Injection from infinite Cartesian product 
Define $(0,1)^{\infty}$ as the infinite Cartesian product $(0,1) \times (0,1) \times \cdots$. Find an injection $ f: (0,1)^{\infty} \to (0,1)$.

I know how to write an injection from $(0,1)^2 \to (0,1)$. Take two fractions, take their decimal expansions that don't contain an infinitely long string of nine's, and then alternate decimal places. I do not know how to extend this to the case of $(0,1)^{\infty}$. I can take a countable set of infinite decimals, disallow an  infinitely long string of nine's, but I can't "alternate" in the same way because then I'll never finish the first digit after the decimal point. If I alternate, e.g., first take the first digit, then the second, then the third, I can't guarantee that the entirety of each decimal is equal.
Help would be appreciated.
 A: For $x \in (0,1)$, and $p\geq 1$, let $x(p)$ be the $p-$th decimal of $x$.
Define $f : (0,1)^{\mathbb{N}} \rightarrow (0,1)$ for any sequence $(x_n)_{n \in \mathbb{N}} \in (0,1)^{\mathbb{N}}$ by
$$\forall p\geq 1, \quad\left[f((x_n)_{n \in \mathbb{N}})\right](p) =  \begin{cases}
      x_{k}(l) &\text{if } p=2^k3^l \text{ for some } k,l \in \mathbb{N} \\
       0 & \text{otherwise} \\
     \end{cases}$$
You can prove that $f$ is an injection.
A: An element of $(0,1)^{\mathbb N}$ is a sequence taking values in $(0,1)$. And an element of $(0,1)$ is a binary sequence, i.e. an element of $\{0,1\}^{\mathbb N}$.
Finally an element of $(0,1)^{\mathbb N}$ can be seen as an element of $\{0,1\}^{\mathbb N \times \mathbb N}$.
Now if $\phi$ is a bijection from $\mathbb N \times \mathbb N$ onto $\mathbb N$, define $\{a_{\phi(n,m)}\}$ as the image of $\{a(n,m) \mid (n,m) \in \mathbb N^2\}$.
A: $$0.\color{red}{a_{11}}0\color{red}{a_{12}a_{21}}0\color{red}{a_{13}a_{22}a_{31}}0\color{red}{a_{14}a_{23}a_{32}a_{41}}0\ldots$$
