Every normed space with countable dimension is separable. I am trying to figure out if this reult from functional analysis is true. "Every normed space with countable Hamel dimension is separable." I know that this hold if the space is of finite dimension and i know that proof using the density of rationals in R, also I am pretty sure that this is true(for the space being of countable dimension) and that the proof of it goes quite similarly as if the space was finite dimensional, with small changes. I just wanted some help for confirmation, not even the proof, just if this is indeed true.
 A: Let $X$ be a normed space with infinite countable Hamel basis $\{e_n\}_{n=1}^\infty.$ Then $$X= \bigcup_{n=1}^\infty X_n,\qquad X_n={\rm span}\{e_1,e_2,\ldots, e_n\},\qquad \dim X_n=n$$
Let $A_n$ be a countable dense subset in $X_n.$ Then $A=\displaystyle\bigcup_{n=1}^\infty A_n$ is a countable set. Moreover $A$ is dense in $X.$ Indeed, every element $x\in X$ belongs to $X_{n_0},$ for some $n_0.$ Hence it can be approximated by the elements of $A_{n_0}\subset A.$
A: Consider a sequence $(e_n)$ that is basis of $\mathrm{V},$ your normed (vector) space. Then, consider the $\mathbf{Q}$-span of this basis, meaning the set of all linear combinations (by definition, finite sums) with scalars in $\mathbf{Q}$ of the basis $(e_n).$ Call this $\mathbf{Q}$-span $\mathrm{D}.$
Claim 1. $\mathrm{D}$ is dense in $\mathrm{V}.$
Proof. Let $\varepsilon > 0.$ Since $(e_n)$ is a basis, then every $v$ can be written as $v = \sum\limits_{i \in \mathrm{F}} \alpha_i e_i$ for some finite set $\mathrm{F} \subset \mathbf{N}.$ You can choose rational numbers $\beta_i$ such that $\delta = \max\limits_{i \in \mathrm{F}} |\beta_i - \alpha_i| \leq \dfrac{\varepsilon}{\sum\limits_{i \in \mathrm{F}} \|e_i\|}.$ Then,
$$
\left\| \sum_{i \in \mathrm{F}} \alpha_i e_i - \sum_{i \in \mathrm{F}} \beta_i e_i \right\| \leq \delta \sum_{i \in \mathrm{F}} \|e_i\| \leq \varepsilon.
$$
Since $\sum\limits_{i \in \mathrm{F}} \beta_i e_i \in \mathrm{D},$ we are done. QED
Claim 2. $\mathrm{D}$ is countable.
Proof. We can construct an injective function $\varphi:\mathrm{D} \to \bigcup\limits_{k = 1}^\infty \mathbf{Q} e_1 \times \cdots \times \mathbf{Q} e_k$ given as follows: every $d \in \mathrm{D}$ has a unique expansion $d = \sum\limits_{i \in \mathrm{F}} r_i e_i,$ where $\mathrm{F} \subset \mathbf{N}$ is finite and the $r_i \neq 0$ are rational. Consider $k = \max \mathrm{F}$ and then $\varphi(d) = (s_1, \ldots, s_k)$ where $s_j = 0$ if $j \notin \mathrm{F}$ and $s_j = r_j$ otherwise. As $\varphi$ is injective, $\mathrm{D}$ is countable (since the codomain of $\varphi$ is countable being a countable union of finite Cartesian products of countable sets). QED
Comments.

*

*Every countable-basis normed space cannot be complete. A fortiori, every infinite-dimensional complete normed space has an uncountable Hamel basis.

*Every countable-basis normed space is akin to some $c_{00}$ and are all-things-considered uninteresting. Indeed, write $\mathrm{V}_n = \langle e_1, \ldots, e_n \rangle,$ so that $\mathrm{V}_n \subset \mathrm{V}_{n+1},$ the dimension of the $\mathrm{V}_n$ increase one by one, and $\mathrm{V} = \bigcup\limits_{n \in \mathbf{N}} \mathrm{V}_n.$ Then, $v \in \mathrm{V}$ means $v \in \mathrm{V}_n$ for some $n,$ if you consider the coordinates relative to the basis $(e_n),$ you see at once that the sequence of coordinates of $v$ is finally zero (i.e. the coordinates of $v$ belong to $c_{00}$).

*Claim 2 does not generalise to uncountable-basis normed spaces while Claim 1 does.

