Do algebraically open sets define a vector space topology?

Question

Let $X$ be a vector space. A vector space topology on $X$ is a topology such that addition and scalar multiplication are continuous. A subset $A$ of $X$ is said to be algebraically open if, for all $a\in A$ and $x\in X$, there exists $\varepsilon>0$ such that $a+(-\varepsilon,\varepsilon)\cdot x\subseteq A$. My question is if the algebraically open sets form a vector space topology on $X$.

Answer 1

The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $\mathbb{K}^n$ with the Euclidean topology, for some $n\in\mathbb{N}$. However, there exist algebraically open subsets of $\mathbb{R}^2$ that are not open by the Euclidean topology.