Suppose $\Omega \subset \mathbb{R}^{n}$ is a convex set (actually the convex hull of some compact subset $K\subset \mathbb{R}^{n}$). If $\Omega$ affinely generates $\mathbb{R}^{n}$, does it follow that $\Omega$ has nonempty interior?
Terminology: $\mathbb{R}^{n}$ is affinely generated by $\Omega$ means that every $x \in \mathbb{R}^{n}$ can be written as $x = \theta_{1}\omega_{1}+\cdots + \theta_{k}\omega_{k}$, where $t_{1},…t_{k}$ are all scalars such that $t_{1}+\cdots + t_{k} = 1$ and $\omega_{1},…,\omega_{k}\in \Omega$.
Best Answer
Yes. Applying a translation, we may assume that $0\in\Omega$, so affinely generating is the same as linearly generating. So, $\Omega$ contains some basis for $\mathbb{R}^n$, and applying an invertible linear transformation, we may assume that $\Omega$ contains the standard basis. So by convexity, $\Omega$ contains the convex hull of the standard basis vectors and $0$, which implies it has nonempty interior (explicitly, that convex hull is the set of $(a_1,\dots,a_n)$ such that each $a_i$ is nonnegative and $\sum a_i\leq 1$, whose interior is the set where each $a_i$ is positive and $\sum a_i<1$.)