I'm trying to see if a function $f : (0, 1] × (0, 1] → (0, 1]$ is bijective, where $0.a_1a_2a_3 . . . $is the decimal expansion of $x ∈ (0, 1]$, and $0.b_1b_2b_3 . . .$ is the decimal expansion of $y ∈ (0, 1]$, where the decimal expansion of $f(x, y) ∈ (0, 1]$ is $0.a_1b_1a_2b_2, . . .$
I know it's injective because if two numbers, say $a$ and $c$, are different, then we have $ϵ=a−c≠0$. Since $ϵ≠0$, it must have a most significant digit. a and c must differ at that digit, or the digit before, but how can I prove it's surjective?
To add clarity: only the non-terminating decimal expansion is used in the definition of the interleaving function.
It is not surjective because of the numbers that have two representations. You need to specify which representation you use on the left. A natural one is to use the terminating form for all numbers that may terminate, so use $0.5000\ldots$ instead of $0.499999\ldots$. No pair of numbers on the left will give a number of the form $0.a_1b_1a_29a_39a_49\ldots$ on the right.
Added: based on the update that the non-terminating version is to be used for all the decimals that have two representations, we cannot get a number on the right of the form $0.a_15a_20a_30a_40a_40\ldots$ because that would require that $y=0.50000\ldots$