# abstract description of the topology on a real vector space defined by the algebraically open sets

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?

This is not true, already in $$\mathbf R^2$$. Indeed, if $$V$$ is finite-dimensional, then the Euclidean topology is the coarsest topology for which all linear¹ maps $$V \to \mathbf R$$ are continuous: choosing a basis $$e_1,\ldots,e_n$$ and taking the corresponding coordinate projections $$\pi_i \colon V \to \mathbf R$$ shows that all boxes $$(a_1,b_1) \times \cdots \times (a_n,b_n)$$ have to be open, and these generate the Euclidean topology.
Example. Let $$V = \mathbf R^2$$, set $$v_0 = (1,0)$$, and inductively choose points $$v_1, v_2, \ldots$$ such that $$\lVert v_i \rVert = 2^{-i}$$ and $$v_i$$ does not lie on the (finitely many) lines $$\overline{v_jv_k}$$ for $$j,k < i$$. Define $$Z = \{v_0,v_1,\ldots\}$$ and $$U = V \setminus Z$$. Then $$U$$ is not open in the Euclidean topology because the $$v_i$$ converge to $$0 \not \in Z$$. But for every line $$\ell \subseteq V$$, the intersection $$\ell \cap Z$$ contains at most $$2$$ points by construction, so $$\ell \cap U$$ is open.
¹It doesn't matter if you say "linear" or "affine linear" here, because any affine linear map differs from a linear one by a translation, and translation $$\mathbf R \to \mathbf R$$ is continuous.