Prove "$ \lim_{n\to\infty}\|x_n\|_1 =0 \iff \lim_{n\to\infty}\|x_n\|_2 =0$" $\implies \|\cdot\|_1$ equivalent to $\|\cdot\|_2$ Let $X$ be a normed linear space and let $\|\cdot\|_i$ be arbitrary norms on $X$. Prove that the two arbitrary norms are equivalent if the following statement is true:
$$\lim_{n\to\infty}\|x_n\|_1 =0 \iff
\lim_{n\to\infty}\|x_n\|_2 =0$$

We want to prove that $\exists m,M$ such that $0<m\leq M$ such that
$$ m\|x\|_1 \leq \|x\|_2  \leq M\|x\|_1 \quad\forall  x\in X$$
By contradiction, suppose $\exists y\in X$ such that $\forall m\leq M \in\mathbb{R}_{>0}$ we have that
$$
m\|y\|_1 > \|y\|_2
\quad
and
\quad
M\|y\|_1 < \|y\|_2
$$
Then we have that
$$
m\|y\|_1 > M\|y\|_1
$$
which is a contradiction. 
QED.
Now I know this must be wrong, because I did not use the statement at all. Where is the mistake and how do I fix this?
 A: The negation of: There exists a $0<m\leq M$ for every $x\in X$ such that
$$m\|x\|_1 \leq \|x\|_2  \leq M\|x\|_1.$$
is: For every $0<m\leq M$ there exists an $x\in X$ such that
$$m\|x\|_1 > \|x\|_2\quad\color{red}{\text{or}}\quad\|x\|_2  > M\|x\|_1.$$
A: You are close. Contradiction is the right approach, but make sure you negate the quantifiers precisely.
That is, if the norms are not equivalent, for all $m$ and $M$, there exists a $y_{m,M} \in X$ depending on $m$ and $M$ such that 
$$m||y_{m,M}||_1 > ||y_{m,M}||_2$$
or
$$||y_{m,M}||_2 > M ||y_{m,M}||_1.$$
Note the order of the quantifiers, their dependencies, and that we have an OR statement, not an AND statement.
So now we need to use the hypothesis. 
First, reformulate our assumed negation. For any interval $[m, M]$, there exists an $y_{m, M}$ such that $$||y_{m,M}||_2/||y_{m,M}||_1 \in [0, m) \cup (M, \infty)$$
Pick the intervals $(1/n,n)$ to get a sequence $x_n$ with 
$$||x_n||_2/||x_n||_1 \in [0, 1/n) \cup (n, \infty)$$
Hence, there is a convergent subsequence such that $||x_n||_2/||x_n||_1$ goes to zero or to infinity. Or alternatively, either
$$||x_n||_2/||x_n||_1 \to 0$$
or
$$||x_n||_1/||x_n||_2 \to 0.$$
In the former case, compare the sequence $y_n=x_n/||x_n||_1$ with respect to the original hypothesis and see how we get a contradiction. In the latter case, do something similar. Do you follow?
A: Denying that $\exists m,M$ such that $0<m\leqslant M$ and that$$(\forall x\in X):m\|x\|_1 \leqslant \|x\|_2  \leqslant M\|x\|_1\tag1$$is to assert for every $m,N>0$ such that $m\leqslant M$, there is some $x$ such that$$m\|x\|_1>\|x\|_2\text{ or }\|x\|_2>M\|x\|_1.$$That is not what you wrote.
If the norms are equivalent, take $m,M>0$ such that $(1)$ holds. If $\lim_n\|x_n\|_1=0$, then $\lim_n\|x_n\|_2=0$ because$$(\forall n\in\mathbb{N}):\|x\|_2\leqslant M\|x_n\|_1.$$And if $\lim_n\|x_n\|_2=0$, then $\lim_n\|x_n\|_1=0$ because$$(\forall n\in\mathbb{N}):\|x\|_1\leqslant\frac1m\|x_n\|_2.$$
Now, suppose that the norms are not equivalent. Consider the set$$S=\left\{\frac{\|x\|_2}{\|x\|_1}\,\middle|\,x\in X\setminus\{0\}\right\}.$$Then either $\inf S=0$ or $\sup S=+\infty$. If $\inf S=0$ then, for each $n\in\mathbb N$, take $y_n\in X\setminus\{0\}$ such that $\frac{\|y\|_2}{\|x\|_1}<\frac1n$. This means that $\|y_n\|_2<\frac1n\|y_n\|_1$. Let $x_n=\frac{y_n}{\|y_n\|_1}$. Then$$\|x_n\|_1=1\text{ and }\|x_n\|_2<\frac1n.$$Therefore, $\lim_n\|x_n\|_2=0$, whereas $\lim_n\|x_n\|_1=0$.
The case in which $\sup S=+\infty$ is similar.
