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.
Let $\Bbb K\in\{\Bbb R,\Bbb C\}.$ Since the scalar multiplication $\Bbb K\times V\to V$ is required to be continuous and $V$ has the discrete topology, the following must hold:
For each $k\in\Bbb K$ and each $v\in V$ there is an $\epsilon>0$ such that for each $l\in B_\epsilon(k)$ we have $lv=kv$.
But as there is always an $l\ne k$ in the ball around $k,$ we get $(l-k)v=0.$ Since $l-k$ is not zero, $v$ must be the zero vector.
This also shows that if $V\ne\{0\},$ then the underlying field has to be discrete.