Standard LP vs Mixed-Integer LP Optimal solution. I have a question about standard LP optimal solution vs Mixed Integer LP optimal solution.

Let $z*$ denote the optimal value for $c \cdot x$ and $z^{int}$ to be the optimal value $c \cdot x$ for the mixed interger LP. Why is it that $z* \ge z^{int}$?
$$
 \max  c \cdot x  \\
\mathrm{s.t.} \quad Ax \leq b\\
x \geq 0 \\$$

$$
 \max  c \cdot x  \\
\mathrm{s.t.} \quad Ax \leq b\\
x \geq 0 \\
x_{i} \in Z, i \in I $$
$z*\ge z^{int}$ because $z*$ is the optimal solution of the relaxation of the associated MILP. In other words it has less constraints, so the feasible space is larger: there are more potential good solutions.

Note that for a minimization problem, you would obviously have $z*\le z^{int}$.
It can be proved the other way that $z^{int}$ can never be greater than $z*$ by contradiction since the space of $z^{int}$ is a subset of the space of $z*$. So, $x^{int}$ corresponding to $z^{int}$ would also be a solution for $z*$ too.