Let $C_1$ and $C_2$ be two proper full dimensional closed convex cones in $\mathbb{R}^n$ that are pointed. Suppose that $C_1\subseteq C_2$ and that the boundary of $C_1$ is contained in the boundary of $C_2$. Then is $C_1=C_2$? Any references for a result of this form would be welcome. I suspect this to be true, and I have some rough ideas about how to prove this, but my arguments seem messy and I am worried that my intuition from low dimensional and finitely generated cases goes wrong in the generality I am considering.

By pointed I mean that $(-C)\cap C=\{0\}$ and by full dimensional I mean that $C$ spans $\mathbb{R}^n$. I am particularly interested when $C_1,C_2$ are not necessarily finitely generated, though partial results in the finitely generated or finitely generated rational case would be welcome.


Assume that $q\in C_2\setminus C_1$. Let $p$ be an interior point of $C_1$. Then the interval $(p,q)$ contains a boundary point of $C_1$ but only interior points of $C_2$. A contradiction.

  • $\begingroup$ Just to make sure I am understanding you correctly, you take $q$ to be in the interior of $C_2$ and not in $C_1$ and then consider the line segment between $p$ and $q$. $\endgroup$
    – slack tide
    Oct 29 '20 at 19:29
  • $\begingroup$ not necessarily interior, $q$ may be a boundary point of $C_2$, the whole interval $pq$ (without endpoints) still consists of interior points of $C_2$ $\endgroup$ Oct 29 '20 at 20:14
  • $\begingroup$ I see. Thanks for taking the time to walk me through this. $\endgroup$
    – slack tide
    Oct 29 '20 at 20:34

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