Let $X_n=\{1,2,...,n\}$. I want to compute the cardinal of $\prod\limits_{n\in\mathbb{N}}X_n:=\{f:\mathbb{N}\rightarrow \bigcup\limits_{n\in\mathbb{N}}X_n: f(n)\in X_n\}$. I believe that its cardinal is $2^{\aleph_0}$.
I have been able to show that \begin{align} \left|\prod\limits_{n\in\mathbb{N}}X_n\right|\geq 2^{\aleph_0}. \end{align}
In order to show that $\left|\prod\limits_{n\in\mathbb{N}}X_n\right|\leq 2^{\aleph_0}$, I am trying to find a one to one map from the set $\{f:\mathbb{N}\rightarrow\bigcup\limits_{n\in\mathbb{N}}X_n: f(n)\in X_n\}$ to the set $\{g:\mathbb{N}\rightarrow \{0,1\}\}$ but I am not able to find it. Is it even possible?
Any hint to find such a map?
Is it correct that $\prod\limits_{n\in\mathbb{N}}X_n$ has cardinality $2^{\aleph_0}$?
Define
$$\varphi:\prod_{n\in\Bbb Z^+}X_n\to\wp(\Bbb Z^+):f\mapsto\left\{2^n3^{f(n)}:n\in\Bbb Z^+\right\}\,.$$
That should be good enough for your purposes, but if you really want a map to ${^{\Bbb Z^+}}\{0,1\}$, you can use $f\mapsto\chi_{\varphi(f)}$, mapping $f$ to the indicator (characteristic) function of $\varphi(f)$. The set $\varphi(f)$ is a fairly standard way of encoding the ordered pairs making up $f$ as single positive integers. (I have used $\Bbb Z^+$ here instead of $\Bbb N$, because for me the latter includes $0$.)