Product topology and standard euclidean topology over $\mathbb{R}^n$ are equivalent I would like to know why the product topology and the standard euclidean topology over $\mathbb{R}^n$ are equivalent.
I already found the question here: 
Showing that the product and metric topology on $\mathbb{R}^n$ are equivalent
But I think in this answer it has only been proven that all norms in $\mathbb{R}^n$ are equivalent. 
I think the more important question is: Why is the product topology induced by this norm:
$$\|x\|_{\rm prod} = \max\{|x_k|, 1\le k \le n\}.$$
Can anybody help me to see this?

Look at the $\epsilon$-Balls generated by thus norm, i.e. at $$
  B^n_\epsilon(x) = \{y \,:\, \|x - y\|_p < \epsilon \} \text{.}
$$
These "balls" are $n$-dimensional rectangles, i.e. $$
  B^n_\epsilon(x) = x + (-\epsilon,\epsilon)^n \text{.}
$$
Now look at the open sets in the product topology on $\mathbb{R}^n$. This topology is generated by the base $$
  \mathcal{B} = \left\{\prod_{k=1}^n O_k \,:\, O_k \text{ open in $\mathbb{R}$}\right\} \text{.}
$$
Let $X \in \mathcal{B}$, and let $O_k$ be the corresponding open sets in $\mathbb{R}$. Then, because the $O_k$ are open, they all contain some $\epsilon$-Ball, i.e there are $x_1,\ldots,x_n$ and $\epsilon_1,\ldots,\epsilon_n$ with $B^1_{\epsilon_i}(x_i) \subset O_i$ for all $1 \leq i \leq n$. But then $$
  B_{\epsilon}^n(x) \subset X \text{ where } x = (x_1,\ldots,x_n) \text{ and } \epsilon = \min\{\epsilon_1,\ldots,\epsilon_n\}.
$$
Let conversely be $B^n_\epsilon(x)$ be some epsilon ball in $\mathbb{R}^n$. By definition, $$
  B^n_{\epsilon}(x) = \prod_{k=1}^n B^1_\epsilon(x_k) \text{ and } B^1_\epsilon(x_k) \text{ is open in $\mathbb{R}$,}
$$
and therefore $$
  B^n_\epsilon(x) \in \mathcal{B} \text{.}
$$
We have thus shown that every $\epsilon$-Ball contains an open set of the product topology, and the every set in the base $\mathcal{B}$ of the product topology contains an $\epsilon$-Ball. This proves that both generate the same topology.
