Show that there is a bijection between $2^{\mathbb{N}}$ and $\mathbb{R}$

Question

I tried to do it with ternary/binary expansions by finding two injections and then use the Cantor-Bernstein-Schröder Theorem, but I wonder if there is some easier method to prove this.

Answer 1

Because of the issue of $0.999\ldots =1$, and more generally that any standard number system representation is non-injective (as a function from sequences of digits to $\Bbb R$), I sincerely doubt there is an easier way. Assuming you already have access to the CBS theorem.

If you do not have access to the theorem, and don't want to include a full proof of it in your proof, then it might be simpler to fix the above issue directly, similar to how you would normally prove that $(0,1]$ and $(0,1)$ are equinumerous, by moving $1$ to $\frac 12$, moving $\frac12$ to $\frac13$, and so on.