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As far as I understand, $2^{\aleph_0}$ is the cardinality of the real numbers (and whether this equals $\aleph_1$ is the continuum hypothesis). But would $2^{2^{\aleph_0}}$ be of a higher cardinality than the cardinality of the real numbers?

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    $\begingroup$ The powerset of the continuum is bigger than the continuum $\endgroup$ May 26, 2016 at 13:59

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Yes, you are right. Generally, we have $2^\kappa > \kappa$ for any cardinal number $\kappa$.

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