Sequences are Cauchy depending on the norm? Assume that we have two Banach spaces $B_1, B_2$ with their underlying sets being identical. Is it possible that a Cauchy sequence in one of the spaces would fail to be Cauchy in the other? I am assuming that it is possible since the Cauchy test directly depends on the norm. However, this seems kind of unnatural, will someone please clarify.
Here are some  details to add to Jonas' answer in the link of my comment above.

Let $V$ be a vector space of Hamel dimension $c$. Let $T$ be a linear bijection from $V$ to $\ell_1$ and let $S$ be a linear bijection from $V$ to $\ell_2$.  This can be done since any two separable infinite dimensional
 Banach spaces have  Hamel dimension $c$. 

Define two norms on $V$ via 
$\Vert x\Vert_T=\Vert Tx\Vert_{\ell_1}$ and 
$\Vert x\Vert_S=\Vert S
x\Vert_{\ell_2}$. Then $(V,\Vert\cdot\Vert_S)$ is isomorphic to $\ell_2$  and  $(V,\Vert\cdot\Vert_T)$ is isomorphic to $\ell_1$. In particular, these two norms are complete. 

Since $\ell_1$ is not isomorphic to $\ell_2$, it follows that these two norms are not equivalent. Per Julien's answer, these two norms have different Cauchy sequences.
It's definitely possible. Consider the real numbers with (1) the usual metric and (2) the discrete metric. Both are complete, yet the only Cauchy sequences in (2) are the eventually constant sequences, while (1) has many more Cauchy sequences.