I know and have been able to prove that, if $L$ is a linear variety of a normed space $X$, i.e. a vector subspace $L$ of $X$ regarded as a vector space, then its closure $\bar{L}$ with respect to the topology induced by the norm is a linear variety too.
I wonder whether that is true for any topological vector space $X$ and its linear varieties, but cannot find any information about it. If that were true, can anybody write or link a proof?
Thank you very much!!!
We can simplify life a little by dealing with the subspace $S = L - \{l\}$ (for any $l \in L$) instead.
So the question becomes if $S$ is a subspace, is $\bar{S}$ a subspace?
The key facts are continuity of scalar multiplication and addition.
Suppose $x \in \bar{S}$ and $\lambda \neq 0$. Suppose $U$ is open with $\lambda x \in U$. Then ${1 \over \lambda } U$ is an open set containing $x$ and so there is some $s \in {1 \over \lambda } U \cap S$ from which it follows that $\lambda s \in U$. Since $U$ was arbitrary, we have $\lambda s \in \bar{S}$.
Now suppose $x,y \in \bar{S}$ and $U$ be open with $x+y \in U$. By continuity of addition, we have open $U_x,U_y$ such that $U_x+U_y \subset U$. Hence we have $s_x \in U_x\cap S, s_y \in U_y\cap S$, and so $s_x+s_y \in U$. Since $U$ was arbitrary, we have $x+y \in \bar{S}$.