Vector space translation continuous implies addition continuous? In here, it is proved if a topology on a vector space makes the addition function continuous, then the translation is also continuous everywhere. My question is whether the inverse is still true:
Let $V$ be a vector space with a topology $\tau$. If, for every $x\in V$, the translation $T_x: V\to V$, $T_x(y):=x+y$, is continuous, is it guaranteed the addition $+: V\times V \to V$ is also continuous?
I know if the translations are continuous everywhere, then they are homeomorphisms, regardless of whether the addition is also continuous. At this point, I'm not assuming anything regarding the scalar multiplications, as in, it's not known whether it is continuous.
 A: Nice question! Here is a counterexample. Let $V$ be $\mathbb{R}$ and endow it with the cofinite topology, in which a set $A \subseteq \mathbb{R}$ is open if and only if it is empty or its complement $\mathbb{R} \setminus A$ is finite. Then every bijection from $\mathbb{R}$ to itself, and in particular translations, are continuous. On the other hand, take the open set $A = \mathbb{R} \setminus \{0\}$. Its preimage under $+ : \mathbb{R} \times \mathbb{R} \mapsto \mathbb{R}$ is $B = (\mathbb{R} \times \mathbb{R}) \setminus D$ where $D = \{(x,-x) : x \in \mathbb{R}\}$ is the diagonal. But $B$ is not open in the product topology: you can check that any nonempty open set in this topology misses only a finite number of points from $D$.
Of course, this is not a particularly nice topology: it is not Hausdorff, for example. Indeed, a topology is Hausdorff if and only if the diagonal is closed in the product topology, so this example uses "non-Hausdorffness" in an essential way. It is possible that there is a positive answer to your question for some nicer class of topologies, though a priori I see no reason for there to be one.
