I am sorry that the following question is elementary. I have not received an answer from my post at [Math Stack Exchange](https://math.stackexchange.com/questions/4676078/is-the-interior-of-the-tensor-product-of-two-convex-cones-equal-to-the-tensor-pr).


In the following question, all cones are convex and contain the origin. Let $C \subset \mathbb{R}^{m}$ be a cone consisting of $m$-dimensional column vectors, and $D \subset \mathbb{R}^{n}$ another cone of $n$-dimensional column vectors.


Given any subsets $A \subset \mathbb{R}^{m}$ and $B\subset \mathbb{R}^{n}$, define $A \otimes B \subset \mathbb{R}^{m \times n}$ as the set of $m \times n$ matrices of the form $x\_1 {y\_1}^T + \dotsb + x\_q {y\_q}^T$, where $x\_1, \dotsc, x\_q \in A$ and $y\_1, \dotsc, y\_q \in B$, $q\geq 1$ is an arbitrary positive integer, and $^T$ denotes the transpose.


Is it true that
$$(C \otimes D)^\circ = C^\circ \otimes D^\circ,$$
where $^\circ$ denotes the interior relative to the respective euclidean topologies?




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Here's an example where the said equality holds. Take $C = {\mathbb{R}\_{\ge 0}}^m$ to be the cone of all $m$-dimensional column vectors whose every entry is a nonnegative real number. Similarly, take $D = {\mathbb{R}\_{\ge 0}}^n$. Then $C \otimes D = {\mathbb{R}\_{\ge 0}}^{m \times n}$ is the cone of $m \times n$ matrices whose every entry is nonnegative. Indeed, it suffices to show that $\mathbf{E}\_{ij}$, the matrix whose $(i, j)$th entry is $1$ and has $0$s everywhere else, lies in $C \otimes D$, because $C \otimes D$ is closed under taking nonnegative linear combinations. But $\mathbf{E}\_{ij} = \vec{e\_i}\vec{e\_j}^T$, where $\vec{e\_i} \in C$ and $\vec{e\_j} \in D$. Therefore $(C \otimes D)^\circ = {\mathbb{R}\_{> 0}}^{m \times n}$ is the set of all $m \times n$ matrices whose every entry is strictly positive.


Noting that $C^\circ = {\mathbb{R}\_{> 0}}^m$ consists of all $m$-dimensional vectors whose entries are all strictly positive real numbers, and similarly $D^\circ = {\mathbb{R}\_{> 0}}^n$, we get $C^\circ \otimes D^\circ = {\mathbb{R}\_{> 0}}^{m \times n}$ also.



Is the interior of the tensor product of two convex cones equal to the tensor product of their respective interiors?
Yes, $(C \otimes D)^\circ = C^\circ \otimes D^\circ$ is correct.


Let us start by proving the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$. To this end, it is enough to show that $C^\circ \otimes D^\circ$ is open. For a given point $z = x\_1 {y\_1}^T + \dotsb + x\_q {y\_q}^T \in C^\circ \otimes D^\circ$, we can use convexity and openness to replace $x\_1$ by a convex combination of $n$ linearly independent points in $C$, and similarly for $y\_1$. In this way, we can assume without loss of generality that the $x\_i$'s span $\mathbb{R}^m$, and similarly the $y\_i$'s span $\mathbb{R}^n$. Therefore by nudging each one of these points a little (without leaving $C^\circ$ or $D^\circ$), we can move in any direction in $C^\circ \otimes D^\circ$, and it follows that the latter set is open.


Furthermore, the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$ is dense, because $C^\circ \otimes D^\circ$ is dense even in $C \otimes D$ by the obvious termwise approximation argument.


To now arrive at the claimed equality, recall that [every convex open set is the interior of its closure](https://math.stackexchange.com/questions/360187/is-convex-open-set-in-mathbbrn-is-regular). This applies in particular to both $C^\circ \otimes D^\circ$ and $(C \otimes D)^\circ$. This gives
$$
C^\circ \otimes D^\circ = \overline{C^\circ \otimes D^\circ}^\circ = \overline{(C \otimes D)^\circ}^\circ = (C \otimes D)^\circ,
$$
where the second step is the density from the previous paragraph.


