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)?


It was proved by G.W. Mackey, in On infinite-dimensional linear spaces, Trans. Amer. Math. Soc. 57 (1945), 155-207, see Theorem I-1, p.158, that an infinite-dimensional Banach space has Hamel dimension at least $\mathfrak{c} = 2^{\aleph_0}$. A short proof can also be found in H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is $c$, Amer. Math. Mon. 80 (1973), 298.

Moreover, a vector space over $\mathbb{R}$ of cardinality $\kappa \gt \mathfrak{c}$ has dimension $\kappa$ by a theorem of Löwig, Über die Dimension linearer Räume, Studia Math. 5 (1934), pp. 18–23.

Added: By combining these two facts we get the crisp statement (as given by Halbeisen and Hungerbühler in the paper Jonas linked to in a comment): “The Hamel dimension of an infinite-dimensional Banach space is equal to its cardinality.”

Finally, $\ell^1(\mathbb{R})$ embeds isometrically into $\ell^\infty(\mathbb{N})$, so its dimension is at most the cardinality of $\ell^\infty(\mathbb{N})$ which is $\mathfrak{c} = \#(\mathbb{R}^{\aleph_0})$.


To answer your question whether a Banach space $X$ of density $\mathfrak{c} = 2^{\aleph_0}$ must have dimension $\mathfrak{c}$ and whether this is a consequence of knowing the dimension of $\ell^1(\mathbb{R})$: yes.

This is because it suffices to pick a dense subset $S$ of cardinality $\mathfrak{c}$ in the unit sphere of $X$, then choose a bijection $\mathbb{R} \to S$ and send the standard basis $(e_t)_{t \in \mathbb{R}}$ of $\ell^1(\mathbb{R})$ to $S$. This map extends to a map $\ell^1(\mathbb{R}) \to X$ which is onto by the Banach–Schauder theorem (usually proved as part of the open mapping theorem: if a continuous linear map sends the unit ball of $Y$ densely into the unit ball of $X$ then it is onto).

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    $\begingroup$ Life in a universe where inf. dim. Banach spaces have a Hamel basis?? Just crazy... What's next? Ultrafilters over $\mathbb N$ and choice function on socks?? $\endgroup$ – Asaf Karagila May 5 '12 at 22:25
  • $\begingroup$ :) ${}{}{}{}{}{}$ $\endgroup$ – t.b. May 5 '12 at 22:26
  • $\begingroup$ @t.b. Could you provide an English version for the German article above? I want to read it, but I can't read German. $\endgroup$ – Ma Joad Aug 11 '19 at 13:44
  • $\begingroup$ The original link to the paper by Halbeisen and Hungerbühler was no longer working, so I replaced ti with people.math.ethz.ch/~halorenz/publications/pdf/hamel.pdf - in case this link stops working at some point, too, here is a link to the Wayback Machine. $\endgroup$ – Martin Sleziak Jul 8 '20 at 8:09

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

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

  2. 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 $c_j=0$.

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    $\begingroup$ I think that it is worth mentioning that this is basically the proof from H. Elton Lacey's paper mentioned in the other answer: jstor.org/stable/2318458 $\endgroup$ – Martin Sleziak Nov 25 '15 at 7:09
  • $\begingroup$ @MartinSleziak: This is true! $\endgroup$ – Yiorgos S. Smyrlis Nov 25 '15 at 7:19

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