Topological group structure of a vector space

Remarks

The map $x \mapsto -x$ is continuous because it is scalar multiplication by $-1$. For each $a \in E$ the translation $x \mapsto x + a$ is a homeomorphism of $E$, with inverse $x \mapsto x - a$. So a set $U$ is a neighbourhood of $a$ if and only if $U - a$ is a neighbourhood of $0$. Example: $\R^n$ with its Euclidean topology is a commutative topological group under addition, since addition and negation are continuous functions of the coordinates. Non-example: give $\R$ the topology whose open sets are $\emptyset$, $\R$ and the intervals $(a, \infty)$. This is not a topological group, because $x \mapsto -x$ is not continuous: the preimage of $(0, \infty)$ is $(-\infty, 0)$, which is not open. The topological group structure gives a uniform structure on $E$; a net $(x_i)$ is Cauchy when $x_i - x_j$ tends to $0$.