TITLE: Sequences are Cauchy depending on the norm?
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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.