Can a vector space be complete for two non-compatible norms? Let $V$ be a vector space and suppose that we have two non-compatible norms on it,
i.e. I distinguish  $E = (V, \|\cdot\|_1)$ from $F = (V, \|\cdot\|_2)$ and I ask 
that
$\not\exists C>0 \; \forall x\in V:\quad  \| x \|_1 \le C \|x \|_2$
and
$\not\exists C>0 \; \forall x\in V:\quad  \| x \|_2 \le C \|x \|_1$.
Is it possible that $E$ and $F$ are both complete? Of course, the identity map $I: E\to F$ is bijective, but from this I cannot infer anything by the open-mapping theorem, do I? (this is no home-work question :)

Yes, that is possible. Consider $E = (\ell^1,\lVert\,\cdot\rVert_1)$. Its algebraic dimension (and its cardinality) is $2^{\aleph_0}$, as is the case for $\ell^\infty$. Thus there is a (discontinuous) linear isomorphism $f\colon \ell^1\to\ell^\infty$. Transport the norm of $\ell^\infty$ to $\ell^1$ via $f$:
$$\lVert x\rVert_2 := \lVert f(x)\rVert_\infty.$$
Then $E$ and $F = (\ell^1,\lVert\,\cdot\,\rVert_2)$ are both complete, and the two norms are incomparable ($\ell^1$ is separable in the $\lVert\,\cdot\,\rVert_1$-norm, $\ell^\infty$ is not separable in the $\lVert\,\cdot\,\rVert_\infty$-norm, so the open mapping theorem guarantees that neither $\lVert\,\cdot\,\rVert_1$ nor $\lVert\,\cdot\,\rVert_2$ generates a finer topology than the other).

but from this I cannot infer anything by the open-mapping theorem

You can infer that any two norms in which $V$ is complete are either equivalent or incomparable. If the two are comparable, the open mapping theorem asserts that the identity is a topological isomorphism, i.e. the two norms are equivalent.
