I am working on this problem from Simmons' "Introduction to Topology and Modern Analysis" (Problem 10-6).
Let $X_1, X_2, \dots, X_n$ be a finite class of metric spaces with metrics $d_1, d_2, \dots, d_n$. If $X = X_1 \times X_2 \times \dots \times X_n$, and $d$ and $\bar{d}$ are metrics defined as follows:
$$d(\{x_i\}, \{y_i\}) = \max d_i(x_i, y_i)$$ $$\bar{d}(\{x_i\}, \{y_i\}) = \sum_{i=1}^{n} d_i(x_i, y_i)$$
then show that the two metric spaces $(X, d)$ and $(X, \bar{d})$ have precisely the same open sets.
(Here, $\{x_i\}$ denotes an n-tuple from $X = X_1 \times X_2 \times \dots \times X_n$)
Here is my approach:
I know a theorem that two metrics are equivalent iff. every open ball in the first metric contains an open ball from the second metric (centred at the same point) and vice-versa.
So, let $B_d(\{x_i\}, r)$ be and open ball in $(X, d)$. Here, $B_d(p, r)$ denotes an open ball with respect to metric $d$, centered at p and with radius $r$. Let $\{y_i\} \in B_d(\{x_i\}, r)$, now, we wish to show that $\{y_i\} \in B_{\bar{d}}(\{x_i\}, s)$ for some $s > 0$ (possibly a function of $r$?).
Now, $$\{y_i\} \in B_d(\{x_i\}, r) \implies \max d_i(x_i, y_i) < r \implies \sum_{i=1}^{n} d_i(x_i, y_i) < nr \\ \implies \{y_i\} \in B_{\bar{d}}(\{x_i\}, nr)$$
Similarly, let $\{y_i\} \in B_{\bar{d}}(\{x_i\}, r)$, then,
$$\{y_i\} \in B_{\bar{d}}(\{x_i\}, r) \implies \sum_{i=1}^{n} d_i(x_i, y_i) < r \implies \max d_i(x_i, y_i) < r \\ \implies \{y_i\} \in B_d(\{x_i\}, r)$$
So, we conclude that the metrics $d$ and $\bar{d}$ are equivalent.
I would really appreciate some comments/correction on this proof. In particular, I would like to know that when we show an open ball (with respect to first metric) having radius $r$ is contained in an open ball (with respect to second metric) having radius $s$, then, can $s$ be a function of $r$ or must it be constant depending only on the center? At this point I haven't yet studied homeomorphisms or Lipschitz equivalence of norms, so please avoid using those concepts.
Note that $\|x\|_\infty \le \|x\|_1 \le n \|x\|_\infty $.
Hence $ {d}(\{x_i\}, \{y_i\}) \le \bar{d}(\{x_i\}, \{y_i\}) \le n{d}(\{x_i\}, \{y_i\}) $
Just to clarify, if $x = (x_1,...,x_n)$ then $d(x,y) = \|(d_1(x_1,y_1),...,d_n(x_n,y_n))\|_\infty$ and $\bar{d}(x,y) = \|(d_1(x_1,y_1),...,d_n(x_n,y_n))\|_1$.