Hey guys, could you please check over my solution and let me know if I made a mistake somewhere?

Problem: Let $\displaystyle A_1, A_2, A_3, ...$ be countable sets. Then show that the Cartesian product $\displaystyle A = A_1 \times A_2 \times A_3 \times ...$ is uncountable. In other words, show that the set of countably infinite sets whose elements are natural numbers is uncountably large.

Approach: I know that this is solvable using a diagonalization argument. But I wanted to try something else. I attempted to surject this to real numbers.

Consider a subset of $\displaystyle A$, say $\displaystyle B$ where $\displaystyle B$ holds all such elements $\displaystyle \{a_1, a_2, a_3, ...\}$ of A where $\displaystyle a_2,a_3,a_4,... \leq 9$.

So I defined a function $\displaystyle f: B \rightarrow \mathbb{R}$ where:

For any element of B, $\displaystyle f$ sends that set to $\displaystyle \sum_{n \in \mathbb{N}} \frac{1}{10}^{n-1} a_n = a_1 + \frac{a_2}{10} + \frac{a_3}{100} + ... a_1.a_2 a_3... \in \mathbb{R}$

This function is surjective because for all real numbers $\displaystyle x \in \mathbb{R}$ whose digits are: $\displaystyle a_1. a_2 a_3 ...$, we can find an element $\displaystyle \{a_1, a_2, a_3, ...\} \in B$

The existence of this surjection implies that $\displaystyle card(B) \geq card(\mathbb{R})$. And we also know that $\displaystyle B \subseteq A$, so that $\displaystyle card(A) \geq card(B)$. By transitivity, $\displaystyle card(A) \geq card(B) \geq card(\mathbb{R})$ which means that $\displaystyle A$ must be uncountable.