Let $||.||_1$, $||.||_2$ be two equivalent norms in a vector space $X$.
a) Show that the Cauchy sequences in $(X, ||.||_1)$ and in $(X, ||.||_2)$ are the same
b) Show that $\lim_{n\to\infty}||x_n-x||_1=0\iff \lim_{n\to\infty}||x_n-x||_2=0$
c) Conclude that $(X, ||.||_1)$ is Banach $\iff$ $(X, ||.||_2)$ is Banach
I didn't understand question a). Is it asking for me to show that every cauchy sequence that converges in the first space is also a Cauchy sequence that converges in the second space?
b) $$\lim_{n\to\infty}||x_n-x||_1=0\iff \forall n_0,\exists n>n_0\implies ||x_n-x||_1<\epsilon$$
But $||.||_1$ and $||.||_2$ being equivalent means that there exists constants $a,b$ such that
$$a||.||_2\le ||.||_1 \le b||.||_2$$
so our condition above means
$$ \forall n_0,\exists n>n_0\implies a||x_n-x||_2 \le ||x_n-x||_1<\epsilon_2$$
and if we take $\epsilon_2 = 2\epsilon$ we have
$$ \forall n_0,\exists n>n_0\implies ||x_n-x||_2 \le \epsilon$$
which implies $\lim_{n\to\infty}||x_n-x||_2=0$
The converse is similar
c) If $(X, ||.||_1)$ is Banach, it means it is a vector normed space that is complete. Which means that every Cauchy sequence in $X$ converges in $X$ using the norm $||.||_1$. By b) we know that a sequence that converges to $x$ with norm $1$ also converges to $x$ with norm $2$. So now I should use a) somehow to say the sequences are the same
You are being asked to show that if $\{x_n\}$ is Cauchy with respect to one of the norms then it is Cauchy with respect to the other.
For instance suppose $\{x_n\}$ is Cauchy with respect to $\|\cdot\|_1$.
Let $\epsilon > 0$ be given.
There exists $N \in \mathbb N$ with the property that $n,m \ge N \implies \|x_n - x_m\|_1 < a \epsilon.$
Thus $n,m \ge N \implies \|x_n - x_m\|_2 < \epsilon$ so that $\{x_n\}$ is Cauchy with respect to $\|\cdot\|_2$.
This is unrelated to convergence since the space could fail at this point to be complete.