Is the core topology on $\mathbb{R}^2$ a group topology? The core topology on $\mathbb{R}^2$ is the final topology induced by the inclusions $i_{v,w} : \mathbb{R} \to \mathbb{R}^2$, $t \mapsto v + t w$ of lines (affine one-dimensional subspaces), where $\mathbb{R}$ carries the Euclidean topology and $v, w \in \mathbb{R}^2$. In other words, a set is open if its intersection with every line is an open subset of the line with its (one-dimensional) Euclidean topology. These open sets are also known as algebraically open. Thus, a set $A \subseteq \mathbb{R}^2$ is algebraically open if and only if for every point $v \in A$ and every direction $w \in \mathbb{R}^2$ there is $\varepsilon > 0$ such that $A$ contains with $v$ also the whole open line segment $v + (-\varepsilon, \varepsilon) w$. Note that the choice of $\varepsilon$ depends on the direction $w$.
The core topology is strictly finer than the Euclidean topology on $\mathbb{R}^2$, but a convex algebraically open set is also open w.r.t. the Euclidean topology. Thus, an algebraically open set that is not Euclidean open must be necessarily non-convex. As an example, the set $A := (\mathbb{R}^2 \setminus S^1) \cup \{ (1,0) \}$ is algebraically open but not Euclidean open: if $L \subseteq \mathbb{R}^2$ is a line then $A \cap L = L \setminus S$ where $|S| \leq 2$, thus $A \cap L$ is an Euclidean open subset of $L$. But there is no two-dimensional (Euclidean) open ball around $(1,0)$ completely contained in $A$.
The core topology is clearly translation-invariant, i.e. the vector addition $+ : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$, $(x,y) \mapsto x + y$ is continuous separately in each of its components. Is vector addition moreover continuous (where the domain carries the product topology)? In other words, is the core topology a group topology? I think, for this purpose it would be nice to have an explicit description of some nice base for the core topology. (I was not able to create a contradiction for the set $A$ above.) 
 A: Let $X$ be a vector space, $\tau$ a topology on $X$ and consider the following compatibility properties of $\tau$ with the vector space structure of $X$:
(i) addition is continuous
(i') addition is continuous separately in its components (i.e. translations are homeomorphisms)
(ii) scalar multiplication is continuous
(ii') scalar multiplication is continuous separately in its components.
The core topology $\tau_c$ on $X$ is known to be the finest topology satisfying (i') and (ii'). The following shows that $\tau_c$ does neither satisfy (i) nor (ii) whenver $\dim(X) \geq 2$.
As Henno Brandsma pointed out in his comment, the core topology $\tau_c$ on $X = \mathbb{R}^2$ is also known as the "radial plane" topology, see e.g. [Willard, "General Topology"]. [S. P. Franklin, Solution to Monthly Problem #5468, American Mathematical Monthly 75 (1968), p. 208] shows that this topology is not regular. Since a group topology is completely regular (it carries a compatible uniform structure induced by the group structure) it follows that $\tau_c$ does not satisfy (i).
Moreover, $\tau_c$ does not satisfy (ii) since the Euclidean topology (on a finite-dimensional space) is known to be the finest topology with (i') and (ii). (The finest topology on a generic vector space having (i') and (ii) coincides with the the finite topology (= the finest topology that induces the Euclidean topology on every finite-dimensional subspace).)
Addendum: The fact that $\tau_c$ is not regular was also already shown by [Klee, "Some finite-dimensional affine topological spaces" (1955)].
