TITLE: Topological vector space with discrete topology is the zero space
QUESTION [11 upvotes]: Hello i have a question about topological vector spaces. To remind the definition of such a space:

A topological vector space is a pair $(X,\tau)$ with $X$ a vector space and $\tau$ a topology on $X$ such that all singeltons are closed sets and the operations are continuous (thus vector addition and scalar multiplication).

I want to prove the following: Suppose $X$ is a topological vector space (over $\mathbb{R}$ or $\mathbb{C}$) with $\tau$ the discrete topology (thus all sets are open). Then $X$ is the zero space (the same also should be hold if the topology is the indiscrete one).
How can i prove this? I think i have to use the continuity of the vector space operation such as this: Suppose $x\in X$ than $x+0=x$, then for each nbhd $V$ of $x\in X$ there are nbhd's $V_1$ and $V_2$ for $x$ and $0$ respectively such that $V_1+V_2\subset V$. But how can we conclude that this $x\in X$  is the zero vector?!
Can someone help me? Thank you for help.

REPLY [7 votes]: Another argument: A topological vector space over $\Bbb{R}$ or $\Bbb{C}$ is path-connected, since the map $t\mapsto(1-t)x+ty$ is continuous. The only connected discrete space consists of one point.