Bijection between $\mathbb R^\mathbb N$ and $\mathbb R$ $\mathbb R^\mathbb N$ is the set of all functions from the naturals to the reals.
I have to prove that $\mathbb R^\mathbb N$ has the same cardinality as $\mathbb R$. I found an injective function from $\mathbb R$ to $\mathbb R^\mathbb N$, but can't seem to find one for $\mathbb R^\mathbb N$ to $\mathbb R$.
 A: It is sufficient to show that there is a bijection between $[0,1]$ and $[0,1]^\mathbb N$. Let $r \in [0,1]$ and take the decimal form $r = 0.a_1 a_2 ...$ where $(a_n)$ is a sequence in $\{0,1,...,9 \}^\mathbb N$. Then take an infinity of disjoints subsequences of disjoints integers $(n^1_k)_{k \in \mathbb N}$, $(n^2_k)_{k \in \mathbb N}$,... You can form an infinity of reals written in decimal form $r_j = 0.a_{n^j_1} a_{n^j_2}...$, one for each sequence $(n^j)_{k \in \mathbb N}$, $j \in \mathbb N$. Here's your bijection :).
A: Just an augmentation to @francis-jamet's answer with a solution that avoids cardinal arithmetic and instead uses CSB: the equality of cardinalities to be shown is $|[0,1]^\mathbb N|=|[0,1]|$. An injection $[0,1]\to [0,1]^\mathbb N$ is clear. To form an injection in the other direction, consider a sequence $(r_i)_{i\in \mathbb N}$ with $r_i\in [0,1]$. Write the $r_i$ in an infinite column and expand each in decimal form without repeating '9's. Traverse the digits as in the proof of the countability of $\mathbb N \times \mathbb N$ to form a single real number in $[0,1]$. This is an injection $[0,1]^{\mathbb N}\to [0,1]$.
A: $\mathbb{R} \cong 2^{\mathbb{N}}$ and $\mathbb{N}\times \mathbb{N} \cong \mathbb{N}$.
So $\mathbb{R}^{\mathbb{N}} \cong ( 
    2 ^      { \mathbb{N}})^{\mathbb{N}} \cong 2^{\mathbb{N} \times \mathbb{N}} \cong 2^{\mathbb{N}} \cong \mathbb{R}$
A: More generally, if $\kappa,\lambda$ are cardinals and $\kappa$ is infinite, then $\mu:=\lambda^{\kappa}$ satisfies $\mu^{\aleph_0}=\mu$. The complete answer is given in Jech's Set theory: The exponential function $x \mapsto x^{\aleph_0}$ on cardinals is constant on each interval $\kappa \leq x \leq \kappa^{\aleph_0}$. This means that if $\lambda<\kappa$ with $\lambda^{\aleph_0} \geq \kappa$, then $\kappa^{\aleph_0}=\lambda^{\aleph_0}$. Now assume that $\lambda^{\aleph_0}<\kappa$ for all $\lambda<\kappa$. If the cofinality of $\kappa$ equals $\aleph_0$, then $\kappa^{\aleph_0} > \kappa$. Otherwise, $\kappa^{\aleph_0}=\kappa$.
Just a remark: Actually from this it follows that $\mathbb{R}^{\mathbb{N}}$ and $\mathbb{R}$ are isomorphic as abelian groups, since the $\mathbb{Q}$-dimensions agree with the cardinalities here. Of course no explicit isomorphism can be written down.
