I wish to prove that in a tvs for any $x,y$ and $W$ neighborhood of $x+y$ there a two neighborhoods $V_x,V_y$ such that
$$ V_x+V_y \subset W, $$
the proof should be simple because if this weren't the case then I could build a convergent sequence $a_j+b_j \in X- W$ (which is closed) such that $a_j+b_j \to x+y$, but this would imply $x+y\in X - W$ which is closed, hence a contradiction, is this correct?
(I assume a similar argument could be applied for the scalar multiplcation, is this correct?)
Thank you
Since $W$ is neighborhood of $x+y$ and since $+$ is continuoues, there is a neighborhoods $W^\star$ if $(x,y)$ such that$$(\forall (x',y')\in W^\star):x'+y'\in W.\tag1$$And since we're dealing with the product topology here, there are neighborhoods $V_x$ and $V_y$ of $x$ and $y$ respectively such that $V_x\times V_y\subset W^\star$. But then $(1)$ implies that $V_x+V_y\subset W$, because if $x'\in V_x$ and $y'\in V_y$, then $(x',y')\in V_x\times V_y\subset W^\star$ and therefore $x'+y'\in W$.