I would like to show the existence of a bijection between $(0,1)$ and $(0,1) \times (0,1)$. Would $f(x)=(x,x)$ work ($x \in (0,1)$)? It seems to be injective, but am I okay in using this even if its domain and range can go outside of $(0,1)$?
Define a second function $g: (0,1) \times (0,1) ↦ (0,1)$ which takes two decimal expansions $0.(x_1)(x_2)(x_3)...$ and $0.(y_1)(y_2)(y_3)...$, the first expansion coming from the first component of $(0,1) \times (0,1)$ and the second coming from the second component and maps them to $0.(x_1)(y_1)(x_2)(y_2)(x_3)(y_3)...$. For example, take $0.04265$ and 0.83169. The function g maps the two decimal expansions to 0.0843216659. This seems like it could be an injective function because it takes two numbers and maps them to a unique number. However, I'm not sure how I would formally prove it...
This implies that, by Cantor-Schroder-Bernstein theorem, there exists a bijection between $(0,1)$ and $(0,1) \times (0,1)$.