# Topological vector space with discrete topology is the zero space

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.

• Is this true? $\mathbb{Z}/2\mathbb{Z}$ is a topological vector space (one dimensional) with discrete topology. – Dan Rust Sep 13 '13 at 10:33
• Also if you use continuity of addition and scalar multiplication? – Local Base Sep 13 '13 at 10:44
• It's a discrete space so any maps from $\mathbb{Z}/2\mathbb{Z}$ and its product with itself are continuous. – Dan Rust Sep 13 '13 at 10:47
• right ... hmmm ... And if you look just to vector spaces over the reals or the complex numbers? – Local Base Sep 13 '13 at 10:50
• That may be different (I'm not sure yet). Feel free to edit your question to only consider those cases of TVSs. – Dan Rust Sep 13 '13 at 10:51

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.

• What exactly is the meaning of $B_\epsilon$ in a topological vector space? And anyway, isn't every topological module over a path-connected ring (with unity) trivially contractible? – Niels J. Diepeveen Aug 29 '15 at 2:59
• $B_\epsilon(k)$ is a ball in the field $\Bbb K$ (which we assume to be either $\Bbb R$ or $\Bbb C$), so this is well-defined @NielsDiepeveen – Stefan Hamcke Aug 29 '15 at 9:20
• Pardon my perhaps stupid question, but I don't understand why "the following" must hold. How does this follow from continuity of scalar multiplication? – Adomas Baliuka Sep 20 '16 at 19:35
• Continuity of the scalar multiplication implies that the map $\mu_v: \mathbb K \to V$ which sends $k$ to $kv$ is continuous, as this is the composition $\mathbb K \to \mathbb K \times \{v\} \hookrightarrow \mathbb K \times V \to V$, where the first map sends $k$ to $(k,v)$, and this holds for any $v\in V$. In particular, continuity at $k$ means that for any neighborhood $W$ around $kv$, there is a neighborhood $U$ around $k$, and thus an $\varepsilon>0$, such that $\mu_v(B_\epsilon(k)) \subseteq \mu_v(U) \subseteq W$. Now simply take $W=\{kv\}$. @AdomasBaliuka – Stefan Hamcke Sep 20 '16 at 21:07

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.

• Nice. The reason this works is that $[0,1]$ inherits the topology of $\mathbb{R} \subseteq \mathbb{C}$, making the map you wrote down continuous. – QuantumSpace Feb 21 '20 at 12:01