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 pass through $x$ has the property that $x \in (\ell \cap B)^\circ$. Here $\ell \cap B$ is a line segment and $(\ell \cap B)^\circ$ denotes its interior in the usual sense.
Thus, we can define a topology on $V$, where a subset $U \subseteq V$ is open if the algebraic interior of $U$ is $U$ itself.
I have a suspicion this topology is the coarsest topology such that every linear function on $V$ is a continuous map. Is my suspicion true?