Let $\lVert\cdot\rVert_1,\lVert\cdot\rVert_2$ be norms on vector space $X$. Prove that they generate the same topology iff they are equivalent.

Question

Note that by "generate the same topology" we mean that any set that is open with respect to $\lVert\cdot\rVert_1$ is also open with respect to $\lVert\cdot\rVert_2$ and vice versa. By "equivalent" we mean that $\exists m,M>0$ such that $m\lVert x \rVert_2\leq \lVert x \rVert_1 \leq M\lVert x \rVert_2$.

This is part of an old preliminary exam in Analysis I'm using to prep for my own prelim. I have already proven that equivalent $\implies$ same topology, but the other direction is causing me trouble. I've been attempting a reductio, supposing that for some open $A$ we have $\inf_{x\in A} \{m>0: m\lVert x \rVert_2 \leq \lVert x \rVert_1\}=0$. I don't see any contradiction here, though.

Answer 1

Let $B_{r,i}(x)$ denote the open ball of radius $r$ around $x$ according to the $i$th norm. Suppose the topologies are equivalent. $B_{1,1}(0)$ is open according to $||\cdot||_2$, so there is some $r>0$ s.t. $B_{r,2}(0)\subseteq B_{1,1}(0)$. Therefore, if $||x||_1 =1$ then $||x||_2\geq r$. So for every $x\neq 0$, $||\frac{x}{||x||_1}||_2\geq r$ hence $||x||_2\geq r||x||_1$. A similar argument yields the other inequality.