Let $X$ be a topological $\mathbf{R}$-vector-space (not necessarily Hausdorff) and $x_0 \neq 0$ a non-zero element of $X$. Then intuitively the sequence $(\frac{1}{n} x_0)_{n \in \mathbf{N}}$ converges to the origin.
This would be easy to prove, if one could show that the origin in any TVS would have a fundamental system of neighborhoods which are absorbing. But is this really correct for any TVS?
Thanks!
The map $\mathbf{R}\times X\to X$, $(r,x)\mapsto x$, is continuous by assumption, with $\mathbf{R}$ given the usual topology. As a consequence, the map $\mathbf{R}\to X$, $r\mapsto rx_0$, is continuous. Since the sequence $1/n$ converges to $0$, you have $$ \lim_{n\to\infty}\frac{1}{n}x_0=0x_0=0 $$ Well, if $X$ is not Hausdorff, $\lim$ is used in the sense that the sequence converges to $0$ (and may converge to other points, actually to all points in the closure of $\{0\}$).