Regarding direct sums in topological vector spaces

Question

If $E=E_0\oplus E_1$ is a topological vector space and $A\subseteq E_0$ open in the induced topology on $E_0$, can I conclude that $A+E_1$ is open in $E$? Possibly if I assume $E$ to be locally convex?

Answer 1

It seems the following.

It depends. If you consider on $E=E_0\oplus E_1$ the product topology from $E_0\times E_1$ then the answer is positive, because the set $A\times E_1$ is open in $E_0\times E_1$. But in the opposite cases the answer may be negative. For instance, suppose that $E_1$ is dense in $E$, $A$ is non-empty and $E_0\ne A-A$. If the set $A+E_1$ is open in $E$ then it is a dense subset of $E$, So $(x+A+E_1)\cap (A+E_1)\ne\varnothing$ for each point $x\in E$. So $E=(A-A)+E_1$, which contradicts to $E$ is a direct sum of $E_0$ and $E_1$ and $E_0\ne A-A$.