# What is the cardinality of the set of all explicit functions?

I was just wondering the number of all explicit functions , whatever it is continuous or discontinuous, or it has a graph or not.

As far as I know, it seems that $\mathbb{R}^{\mathbb{R}}$ which denotes all functions from reals from reals has a cardinality of $\aleph_2$, with my attempt like $$card(\mathbb{R}^{\mathbb{R}})=\aleph_1 ^{\aleph_1 } =2^{\aleph_0 \cdot 2^{\aleph_0 }}<2^{2^{2^{\aleph }}}=\aleph_3$$ If we accepted GCH (the Generalized Continuum Hypothesis) the result followed trivially. But when it comes to "all" functions with no description of domain, I have literally no idea.

• Isn't $\aleph_0 \cdot 2^{\aleph_0}= 2^{\aleph_0}$? – Ranc May 23 '17 at 6:38
• $|\mathbb{R}|$ is $2^{\aleph_0}$. The statement $|\mathbb{R}| = \aleph_1$ is the continuum hypothesis, which is indecidable in ZFC. – nombre May 23 '17 at 14:25