Equivalence of norms and a contradiction

Question

I know the following theorem:

Let $V$ be a real (or complex) finite dimensional vector space. Then all the norms on $V$ are equivalent.

Take $V=\mathbb{R}^n$ on $\mathbb{R}$ and the metric $d_m(x,y)=\min(1,d(x,y))$, where $d$ is the usual euclidean metric.

Since $d_m$ is a metric it induces a norm $ \left\lVert\cdot \right\lVert_m=\sqrt{d_m(\cdot, \cdot)}$, whereas $d$ induces the usual euclidean norm $ \left\lVert\cdot \right\lVert$. Suppose there exist $C>0$ such that, for every $x \in \mathbb{R}^n$ we have $\left\lVert x \right\lVert\leq C\left\lVert x\right\lVert_m$. Take $x_n=(n,0,...,0)$ and we have an absurd.

What did I get wrong?

Answer 1

Note that your definition of $\|\cdot\|_m$ makes no sense, as it is always zero; I'll assume that the intention was $$\|x\|_m=\sqrt{d_m(x,0)}$$

What did I get wrong?

Since dm is a metric it induces a norm

Using your own example, $$ \|x_n\|_m=\sqrt{d_m(x_n,0)}=1\ne n\,\|x_1\|. $$ Your "norm" is not even close being a norm.