The algebraic interior of a convex set is not empty if the affine hull is the whole space

convex-analysisconvex-geometrylinear algebratopological-vector-spaces

Definitions. Let $X$ be a subset of a linear space $S.$

  1. The affine hull of $X$ is the intersection of all affine spaces (translations of subspaces) that contain $X.$ It is easy to prove that the affine hull of $C$ takes the form $t + H$ where $t \in C$ and $H$ is a vector subspace.
  2. The algebraic interior (relative to the vector space $S$) of $X$ is the set of points $x \in X$ such that for every point $y \in S,$ there is a $\delta = \delta_x > 0$ obeying $x + [0, \delta]y \subset X.$ If $X$ is a convex set (meaning that $x,y \in X$ implies $[x,y] =x + [0,1](y-x) \subset X$) then the algebraic interior is the set of $x \in X$ such that for every $y \in S,$ there exists a $\delta > 0$ such that $x + \delta y \in X.$

Prove or disprove: Let $C$ be a convex subset of $S.$ If the affine hull of $C$ is $S,$ then $C$ has non empty algebraic interior (the reciprocal is easily shown to be true).

In the question: An example of a convex set with empty interior and whose convex hull is the whole space they asked a similar question except there they are considering the usual (topological) interior. More explicitly, if $S$ is a topological vector space (TVS), then the topological interior $X$ is contained in the algebraic interior. In particular, the algebraic interior of $X$ contains the union of $\mathop{\mathrm{int}}\limits_{\tau} (X)$ as $\tau$ runs through all topologies that make $S$ a TVS. That question asks whether the affine hull of $X$ being the whole $S$ is sufficient so that every topology $\tau$ will have $\mathop{\mathrm{int}}\limits_{\tau} (X) \neq \varnothing.$ This question will be answered positively if for some topology $\tau,$ we have $\mathop{\mathrm{int}}\limits_{\tau} (X) \neq \varnothing$ (the possibility that every topological interior is empty and yet the algebraic interior is not empty is open).

Best Answer

I'm going to disprove the claim by providing a counter-example.

  • Let $S=\{(x_1,x_2,\ldots)\in\Bbb R^\infty: x_n=0\text{ for almost all }n\in\Bbb N\}$. This is a linear space.
  • Let $C=\{x\in S:x_n\geq 0\text{ for all }n\in\Bbb N\}$.
  • Let $x\in S$ be any vector. Then $x=u-v$ for some $u,v\in C$ ($u$ and $v$ can be chosen as a positive and negative part of $x$, respectively). Moreover $0\in C$. Therefore $x\in\mathrm{aff}\,C$. This shows that $\mathrm{aff}\,C=S$.
  • Fix arbitrary $x=(x_1,x_2,\ldots,x_n,0,0,\ldots)\in C$. Let $y\in S$ be $y=-e_{n+1} = (0,0,\ldots,0,-1,0,0,\ldots)$ (minus one on the position $n+1$). Observe that for all $\delta>0$ we have $x+\delta y\notin C$. Therefore $x$ isn't an interior point. This shows that the interior of $C$ is empty.
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