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.
Best Answer
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.