Let $(X_i,||\cdot||_i)$ be normed spaces where $i=1,2,…,n \in \Bbb Z_+$.
Let $X$ be the product space of these normed spaces equipped with product topology.
For $x=(x_1,x_2,…,x_n) \in X$ where $x_i \in X_i$ $(i=1,2,…,n)$ set:
$$||x|| := \max \{||x_1||_1,||x_2||_2,…,||x_n||_n\}$$
How to prove that this norm $||\cdot||$ generates the product topology on $X$?
Best Answer
This is true for arbitrary metric spaces of which normed spaces are just a special case.
For let $U = U_1 \times \cdots \times U_n$ be a basis element of the product space $X$ where each $U_k$ is open in $X_k$. And let $x = (x_1, \ldots, x_n)$ be a point in $U$. Then you can show that there is an open $d$-ball around $x$ contained inside $U$.
Indeed note that as each $U_k$ is open there are open $d_k$-balls $B_{d_k}(x_k, r_k) \subseteq U_k$ around each component $x_k$ in $X_k$. Now let $r = \min\{r_1, \ldots, r_n\}$. Then you can verify that $B_d(x, r)$ is an open $d$-ball around $x$ in $X$ such that $B_d(x, r) \subseteq U$.
Conversely let $B_d(x, r)$ be an open $d$-ball around $x$ in $X$. You can show that there is a basis element $U = U_1 \times \cdots \times U_n$ around $x$ that is contained inside that open ball.
Indeed just let $U = B_{d_1}(x_1, r) \times \cdots \times B_{d_n}(x_n, r)$ and we are done.
Hence every open set with respect to the product topology on $X$ is also an open set with respect to the above $d$-metric on $X$. And every open set with respect to the same $d$-metric on $X$ is also an open set with respect to the product topology on $X$ as expected.