Product topology and standard euclidean topology over $\mathbb{R}^n$ are equivalent

Question

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?

Answer 1

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.

Answer 2

The topology of $\mathbb{R}$ has the set of open intervals in $\mathbb{R}$ as a basis. Since the product topology on $\mathbb{R}^n$ has $\{U_1\times U_2\times...\times U_n|U_1,...,U_n\text{are open in $\mathbb{R}$}\}$ as a basis, one can show that $\{]a_1,b_1[\times ]a_2,b_2[\times...\times ]a_n,b_n[\,|a_1,..,a_n,b_1,..,b_n\in\mathbb{R}, a_1<b_1,...,a_n<b_n\}$ is a basis for $\mathbb{R}^n$. Since the metirc topology has the open balls of the metric $||\,\,||_{prod}$ as a basis. You just need to show that an open ball of the metric $||\,\,||_{prod}$ is just a product of open intervals. Hence, the basis of the product topology contains the basis of the metric topology. Now you just need to show that the basis of the product topology is a set of open sets with respect to the metric topology induced by $||\,\,||_{prod}$