# Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$

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).

• 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?? – Asaf Karagila May 5 '12 at 22:25
• :) ${}{}{}{}{}{}$ – t.b. May 5 '12 at 22:26
• @t.b. Could you provide an English version for the German article above? I want to read it, but I can't read German. – Ma Joad Aug 11 '19 at 13:44
• 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. – Martin Sleziak Jul 8 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 $$c_j=0$$.

• 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 – Martin Sleziak Nov 25 '15 at 7:09
• @MartinSleziak: This is true! – Yiorgos S. Smyrlis Nov 25 '15 at 7:19