TITLE: Is this condition sufficient to determine the linear space is of finite dimension?
QUESTION [5 upvotes]: From the Banach theory we knew that:
1) A linear space(a vector space endowed with its vector topology) $X$ of finite dimesion $dimX=n$ has the following property: If ${\left\|  \bullet  \right\|_1}$ and ${\left\|  \bullet  \right\|_2}$ are two norms defined on $X$, then they are equivalent in the sense that $\exists C_{1},C_{2} \succ 0$ and $C_{1}{\left\|  x  \right\|_1}\leq {\left\|  x  \right\|_2} \leq C_{2}{\left\|  x  \right\|_1}$ for any $x\in X$.
Consider the converse problem now:
2) If any two norms defined on a linear space $X$ are equivalent, then can we be sure that $dimX \prec\infty$?
Moreover, we also know from 1) that any linear space $X$ of finite dimension is complete. Now consider the converse problem:
3)  If any norm defined on a linear space $X$ makes the space complete, then can we be sure that $dimX \prec\infty$?
Is this an open problem? 
Is there any reference or paper on some sort of problem like this? Or are there any easy counter-example?

REPLY [4 votes]: Both conditions imply the finite-dimensionality of $X$. If $\dim X = \kappa$ for an infinite cardinal $\kappa$, you can embed $X$ as a dense subspace into $\ell^p(\kappa) = \left\{f \colon\kappa\to\mathbb{K} : \sum_{k\in\kappa} \lvert f(k)\rvert^p < \infty\right\}$ for $1 \leqslant p < \infty$ by mapping each basis element to the "standard basis" element of the same index in $\ell^p(\kappa)$, and for example the induced $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ norms are not equivalent, since the completion in one is reflexive ($\ell^2(\kappa)$ is a Hilbert space), while the completion in the other isn't ($\ell^1(\kappa)^\ast \cong \ell^\infty(\kappa)$, and $\ell^\infty(\kappa)^\ast \not\cong \ell^1(\kappa)$ are shown similarly to the case $\kappa = \mathbb{N}$).
These embeddings also give examples of norms on infinite-dimensional spaces with respect to which the space is not complete.
The norms are naturally given by
$$\lVert f\rVert_p = \left(\sum_{k\in\kappa} \lvert f(k)\rvert^p\right)^{1/p},$$
where the sum over the (possibly) uncountably many non-negative terms $\lvert f(k)\rvert^p$ is defined as
$$\sum_{k\in\kappa} \lvert f(k)\rvert^p := \sup \left\{\sum_{k\in F} \lvert f(k)\rvert^p : F \subset \kappa \text{ finite}\right\}.$$
Of course, if the sum is finite, at most countably many terms of the sum are non-zero.