The two uncountable sets. What is the purpose of this exercise? In Exercise 5 we concluded that all sets except $D$ and $E$ are countably infinite, hence, they all have the same cardinality. However, we also showed that $D$ and $E$ are uncountable, which, by itself, does not imply they have the same cardinality. So, here we show that, in fact, they do.

There is an obvious injection from $E$ to $D$ (each function into $\{0,1\}$ can be considered as a function into $\mathbb{Z}_{+}$ ). To construct the opposite injection note that each function in $D$ can be considered as an infinite countable sequence of positive integers, while a function in $E$ can be considered as an infinite countable sequence on $0$ ’s and $1$ ’s. For each sequence $(n_{1},n_{2},\ldots)$ of positive integers in $D$ we assign the sequence of $0$ ’s and $1$ ’s such that first there are $f(1)$ 0’s then 1 then $f(2)$ 0’s then 1 etc.: $(\underbrace{0,\ldots,0}_{n_{1}},1,\underbrace{0,\ldots,0}_{n_{2}},1,\ldots)$ . Having the two injections, according to Exercise 6(b), we conclude that the sets $D$ and $E$ have the same cardinality.