1. ## compatible

Call a metric $d$ compatible on $\mathbb{R}^{n}$ with the vector space structure if $d(x+z, y+z) = d(x,y)$ for all $x,y,z \in \mathbb{R}^{n}$ and $d (\lambda x, \lambda y) = \lambda d(x,y)$ for $\lambda \in [0, \infty)$. Prove that any such metric puts the same topology on $\mathbb{R}^{n}$ as the Euclidean metric.

2. So when metric satisfies the conditions we can define the norm the following way: $\|x\| = d(x, 0)$. Hence $\|x-y\| = d(x, y)$

Norms on finite dimensional vector space are equivalent. So the metric puts the same topology as euclidian metric (which is generated by euclidian norm).

So the point is really to prove that all norms are equivalent (on finite dimensional space). To do that we can prove that every norm is equivalent to maximum norm defined: $\|x\|_1 = \max_{1 \leq i \leq n}\{x_i\}$.

You can easily show that: $\|x\| \leq C \|x\|_1$ (for every x). Hence convergence in $\| \cdot \|_1$ implies convergence in $\| \cdot \|$.

So let's assume $\| \cdot \|$ is not equivalent with $\| \cdot \|_1$. That means that for every natural number we can find $x_n$ such that:
$n\|x_n\| < \|x_n\|_1$

Let's define the sequence: $y_n = \frac{x_n}{\|x_n\|_1}$. From the above inequality we see that $\|y_n\| \rightarrow 0$,

What is more: $\|y_n\|_1 = 1$ hence the sequence lies on the unit sphere. In finite dimensional space unit sphere is compact. Hence $y_n$ has convergent subsequence $y_{k_n} \rightarrow y$.

We can see that $y \neq 0$ because $\|y\|_1 = \lim \|y_n\|_1 = 1$. But convergence in $\| \cdot \|_1$ implies convergence in $\| \cdot \|$. Hence $y = 0$. It's a contradiction.

Thus all norms in finite-dimensional space are equivalent.