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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 $d$?

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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.

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  • $\begingroup$ 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. $\endgroup$ Oct 3, 2021 at 1:11

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