Countably infinite product of countably infinite sets has cardinality of the continuum How to prove that the countably infinite product of countably infinite sets has cardinality of the continuum?
I know that it is uncountable thus the only thing to prove is the existence of a one-one function from the set $\prod_{n \in \mathbb{N}}{\mathbb{N}}$ to $\mathbb{R}$ .
Thanks
 A: There are two things which need to be proved:


*

*$\mathbb{N}^\mathbb{N}$ has size at least that of $\mathbb{R}$. To each real in $(0, 1)$, we can associate an infinite string of zeroes and ones - its binary expansion. This (ignoring the expansions which are eventually all "$1$"s) yields an injection from $(0, 1)$ into $\mathbb{N}^\mathbb{N}$. Note that it is not enough to merely observe that $\mathbb{N}^\mathbb{N}$ is uncountable - it is consistent that there are uncountable sets of size strictly less than that of $\mathbb{R}$.

*$\mathbb{N}^\mathbb{N}$ injects into $\mathbb{R}$. This is slightly more complicated. If you understand why $\mathbb{N}^\mathbb{N}$ and $2^\mathbb{N}$ have the same cardinality, it's enough to observe that the map defined above had range $2^\mathbb{N}$; if you haven't seen that yet, then here's a straightforward (if somewhat unnatural) injection: given $\alpha=(a_i)_{i\in\mathbb{N}}\in\mathbb{N}^\mathbb{N}$, let $f(\alpha)$ be the real with binary expansion $$0.0...010...010...01...$$ where the $i$th block of zeroes has length $a_i+1$.
