Finding a norm making a subspace dense

Question

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.

Answer 1

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 $V$.