Topologically equivalent metrics on a vector space need not come from equivalent norms

By [L1], the metric $d^{\mathrm{\prime }}(x,y):=\min \{d(x,y),1\}$ has the same topology as the original norm metric $d$.

Suppose $d^{\mathrm{\prime }}$ were induced by some norm $M$ on $V$. Choose a nonzero vector $v$; then $M(v)>0$, so [L2] gives $d^{\mathrm{\prime }}(tv,0)=M(tv)=\mathrm{\mid }t\mathrm{\mid }M(v)$ for every scalar $t$.

The right-hand side of step 2.1 is unbounded as $\mathrm{\mid }t\mathrm{\mid }\to \mathrm{\infty }$, but by definition $d^{\mathrm{\prime }}(tv,0)\le 1$ for every $t$. This contradiction shows that $d^{\mathrm{\prime }}$ is not induced by any norm at all.

Hence $d^{\mathrm{\prime }}$ is topologically equivalent to a norm metric while failing even to come from a norm, so it certainly does not come from a norm equivalent to the original one. This refutes the statement.