Relative interior commutes with cartesian product I have been reading Convex Analysis by Rockafellar. One of the proofs made use of the fact that $\text{ri}(A\times B) = \text{ri}(A) \times \text{ri}(B)$, where $A$ and $B$ are convex sets, $\times$ denotes the Cartesian product of sets, and $\text{ri}(\cdot)$ denotes relative interior.
The book says that this is an elementary result and does not offer a proof, but I fail to see how one can prove it easily.
What am I missing here?
 A: I'll give it a try, using the characterization of the relative interior
for convex sets $C$ from Wikipedia:
$$
\DeclareMathOperator{\ri}{ri}
\ri(C) = \{ c \in C \mid \forall z \in C \, \exists \lambda > 1 : \lambda \, c + (1-\lambda)\, z \in C\} \, .
$$
Note that for fixed $c, z \in C$, $\{ \lambda \, c + (1-\lambda)\, z \mid \lambda >1 \}$ is a ray extending from $c$ to the
"opposite direction" of $z$. From the convexity of $C$ it
follows that if $ \lambda \, c + (1-\lambda)\, z \in C$ holds for some
$\lambda > 1$ then this remains true for all smaller values $\lambda' > 1$.
Now assume that $(a, b)\in \ri(A) \times \ri(B) $.
Let $x \in A$ and $y \in B$. It follows that
$$\begin{align}
 &\exists \lambda > 1 : \lambda \, a + (1-\lambda)\, x \in A \quad \text{ and } \\
 &\exists \mu > 1 : \mu\,  b + (1-\mu)\, y \in B \, \quad .
\end{align}$$
As seen above, both relations still hold if $\lambda$ and $\mu$ are decreased to $\tau = \min(\lambda, \mu)$. Then
$$
   \tau \, (a, b) + (1 - \tau) \, (x, y) \in A \times B \, .
$$
Since $(x, y) \in A \times B$ was arbitrary, this proves that
$(a, b) \in \ri(A \times B)$.
The reverse direction works similarly.
