Finding a norm making a subspace dense Suppose $V$ is a (real or complex) vector space and $W$ is a subspace of $V$. 

Under what conditions is there a norm on $V$ making $W$ a dense subspace of $V$?

That $V$ and $W$ have the same dimension sounds like a reasonable sufficient condition. 
 A: This is possible when the Hamel dimensions of $V$ and $W$ are the same infinite cardinal.
This is not necessary but only sufficient, since countable dimensional space can be dense in an infinite dimensional space: consider the Hamel span of a countable Hilbert basis.
Let us assume this and denote by $B$ and $C$ the bases of $V$ and $W$ so that $C\subset B$.
Partition $C$ into a collection of infinite sets indexed by the difference $D=B\setminus C$.
That is, we have
$$
C=\bigcup_{d\in D}C_d,
$$
where each $C_d$ is infinite and $C_d\cap C_{d'}=\emptyset$ whenever $d\neq d'$.
For every $d\in D$, let $Hilb(d)$ be the Hilbert space with orthonormal basis $C_d$.
We can extend this Hilbert basis to a Hamel basis.
Pick one of the new basis vectors and identify it with $d$.
Consider now the vector space
$$
X=\oplus_{d\in D}Hilb(d).
$$
This is a normed space.
The Hamel span of $C_d\cup\{d\}$ is contained in $Hilb(d)$ and the space $V$ has the Hamel basis $B=\bigcup_{d\in D}C_d\cup\{d\}$.
This makes $V$ naturally a subspace of $X$ and gives it a norm.
The Hamel span of $C_d$ is dense in $Hilb(d)$, so $W$ is dense in $X$.
Therefore $W$ is dense in $V$.
