# Proving that $d'$ induces the usual topology on $\mathbb R^n$

Define $$d'(x, y)= \left[\sum_{i = 0}^{n}\left|x_i - y_i\right|^p\right]^{\frac 1p}$$ for $$x, y \in \mathbb R^n$$. Assume that $$d'$$ is a metric. Show that it induces the usual topology on $$\mathbb R^n$$.

So here's my idea:

1. First prove that the metric topology induced by $$d'$$ is the same as metric topology induced by square metric $$\rho$$.
2. Since square metric $$\rho$$ induces the same topology as Euclidean metric $$d$$ (there is a proof of this in the book I am using), $$d'$$ induces the usual topology.

Let $$x, y$$ be two points in $$\mathbb R^n$$, we have : $$\rho\left(x, y\right) \le d'\left(x, y\right) \le \sqrt[p]{n}\space\rho\left(x, y\right)$$ Therefore given $$\epsilon \gt 0$$, we have: $$B_{d'}\left(x, \epsilon\right) \subseteq B_\rho\left(x, \epsilon\right)$$ and $$B_\rho\left(x, \frac {\epsilon} {\sqrt[p]{n}}\right) \subseteq B_{d'}\left(x, \epsilon\right)$$ Therefore topology induced by $$d'$$ is the same as topology induced by $$\rho$$. And since the $$\rho$$-topology is the same as topology defined by Euclidean metric $$d$$, we have $$d'$$ induces the usual topology.

Am I doing this right? And how can I directly prove that $$d'$$ induces the same topology as $$d$$?

Your approach is correct. If you wanted a formal proof you would only need to add the algebra that shows $$\rho(x,y) \leq d'(x,y) \leq \sqrt[p]{n} \rho(x,y)$$. I assume that when you say $$d$$ you mean the metric $$d:\mathbb{R}^n \rightarrow \mathbb{R}$$ that is defined by $$d(x,y) = \left[ \sum_{i=1}^n |x_i - y_i|^2 \right]^{1/2}$$. As for showing that $$d'$$ induces the same topology as $$d$$ directly, there is nothing wrong with showing that it induces the same topology as $$\rho$$. I think that it gets very messy to do this directly, even for the case $$n=2$$. After all, how would you compare $$(|x_1 - y_1|^2 + |x_2 - y_2|^2)^{1/2}$$ to $$(|x_1 - y_1|^p + |x_2 - y_2|^p)^{1/p}$$ without the intermediate step of comparing them to $$\rho(x,y)$$? The reason that the $$\rho$$ metric is defined is because it is easy to compare to many other metrics that you would want to use on the Euclidean spaces.
• Thank you for the answer. Yeah, $d$ is defined like you mentioned. My book didn't specific what usual topology is so I just assumed it's topology induced by Euclidean metric. Oct 3, 2021 at 1:11