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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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+1 because I was about to ask the same question! –  user23211 May 15 '12 at 21:35
    
See also this question. –  Andres Caicedo Jun 24 '13 at 2:44
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1 Answer 1

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).

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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
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