Which of the following sets are equinumerous with $\mathbb{N}^\mathbb{N}$?
(i) $\mathbb{N} $
(ii) $\mathbb{R}$
(iii) $2^\mathbb{R}$
(iv) $\mathbb{N} \times \mathbb{R}$
(v) $\mathbb{R}^\mathbb{R}$
(vi) $\mathbb{R}^\mathbb{N}$
(vii) The set of permutations of $\mathbb{N}$
It seems that only choice (i) is equinumerous with $\mathbb{N}^\mathbb{N}$ but I am not sure. Could anyone help explain?
For the proofs I'll use $|F|$ as the cardinal of $F$ and $\mathbb N=\{1,2,3,..\}$
(i) No. In general $|2^F|>|F|$ for any set $F$.
(ii) Yes. For $\mathbb R$ define $\mathbb R_n$ as $[n-1,n)$ and for a set of the form $[a,b)$ define $[a,b)_n$ as $\left[b-\frac{b-a}{2^{n-1}},b-\frac{b-a}{2^n}\right)$, notice that this is an ordered partition of $[a,b)$, now we define $\mathbb R_{(n_1,n_2)}$ as $(\mathbb R_{n_1})_{n_2}$ and inductively $\mathbb R_{(n_1,...,n_k)}$ as $(\mathbb R_{(n_1,...,n_{k-1})})_{n_k}$ for every $r=(n_1,n_2,...)\in\mathbb N^\mathbb N$ define $b_{r,k}=(n_1,...,n_k)$ and $f:\mathbb N^\mathbb N\to\mathbb R/f(r)=\bigcap_{k\in\mathbb N}\mathbb R_{b_{r,k}}$ (notice that there always exist a point and only one point in $\bigcap_{k\in\mathbb N}\mathbb R_{b_{r,k}}$). $f$ is clearly inyective, so $|\mathbb N^\mathbb N|\leq|\mathbb R|$. Now $2^{\mathbb N}\subset\mathbb N^\mathbb N$ so $|2^\mathbb N|\leq|\mathbb N^\mathbb N|$, define $f:2^\mathbb N\to\mathbb [0,1]$ as $f((n_i)_{i\in\mathbb n})=\sum_{i\in\mathbb N}\frac{n_i}{2^i}$ (where $n_i\in\{0,1\}$), this function is surjective (binary representation) so $|\mathbb N^\mathbb N|\leq|\mathbb R|=|[0,1]|\leq|2^\mathbb N|\leq|\mathbb N^\mathbb N|$ so $|\mathbb N^\mathbb N|=| 2^\mathbb N|=|\mathbb R|$.
(iii) No. $|2^\mathbb R|>|\mathbb R|=|\mathbb N^\mathbb N|.$
(iv) Yes. Clearly $|\mathbb R|\leq|\mathbb R\times\mathbb N|$, now remember $|[0,1)|=\mathbb R$ so $|[0,1)\times\mathbb N|=|\mathbb R\times\mathbb N|$. Take $f:[0,1)\times \mathbb N\to\mathbb R/f(x,n)=x+n$, this is clearly inyective, so $|\mathbb N^\mathbb N|=|\mathbb R|=|\mathbb R\times\mathbb N|$.
(v) No. $|\mathbb R^\mathbb R|\geq|2^\mathbb R|>|\mathbb N^\mathbb N|$.
(vi) Yes. Clearly $|\mathbb R|\leq|\mathbb R^\mathbb N|$. Take $[0,1)$ and for each $x\in[0,1)$ take its unique binary representation $x=0,x_1x_2...$ such that $x_n\in\{0,1\}$ and there isn't an $N$ such that $x_n=1$ for all $n\geq N$. Now define $y_n^k=x_{{k+n\choose 2}-(n-1)}$, $y_n=0,y_n^1y_n^2...$ and $f:[0,1)\to\mathbb [0,1]^\mathbb N/f(0,x_1x_2,...)=(y_1,y_2,...)$. To help you visualize what I'm doing here: $$0,x_1\color{blue}{x_2}x_3\color{red}{x_4}\color{blue}{x_5}x_6\color{yellow}{x_7}\color{red}{x_8}\color{blue}{x_9}x_{10}...\to(y_1,\color{blue}{y_2},\color{red}{y_3},\color{yellow}{y_4},...)$$ Notice that $[0,1)^\mathbb N\subset f([0,1))$ so $|\mathbb R^\mathbb N|=|[0,1)^\mathbb N|\leq|[0,1)|=|\mathbb R|=|\mathbb N^\mathbb N|$.
(vii) Yes. The set of permutations of $\mathbb N$ ($\mathbb N!$) is a subset of $\mathbb N^\mathbb N$ so $|\mathbb N!|\leq|\mathbb N^\mathbb N|$. Now take $f:\mathbb N\to\{0,1\}$ and define $g_f:\mathbb N\to\mathbb N$ as follows: if $f(n)=0$ then $g_f(2n-1)=2n-1$ and $g_f(2n)=2n$, and if $f(n)=0$ then $g_f(2n-1)=2n$ and $g_f(2n)=2n-1$, the map $f\to g_f$ is inyective and $g_f\in \mathbb N!$, so $|2^\mathbb N|\leq|\mathbb N!|\leq|\mathbb N^\mathbb N|$ which implies $|\mathbb N^\mathbb N|=|\mathbb N!|$.