Prove Minimum AABB Construction
Suppose that the points are $(x_1,y_1),\ldots,(x_n,y_n)$. You are looking for the smallest (in some sense) bounding box $[X_m,X_M] \times [Y_m,Y_M]$ containing all your points. For every $i$, we must have $X_m \leq x_i \leq X_M$ and $Y_m \leq y_i \leq Y_M$. In particular,
$$
\begin{align*}
X_m \leq \min_i x_i =: x_{\min}, \\
X_M \geq \max_i x_i =: x_{\max}, \\
Y_m \leq \min_i y_i =: y_{\min}, \\
Y_M \geq \max_i y_i =: y_{\max}.
\end{align*}
$$
Summarizing, this shows that every bounding box must satisfy
$$
X_m \leq x_{\min} \leq x_{\max} \leq X_M, \\
Y_m \leq y_{\min} \leq y_{\max} \leq Y_M,
$$
that is, it must contain $B := [x_{\min},x_{\max}] \times [y_{\min},y_{\max}]$. Conversely, $B$ is itself a bounding box, since $x_{\min} \leq x_i \leq x_{\max}$ and $y_{\min} \leq y_i \leq y_{\max}$ for all $i$. We conclude the following fact:


  $B$ is the intersection of all bounding boxes.