Whenever possible, I like to present Cantor's diagonal proof of the uncountability of the reals to my undergraduates. For simplicity, I usually restrict to showing that the subset
$$
A = \{x \in [0,1) \mid \text{ the decimal representation of $x$ uses only 0's and 1's} \}
$$
is already uncountable. I was thinking recently that it would be nice to add a quick proof that $A$ is actually of precisely the same cardinality as $\mathbb{R}$. That is, I would like to:


**Demonstrate a bijection between $A$ and $\mathbb{R}$.**


My first instinct was to use find an injection from $A$ into $\mathbb{R}$ and vice versa, then appeal to Cantor-Bernstein to say that a bijection exists (even if we don't know how to construct it). The identity map suffices from $A$ into $\mathbb{R}$. For the other direction, I thought of something like "for $x \in \mathbb{R}$, map $x$ to its binary representation, disregarding the decimal point". I'm afraid this function fails to be injective, however. For example, 1 (base 10) can be represented as $.\overline{1}$ (base 2), and so 2 (base 10) can be represented as $1.\overline{1}$ (base 2). Thus, 1 and 2 (base 10) will have the same image under my map.


Any methods (not necessarily the one I've attempted to start here) are most welcome. I will accept as "correct" the method which demonstrates the bijection with the greatest level of clarity.



A Bijection Between the Reals and Infinite Binary Strings
Just for the fun, we can use continued fractions to map the sequences of positive integers injectively to [0,1] the sequence may end with $\infty$ meaning that we get a finite fraction (a rational number). Now, the mapping 01010111100110110... to 001010111100110110... to 2,1,1,1,1,4,2,2,1,2,... is a clear bijection (add zero to the beginning and start counting group lengths). This can, probably, be upgraded to get $\mathbb R$ as the image but, IMHO, $[0,1]$ is good enough too and we have no problem with multiple decimal representations. 


