TITLE: Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$
QUESTION [31 upvotes]: What is the cardinality of a Hamel basis of $\ell_1(\mathbb R)$? Is it deducible in ZFC that it is seemingly continuum? Does it follow from this that each Banach space of density $\leqslant 2^{\aleph_0}$ has a Hamel basis of cardinality continuum (OK, I do know it cannot be smaller for an inf.-dim. Banach space)?

REPLY [11 votes]: Theorem. Let $X$ be an infinite dimensional Banach space. Then $\,\mathrm{dim}\,X\ge 2^{\aleph_0}$.
Sketch of Proof. Based on M.G. McKay's proof. Let $\{w_n : n\in\mathbb N\}\subset X$ be a linearly independent set.
Step A. Using Hahn-Banach, we shall construct another linearly independent set
$\{v_n : n\in\mathbb N\}\subset X$, and a set of linear functionals $\{v^*_n:n\in\mathbb N\}\subset X^*$, such that
$$
\mathrm{span}\{v_1,\ldots,v_n\}=\mathrm{span}\{w_1,\ldots,w_n\}, \quad
\text{for all $n\in\mathbb N$,}
$$
$\|v_i^*\|=1$, for all $i\in\mathbb N$, and
$$
v_i^*(v_j)=\delta_{ij}, \quad \text{for all $i,j\in \mathbb N$.}
$$
This is done inductively. Define $v_1=w_1/\|w_1\|$, and $v_1^*(v_1)=1$,
and extend, using Hahn-Banach to $X$, so that $\|v_1^*\|=1$. Assume that
$v_1,\ldots,v_k$ and $v_1^*,\ldots,v_1^*$, have been defined so that
$$
\mathrm{span}\{v_1,\ldots,v_k\}=\mathrm{span}\{w_1,\ldots,w_k\},\quad
\|v_i^*\|=1\,\,\text{and}\,\,\,v_i^*(v_j)=\delta_{ij}, \quad 
\text{for all $\,i,j=1,\ldots,k.$}
$$
Then let
$$
v_{k+1}=w_{k+1}-\sum_{j=1}^k v_j^*(w_{k+1})v_j.
$$
Clearly, $\,v_{k+1}\in \bigcap_{j=1}^k\mathrm{ker}\,v_j^*$. Next, we
define the functional $v_{k+1}^*$,
so that $v_{k+1}^*(v_j)=\delta_{k+1,j}$, for $j=1,\ldots,k+1$, and extend it via Hahn-Banach to $X$, and in order to keep its norm unit we suitably rescale $v_{k+1}$.
Step B. It is possible to define a subset $\mathcal S$ of $\mathcal P(\mathbb Q)$, such that

$|\mathcal S|=|\mathcal P(\mathbb Q)|=2^{\aleph_0}$, and

If $A,B\in\mathcal S$, and $A\ne B$, then $\,\rvert A\cap B\rvert<\aleph_0$.


For example, if for every $r\in\mathbb R\setminus\mathbb Q$, we set $A_r\in\mathcal P(\mathbb Q)$ the set of the elements of a sequence of rationals converging to $r$, then $A_r\cap A_{r'}$ is a finite set, whenever $r\ne r'$.
Next, let $\mathbb Q=\{q_n\}_{n\in\mathbb N}$, and set
$$
u_r=\sum_{q_n\in A_r}2^{-n}v_n, \quad r\in\mathbb R\setminus\mathbb Q.
$$
It is readily shown that the set
$U=\{u_r:r\in\mathbb R\setminus\mathbb Q\}\subset X$ is  equi-numerous to $\mathbb R$. It remains to show that $U$ linearly independent. Let $u_{r_1},\ldots, u_{r_k}\in U$, and assume that
$$
c_1u_{u_{r_1}}+\cdots+c_ku_{u_{r_k}}=0, 
\quad\text{for some $c_1,\ldots,c_k\in\mathbb R$.}
$$
For an arbitrary $j=1,\ldots,k$, since $A_{r_j}\cap A_{r_i}$ is finite whenever $i\ne j$, then $A_{r_j}\setminus\bigcup_{i\ne j}A_{r_i}\ne\varnothing$. Let $q_\ell\in A_{r_j}\setminus\bigcup_{i\ne j}A_{r_i}$. Then $v^*_\ell(u_{r_i})=0$, for all $i\ne j$, while  $v_\ell^*(v_j)=2^{-\ell}$, and hence
$$
0=v^*_\ell\big(c_1u_{u_{r_1}}+\cdots+c_ku_{u_{r_k}}\big)=c_jv^*_\ell(u_{r_j})
=2^{-\ell}c_j,
$$
and thus $c_j=0$.