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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}} $ .

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  • $\begingroup$ Can please someone format this question on LateX or MathJax (as I am completely unable of doing it ) ??? $\endgroup$ – Rabi Kumar Chakraborty Jul 2 at 3:49
  • $\begingroup$ @Asaf Karaglia,Thank you Sir, for editing. $\endgroup$ – Rabi Kumar Chakraborty Jul 2 at 11:59
  • $\begingroup$ I've only removed the irrelevant large cardinals tag. The rest was done by others. $\endgroup$ – Asaf Karagila Jul 2 at 11:59
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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.)


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 $\beth_2 = \aleph_2$.

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  • $\begingroup$ GCH is irrelevant: both sets are size $2^\mathfrak{c}$. $\endgroup$ – Henno Brandsma Jul 2 at 4:16
  • $\begingroup$ True. GCH is only for writing it as an equivalent $\aleph$ number $\endgroup$ – Saswat Padhi Jul 2 at 4:17
  • $\begingroup$ It only confuses matters. Don't mention it at all. $\endgroup$ – Henno Brandsma Jul 2 at 4:18
  • $\begingroup$ I moved it to a footnote below the main answer. $\endgroup$ – Saswat Padhi Jul 2 at 4:20
  • $\begingroup$ You're mixing notations pretty badly. You have $\frak c$ and $2^{\aleph_0}$, and you bring up $\beth$ numbers into the picture... $\endgroup$ – Asaf Karagila Jul 2 at 8:14

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