Because I know the cardinality of all the functions$f:[0,1]\to \mathbb{R}$ is $2^{c}$（c is the cardinality of continuum).I wonder whether this holds generally.
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The cardinality of all functions ${0,1}\to\mathbb{R}$ is $c^2=c$. The cardinality of all functions from $\mathbb{R}\to {0,1}$ is $2^c$. As for what you truly want to ask, the answer is affirmative. – Ningxin Jul 15 '19 at 08:08
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Yes. And this was asked before, so please search the site before asking. – Asaf Karagila Jul 15 '19 at 08:11
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@QiyuWen: Nowhere in the question do functions ${0,1}\to\mathbb R$ appear. Note that $[0,1]={x\in\mathbb R:0\le x\le 1} \ne {0,1}$. – celtschk Jul 15 '19 at 08:14
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@celtschk I did notice that, but I considered it a typo. – Ningxin Jul 15 '19 at 08:34
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The axiom of choice implies $\kappa^2=\kappa$ for transfinite cardinals $\kappa$. Then$$2^\kappa\le\kappa^\kappa\le(2^\kappa)^\kappa=2^{\kappa^2}=2^\kappa.$$By the Schröder–Bernstein theorem, we're done.
J.G.
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