One dimensional vector space and not Hausdorff I read that all vector spaces that do not have the Hausdorff property and are one-dimensional need to have the trivial topology. I am not quite sure how to approach this problem, but I would like to get an idea, how we can show this?-Somehow I assume that this problem should not be too hard, so I would be glad about any possible hint.
Thus, either full solutions or ideas are highly appreciated.
 A: Suppose $E$ is a topological vector space over a field $K$ who does not have trivial topology, i.e there is an open set $O$ different from $E$ and $\emptyset$. Choose $x\in O$ and $y\notin O$. Consider, for all $k\in K^{*}$, the continuous function $f_{k}:v\mapsto x+k^{-1}(v-x)$. Then $O_{k}=f_{k}^{-1}(O)$ is open, contains $x$, and does not contain $x+k(y-x)$ since $f_{k}(x+k(y-x))=y$.
If $E$ is one-dimensional, every $z\in E\setminus\{x\}$ can be written $x+k_{0}(y-x)$ for some $k_{0}$, and then $O_{k_{0}}$ separates $x$ from $z$. This shows $x$ can be separated from every other point in $E$. To show that for general $x'\in E$ do the reasoning above again with $O'= O+(x'-x)$, which is open again by continuity of addition.
A: Let's make the framework explicit. We have a topological field $K$ and a topological vector space $V$; this means that the addition operations $K\times K\to K$ and $V\times V\to V$, as well as the multiplication operations $K\times K\to K$ and $K\times V\to V$, are continuous. Also taking opposites in $K$ and $V$ as well as $\alpha\to\alpha^{-1}$ defined on $K\setminus\{0\}$ must be continuous.
By continuity, the neighborhoods of $x\in V$ are the subsets of the form $x+U$, where $U$ is a neighborhood of $0$ (the map $v\mapsto x+v$ is a homeomorphism).
If $\{0\}$ is closed in $V$, then $V$ is Hausdorff. Indeed, let $x\ne 0$ and consider a neighborhood of $0$ such that $x\notin U$. Then there exists a neighborhood $U'$ of $0$ such that $U'-U'\subseteq U$. If $v\in U'\cap(x+U')$, then $v=x+v'$ for some $v'\in U'$, so $x=v-v'\in U'-U'$, which is a contradiction.
Thus $U'\cap(x+U')=\emptyset$. Using translations, we have proved that $V$ is Hausdorff.
Now, again by continuity, the closure of a subspace is again a subspace (use nets, as it's easier).
If $V$ is one dimensional and $\{0\}$ is not closed, its closure must be $V$. Thus the only closed set containing $0$ is $V$. So the only open set not containing $0$ is the empty set. If $A$ is a proper open set and $v\notin A$, then $0\notin -v+A$, so $-v+A=\emptyset$ and $A=\emptyset$.
