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

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 $$\mathbb{R}^{\mathbb{R}}$$ .

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

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 $$\beth_2 = \aleph_2$$.

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