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](https://math.stackexchange.com/q/4622368/1141296) earlier.



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


So it suffices to find a set that is open in your sense but not in 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.


(For the algebraic geometers in the room, this construction is remarkably similar to an example I laid out in [this answer](https://mathoverflow.net/a/423827/82179): being closed cannot be checked on each curve.)




---


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.


