Hausdorff Topology Producing Density Let $X$ be a infinite-dimensional vector space with countable basis and $D$ be a given infinite subset of $X$.  What are the minimal requirements on $D$ for the existence of a topology $T$ on $X$ such that

*

*$(X,T)$ is Hausdorff

*D is dense in X

*$(X,T)$ is a Topological vector space.

 A: The first condition is that $D$ should not span a finite dimensional sub-space, because finite dimensional sub-spaces of Hausdorff vector spaces are all closed. Hence you need infinitely many linearly independent elements of $D$. This is the only condition. If $D$ has an infinite linearly independent subset then there is Hausdorff topology on $X$ in which $D$ is dense.
To show this lets throw out any dependencies and assume that $D$ is a countable linearly independent subset. Additionally I will assume that $D$ spans $X$, ie is a basis, which will simplify the argument a little but is not needed.
For notation $c_0(\Bbb N)$ is the space of complex sequences converging to $0$ endowed with the norm $\|v\|=\sup_{n\in\Bbb N}|(v)_n|$. We will embed $X$ into $c_0(\Bbb N)$ in a way that makes $D$ dense. We begin with a lemma:

Lemma Let $K$ be a countable linear independent sub-set of $c_0(\Bbb N)$ and $v\in c_0(\Bbb N)$. Then there is a sequence $v_n\in c_0(\Bbb N)$ with $v_n\to v$ and $\{v_n\mid n\in\Bbb N\}\cup K$ is linearly independent.

You could for example let $(v_n)_k=(v)_k$ for $k=1,...,n$ and then have a "tail" of magnitude $1/n$. There is a lot of "space" in choosing this tail so that everything remains linearly independent, simply because the "infinite" elements of $c_0(\Bbb N)$ span a vector space of uncountable dimension.
Next suppose $D$ is enumerated and let $L= \mathrm{span}_{\Bbb Q}\{d_n\mid n\in\Bbb N\}$. This is a countable set because it is a countable union of countable sets:
$$L=\bigcup_{N=1}^\infty \mathrm{span}_{\Bbb Q}\{ d_n\mid 1≤n≤N\}.$$
Additionally it will be will be dense in $X$ for any Hausdorff topology on $X$, this is because if $x= a_1 d_{n_1}+...+a_N d_{n_N}$ then $\mathrm{span}\{ d_1,...,d_{n_N}\}$ is a finite dimensional Hausdorff vector space and as such if $q_{i,k}\to a_i$ as $k\to\infty$ you have that $\sum_i q_{i,k} d_{n_i}\to \sum_i a_i d_{n_i}=x$, where $\sum_i q_{i,k}d_{n_i}\in L$.
Now we will be keeping track of cardinalities. Let $i_0:\Bbb N\to \Bbb Q$ be an enumeration. Next let $v$ be an element of $c_0(\Bbb N)$. Inductively use the Lemma to get for each $n\in\Bbb N$ a countable set of vectors $V_{0,n}$ so that $\bigcup_{n≤N}V_{0,n}$ is linearly independent and $V_n$ has a sequence converging to $i_0(n)\cdot v$.
Now let $i_1:\Bbb N\to \bigcup_{n\in\Bbb N}\Bbb Q\cdot V_{0,n}$ be an enumeration. By induction we again build countable sets $V_{1,n}$ so that $V_{1,n}$ contains a sequence approximating $i_1(k)$ and so that
$$\bigcup_{n\in\Bbb N}V_{0,n}\cup \bigcup_{n≤N}V_{1,n}$$
is linearly independent. Continue in this way, you want $V_{k,n}$ to be able to approximate the $n$-th element of $\bigcup_{n\in\Bbb N}\Bbb Q\cdot V_{k-1,n}$ for $\bigcup_{l<k, j\in\Bbb N}V_{l,j}\cup\bigcup_{j≤n}V_{k,j}$ to be linearly independent.
In the end
$$V=\bigcup_{n,k\in\Bbb N}V_{n,k}$$
will be a countable collection of linearly independent vectors in $c_0(\Bbb N)$ so that for every $v\in V$ and $q\in\Bbb Q$ there is a sequence in $V$ approximating $q\cdot v$. Now let $i: D\to V$ be a bijection. Then
$$T: X\to c_0(\Bbb N), \qquad \sum_{d\in D}a_d\ d\mapsto \sum_{d\in D}a_d i(d),$$
which is a linear bijection since the elements of $V$ are all independent. Pull back the norm of $c_0$ to $X$ via $T$ to get a Hausdorff topology on $X$ in which $D$ is dense.

In the event that $D$ should not be a basis this procedure still works. You must just also approximate the the $\Bbb Q$ span of the missing basis elements by elements of $D$, this will not break any countability constructions.
