If we look at $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a Hamel basis $B$ has cardinality $\mathcal{c}$.
What about the cardinality of the set of Hamel bases? Is it simply $2^{\mathcal{c}}$?
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This has nothing to do with the Continuum Hypothesis (see below).
Yes, the cardinality of the set of all Hamel bases for $\Bbb R$ over $\Bbb Q$ is $2^c$. It's certainly no larger than $2^c$, and for the other inequality you could use a cheap trick: If $B$ is a basis and $f:B\to\{1,-1\}$ then $\{f(b)b:b\in B\}$ is a Hamel basis.
Below Proof that $|B|=c$: Say $|B|=\kappa$. Let $\mathcal F$ be the set of finite subsets of $B$. Then $$|\mathcal F|\le\kappa+\kappa^2+\dots =\kappa+\kappa+\dots=\kappa.$$ For each $F\in\mathcal F$ the number of functions from $F$ to $\Bbb Q$ is $\aleph_0^{|F|}=\aleph_0$. Hence $$c\le\aleph_0\kappa=\kappa.$$
David C. Ullrich
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I've understood, that proving cardinality of $B$ to be $\mathcal{c}$ requires Continuum Hypothesis. Showing that $|B|>|\mathbb{N}|$ is easy. What about $|B|=\mathcal{c}$. – Rasmus Erlemann May 15 '16 at 14:54
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@REr See edit... – David C. Ullrich May 15 '16 at 15:06
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1@REr: It's generally easy to prove that if $V$ is a vector space over $F$ and $|V|>|F|$, then $\dim_FV=|V|$. See http://math.stackexchange.com/a/1267596/622 – Asaf Karagila May 15 '16 at 15:13