Proposition 10

*Let $(E_i)_{1 \leq i \leq n}$ be a finite family of normed spaces over a non-discrete valued division ring $K$, and let $E = \prod_{i=1}^n E_i$ be the product vector space. If for each $x = (x_i) \in E$ we put $||x|| = \sup_{1 \leq i \leq n} ||x_i||$, then the function $||x||$ is a norm on $E$, and the topology it defines on $E$ is the product of the topologies of the $E_i$.

For if $x = (x_i)$ and $y = (y_i)$, we have $x + y = (x_i + y_i)$ and therefore

On the other hand, it is clear that $||tx|| = |t|.||x||$, and that if $||x|| = 0$, then $||x_i|| = 0$ and therefore $x_i = 0$ for $1 \leq i \leq n$, so that $x = 0$; hence $||x||$ is a norm on $E$. Also the relation $||x|| < a$ is equivalent to the $n$ relations $||x_i|| < a$, and therefore the norm $||x||$ defines the product topology on $E$.

Similarly we can show that the functions $\sum_{i=1}^n ||x_i||$ and $\sqrt{\sum_{i=1}^n ||x_i||^2}$ are norms on $E$; the inequalities (2) show that all three norms are equivalent.

In particular, in the (left or right) vector space $K^n$, if we put

for $x = (x_i)_{1 \leq i \leq n}$, then the three functions $p_1, p_2, p_3$ are equivalent norms which define on $K^n$ the topology which is the product of the topologies of the factors $K$.