Here is the statement I want to prove:
Let $X$ be a topological vector space over $\mathbb{K}$. Let $(x_k)$ and $(y_k)$ be sequences in $X$. I assume that $x$ is a limit of $(x_k)$ and $y$ is a limit of $(y_k)$. Then $x+y$ is a limit of $(x_k+y_k)$.
Remarks.
- I do not assume unicity of the limit.
- If it matters I can assume that $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$.
Here is my attempt.
Let $A$ be an open set containing $x+y$. The part $(A-(x+y))/2$ is an open set containing $0$. If $n$ is large enough, we have
$x_n-x\in \frac{A-(x+y)}{2}$
and
$y_n-x\in \frac{A-(x+y)}{2}$.
In other terms, there exist $a_n\in A$ and $b_n\in A$ such that
$x_n-x\in \frac{a_n-(x+y)}{2}$
and
$x_n-x\in \frac{b_n-(x+y)}{2}$.
Now we have
$x_n+y_n=\frac{a_n+b_n}{2}-(x+y)$.
If $A$ is convex, the result is proven.
Thus I have two questions.
- If $X$ is a topological vector space, there exist a basis of topology around $0$ made of convex parts ?
- If not, how to prove the $x+y$ is a limit of $(x_k+y_k)$ ?