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Find cardinality all functions $\mathbb{N}\to P(\mathbb{N})$
$P(\mathbb{N})$ is set of all subsets of $\mathbb{N}$.
My question is about if I can use cardinal arithmetic in following way:
$|P(\mathbb{N})^{\mathbb{N}}| = |P(\mathbb{N})|^{|\mathbb{N}|}={\mathfrak c}^{\aleph_{0}}=\mathfrak{c}$, where $\mathfrak{c}$ is the cardinality of the continuum.

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    $\begingroup$ why do you have ${\mathfrak c}^{\aleph_{0}}=\mathfrak{c}$? $\endgroup$ – Jorge Fernández Hidalgo Aug 2 '16 at 16:28
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    $\begingroup$ Yes, the identities you wrote are true, but obviously all three equalities seem in need of justification. Also, what is your definition of the continuum? $\endgroup$ – Andrés E. Caicedo Aug 2 '16 at 16:31
  • $\begingroup$ On my lecture we assumed that ${\mathfrak c}^{\aleph_{0}}=\mathfrak{c}$ Main thing for me if it is true that: $|A^B| = |A|^{|B|}$ where $A^B$ is set of all functions $B\to A$ $\endgroup$ – user343207 Aug 2 '16 at 17:30
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    $\begingroup$ Yes, it's true that $| A^B | = |A|^|B|$ - by definition. $\endgroup$ – Stefan Mesken Aug 2 '16 at 23:24

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