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)?
Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$
-
0See also [this question](http://math.stackexchange.com/q/427834/462). – 2013-06-24
2 Answers
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 $\mathfrak{c} = \#(\mathbb{R}^{\aleph_0})$.
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).
-
0:) ${}{}{}{}{}{}$ – 2012-05-05
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 $v_{k+1}=w_{k+1}-\sum_{j=1}^k v_j^*(w_{k+1})v_j\in \bigcap_{j=1}^k\mathrm{ker}\,v_j^*$, and 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 to keep its norm unit we suitably rescale $v_{k+1}$.
Step B. Let $\mathcal S\subset\mathbb P(\mathbb N)$, such that that: If $A,B\in\mathcal S$, and $A\ne B$, then $\,\rvert A\cap B\rvert<\aleph_0$, while $\lvert\mathcal S\rvert=2^{\aleph_0}$. Then define $ v_A=\sum_{n\in A}2^{-n}v_n, \quad A\in\mathcal S. $ It is readily shown that the set $\{v_A:A\in\mathcal S\}\subset X$ is linearly independent and equi-numerous to $\mathbb R$.
-
0@MartinSleziak: This is true! – 2015-11-25