+2 $$  
![](https://cdn.mathpix.com/cropped/2024_11_19_1799270e91f74486a977g-05.jpg?height=812&width=806&top_left_y=242&top_left_x=628)  
To prove that $2 n+2$ rectangles are sufficient, assume that the number of distinct $y$-coordinates is at least the number of distinct $x$-coordinates. Let  
$$ 0=x_{0}<x_{1}<\cdots<x_{m} \leq 1 $$  
be the distinct $x$-coordinates, and  
$$ 0=y_{0}<y_{1}<\cdots<y_{n} \leq 1 $$  
be the distinct $y$-coordinates, where $m \leq n$. We can cover the points with the following rectangles:  
- For each $i$ from $1$ to $m$, cover the points $(x_i, y_j)$ for $j=1, 2, \ldots, n$ with a rectangle of width $\varepsilon$ and height $y_{j+1} - y_{j-1}$.
- For each $j$ from $1$ to $n$, cover the points $(x_i, y_j)$ for $i=1, 2, \ldots, m$ with a rectangle of width $x_{i+1} - x_{i-1}$ and height $\varepsilon$.
- Cover the points $(0, y_j)$ for $j=1, 2, \ldots, n$ with a rectangle of width $\varepsilon$ and height $y_{j+1} - y_{j-1}$.
- Cover the points $(x_i, 0)$ for $i=1, 2, \ldots, m$ with a rectangle of width $x_{i+1} - x_{i-1}$ and height $\varepsilon$.

This gives a total of  
$$ (m+n+2)+(n-m)=2 n+2 $$  
rectangles.