Remark 2

From 1) it follows that, if B is a precompact set of K and H is a precompact set of E, then the mapping $(\lambda, x) \mapsto \lambda x$ restricted to B $\times$ H is uniformly continuous. For, if V is a neighbourhood of 0 in E then there are neighbourhoods T of 0 in K, and U of 0 in E such that TH + BU $\subset$ V. Since we can write $\lambda x - \lambda' x' = (\lambda - \lambda') x + \lambda'(x - x')$, we see that for $\lambda, \lambda'$ in B, x, x' in H, $\lambda - \lambda' \in T$ and $x - x' \in U$, we have $\lambda x - \lambda' x' \in V$, which proves our assertion.

A topological vector space is called complete if, considering its uniform structure, it is a complete uniform space.