On the cardinality of $\mathbb{N} ^{\mathbb{R}}$ and $\mathbb{R}^{\mathbb{R}}$.

Question

Actually, I have been trying to find out the cardinality of these two sets stated above. Obviously, $\mid 2^{\mathbb{R}} \mid \leq \mid \mathbb{N}^{\mathbb{R}} \mid$ , but what is the atmost value of the cardinality of the set in right hand side ??? I was thinking to differentiate between two functions from $[0,1]$ to $\mathbb{N}$ using the decimal representation of reals in $[0,1]$ , but couldn't proceed . And, please I want some hints about $\mathbb{R}^{\mathbb{R}} $ .

Answer 1

Let $|\mathbb{N}| = \aleph_0$ and $|\mathbb{R}| = 2^{\aleph_0} = \mathfrak{c}$. Using cardinal arithmetic, we have

$$ |\mathbb{N}^\mathbb{R}| = |\mathbb{N}|^{|\mathbb{R}|} = {\aleph_0}^\mathfrak{c} = {\aleph_0}^{2^{\aleph_0}} = 2^{2^{\aleph_0}} = \beth_2 $$

and

$$ |\mathbb{R}^\mathbb{R}| = |\mathbb{R}|^{|\mathbb{R}|} = \mathfrak{c}^\mathfrak{c} = (2^{\aleph_0})^\mathfrak{c} = 2^{\mathfrak{c}\,\aleph_0} = 2^\mathfrak{c} = 2^{2^{\aleph_0}} = \beth_2 $$

(also see this thread for exponentiation of cardinals.)

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If the generalized continuum hypothesis is true, then all of the different cardinalities above can be written as $\aleph$ numbers: $\mathfrak{c} = \aleph_1$ and $\beth_2 = \aleph_2$.