TITLE: Showing that the product and metric topology on $\mathbb{R}^n$ are equivalent
QUESTION [8 upvotes]: I'm new to topology, and can't figure out why the metric and product topologies over $\mathbb{R}^n$ are equivalent.  Could someone please show me how to prove this?

REPLY [7 votes]: The product topology is induced by this norm.
$$\|x\|_{\rm prod} = \max\{|x_k|, 1\le k \le n\}$$
Let us use $\|\cdot\|$ for the Euclidean norm.  Then
$$\|x\| = \left(\sum_{k=1}^n x_k^2\right)^{1/2}\le \left(\sum_{k=1}^n \|x\|_{\rm prod}^2\right)^{1/2} = \|x\|_{\rm prod}\sqrt{n}.$$
Now for a reverse inequality.
We have 
$$|x_k |\le  \left(\sum_{k=1}^n x_k^2\right)^{1/2}, \qquad 1\le k \le n,$$
so $$\|x\|_{\rm prod} \le \|x\|.$$
The norms are equivalent.