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The theorem states that all norms in $\Bbb R^n$ generate the same open sets. How can this be the case when we are talking about norms in terms of a distance/length?

Moreover, what do we mean here with "generate the same open sets"?

How would we go about to prove this theorem?

I have just started taking Advanced Math courses and thus I am a beginner. I would appreciate any help on the idea behind this theorem.

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    $\begingroup$ Topology doesn't have the notion of distance or angle. For example, a square and a circle are the same because we can define a homeomorphism from one to the other. $\endgroup$
    – John Douma
    Commented Nov 16, 2022 at 17:22
  • $\begingroup$ the idea of an open set in a finite dimensional normed space is around every point in the set, you can draw a small ball and stay inside that set. The theorem says any way we can define the meaning of the word length, we can still do that $\endgroup$
    – Alan
    Commented Nov 16, 2022 at 17:52

2 Answers 2

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When talking about open sets we are talking about topology. Now, every norm induce a metric and every metric induce a topology. This topology is called the topology induced by the metric and the open sets (the elements in the topology) consist of the open balls in the metric.

To reach your result you need to know that any two norms are equivalent in finite dimensional spaces. Since $\mathbb{R}^n$ is of dimension $n \in \mathbb{N}$, it follows that all norms are equivalent. This means that the open balls in the metric space induced by the norm via $d(x,y) = ||x-y||$ for any $x,y$ in the space, are all equivalent up to some constant. You can deduce already that the open sets in the topology will be equivalent since the balls are equivalent.

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Given a normed space $X$ and two norms $||\cdot||_1$ and $||\cdot||_2$ we say that they are equivalent if they generate the same topology. This can be proved to be equivalent to the following:

There exist $\lambda, \mu >0$ such that $\lambda||x||_1 \leq||x||_2 \leq \mu ||x||_1 \ \forall x \in X$

Having this, we can proof that all norms are equivalent, and therefore they generate the same topology.

Now we prove that every norm is equivalent to the norm of the sum $||\cdot||_1$. Let $\mathcal{B}$ be the usual base of $\mathbb{R}^n$, and set $\rho=\max\{e_k:e_k \in \mathcal{B}\}$. Given that $||\cdot||$ is a norm we have

$$ ||x||=||\sum_{k=1}^{n}x_ke_k||\leq\sum_{k=1}^n|x_k|||e_k||\leq\rho||x||_1 $$ Now, for the other inequality we consider $S=\{x \in \mathbb{R}^n:||x||_1=1\}$, which is closed and bounded and therfore compact (recall that we are in $\mathbb{R}^n$). In addition, the norm is a continuous function, so it attains its minimum in $S$. That is, $\lambda=\min\{||x||: x \in S\}>0$. For $x \in \mathbb{R}^n\setminus\{0\}$ we have $$ \frac{x}{||x||_1}\in S \Rightarrow \lambda \leq ||\frac{x}{||x||_1}|| \Rightarrow \lambda||x||_1\leq||x|| $$

We therefore have that $\lambda||x||_1\leq||x||\leq\rho||x||_1$ and every norm in $\mathbb{R}^n$ generates the same topology

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