It is known that each $x,y \in (0,1)$ has a unique decimal expansion $ x = 0.x_1x_2x_3\ldots $ and $y = 0.y_1y_2y_3\ldots$ with the requirement that there are infinitely many indices different from $9$. Show that the following function $f : (0,1)×(0,1) → (0,1)$ is injective: $f(x,y) = 0.x_1y_1x_2y_2\ldots$ Is $f$ also surjective? Using the Schröder-Bernstein theorem show that $(0,1)×(0,1)$ and $(0,1)$ have the same cardinality.
For the injectivity part of the question, this is what I have so far. I have defined 2 sets $S=\{x_n : n \in \mathbb{N}\}$ and $T=\{y_n : n \in \mathbb{N}\}$. Then I have said $f(x,y)=\sum_{n\geq0}x_n10^{-2n+1}+y_n10^{-2n}$. Then I have concluded that since $x$ has a unique decimal expansion and so does $y$ then it follows that the sum of 2 unique expansions is itself unique, but I feel that this is missing a great deal of rigour and would like to know if it's even right or how to improve it. Also, I am unsure as to how to deal with the surjectivity part of the question. Any help would be greatly appreciated.
I guess that you mean $x = 0.x_1x_2x_3 . . .$ and $y = 0.y_1y_2y_3 . . .$
Injectivity is easy. If you have two different points $(u,v)$ and $(x,y)$ then either $u \neq x$ or $v \neq y$. If it is $u \neq x$ then they have at least one different digit $u_i \neq x_i$. So, you can point to a differing digit in the images of these points under your map. Similarly if $v \neq y$.
However, the map is not surjective. Nothing maps to $0.09090909 . . .$.
So, you need to think of of another map which is surjective. That should be very easy.