Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be equivalent norms on a normed field. Then
(i) $\|x\|_1<1$ iff $\|x\|_2<1$; $\|x\|_1>1$ iff $\|x\|_2>1;$
(ii) $\|x\|=1$ iff $\|x\|_2=1$.
I want to prove this using the following definition,
Two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent if there exists $a,b>0$ such that $a\|\cdot\|_1\leq \|\cdot\|_2\leq b\|\cdot\|_1$. It is equivalent to saying that $1/a\|\cdot\|_2\leq \|\cdot\|_1\leq 1/b\|\cdot\|_2$.
Suppose $\|x\|_2<1$. Then $\|x\|_1< 1/b$. It is need not be true that $1/b<1$. Please help me!
Hint: $$||x||_1 < 1 \iff ||x||_1^n \to 0 \iff ||x^n||_1 \to 0 \iff ||x^n||_2 \to 0 \iff ||x||_2^n \to 0 \iff ||x||_2 <1$$
Obs: Note that both norms are multiplicative, that is
$$||x^n||= ||x||^n$$