Intuition for the construction of the product topology and its equivalence to the euclidian metric

Question

While I have been provided a proof for the previous statement, I still cannot fully grasp why the euclidian metric [ $d(x,y)=((x_1-y_1)^2+...(x_{n}-y_{n})^2)^{1/2}$] generates the same topology as the product topology.

I'm requesting a description for the basis elements for the product topology, as well as an alternative definition from those provided in wikipedia and Munkres' topoology. I do not understand how it is constructed, and why it generates the same topology as the euclidian metric.

Answer 1

Dont't think of finest or coarsest topologies with certain properties. Instead just look at the problem in an intuitive geometric way.

Given that the topology on ${\mathbb R}$ is "generated" by open intervals $\ ]a,b[\ $ the product topology on ${\mathbb R}^n$ is generated by open boxes, i.e. cartesian products $\prod_{i=1}^n\>]a_i,b_i[\>$ of open intervals. A set is open, if it is the union of such boxes. Since any finite intersection of such boxes is (empty or) again such a box it follows that we have indeed a topology on ${\mathbb R}^n$.

In order to prove that this topology coincides with the topology coming from the euclidean metric it is sufficient to show that any euclidean neighborhood of any point $p$ (it is sufficient to consider $p=0$) is also a box neighborhood of $p$, and vice versa.

I'll do one half of the proof for you: Given an euclidean neighborhood $U$ of $0$ there is an $\epsilon>0$ such that $0\in U_\epsilon(0)\subset U$. The open box $$B:=\left\{x=(x_1,\ldots, x_n)\>\biggm|\>|x_k|<{\epsilon\over\sqrt{n}} \ (1\leq k\leq n)\right\}$$ is a neighborhood of $0$ with respect to the product topology. For each $x\in B$ one has $|x|^2<n{\displaystyle{\epsilon^2\over n}}$, or $|x|<\epsilon$. This shows that in fact $$0\in B\subset U_\epsilon(0)\subset U\ ;$$ in other words: $U$ is also a neighborhood of $0$ with respect to the box topology. Since $U$ was arbitrary we are done.

Answer 2

This particular problem is due I think to an excessive focus on topology at the expense of geometry, which sheds more light on the matter. The point here is that any two norms on a finite-dimensional vector space are equivalent. It follows in particular that the box topology is the same as the topology defined by the Euclidean metric.