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)?
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See also [this question](http://math.stackexchange.com/q/427834/462). – Andrés E. Caicedo Jun 24 '13 at 02:44
2 Answers
It was proved by G.W. Mackey, in On infinitedimensional linear spaces, Trans. Amer. Math. Soc. 57 (1945), 155207, see Theorem I1, p.158, that an infinitedimensional 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 infinitedimensional 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 $\mathfrak{c} = \#(\mathbb{R}^{\aleph_0})$.
Added:
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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11Life 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 05 '12 at 22:25


@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 https://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](http://web.archive.org/web/20200708080750/https://people.math.ethz.ch/~halorenz/publications/pdf/hamel.pdf). – Martin Sleziak Jul 08 '20 at 08:09
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 HahnBanach, 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 HahnBanach 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 HahnBanach 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 equinumerous 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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3I think that it is worth mentioning that this is basically the proof from H. Elton Lacey's paper mentioned in the other answer: http://www.jstor.org/stable/2318458 – Martin Sleziak Nov 25 '15 at 07:09
