Let $V$ be a real vector space. Given a subset $A \subseteq V$, say that a point $x \in A$ lies in the *algebraic interior* of $A$ if every affine line $\ell$ that passes through $x$ has the property that $x \in (\ell \cap B)^\circ$. Here $\ell \cap B$ is a subinterval of $\ell \cong \mathbb{R}$, so we defined its interior $(\ell \cap B)^\circ$ by equipping $\ell$ with the Euclidean topology.

Say that a subset $U \subseteq V$ is *algebraically open* if the algebraic interior of $U$ is $U$ itself. We thereby get a topology on $V$.

Any affine functional $\pi: V \to \mathbb{R}$ is continuous with respect to this topology. Is this topology defined by the algebraically open subsets the coarsest topology such that every affine functional on $V$ is continuous map?

I had asked this question on Math Stack Exchange earlier.