Cartesian Product of Intervals is Convex Set

Theorem
Let $n \in \N$.

For all $k \in \set {1, \ldots, n}$, let $\Bbb I_k$ be a real interval of any of the real interval types.

Then the cartesian product $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.

Proof
Let $\mathbf x, \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$ with:

where $x_k, y_k \in \Bbb I_k$ for all $k \in \set {1, \ldots, n}$.

Let $t \in \closedint 0 1$.

Suppose that $x_k \le y_k$.

It follows that:

By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.

If instead $x_k \ge y_k$, it follows by similar calculations that:
 * $x_k \ge t x_k + \paren {1-t} y_k \ge y_k$

By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.

It follows that $t \mathbf x + \paren {1-t} \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$.

By definition of convex set, it follows that $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.