Salvaging a cardinal proof I'm trying to prove that $\mathbb{R}^n$ has a bijection with $\mathbb{R}$. I found a way of converting an $n$-tuple of reals into a real by interleaving the digits:
      1  2.  3  5  6…
  1  3  7 . 2  7  3 …
    9  3  .1  5  8  …
⇒ 10931372.123575836…

However this fails to work for special cases like $0.9999...=1$, or at least fails to provide immediate proof of one-to-one-ness in such cases. How can I rigorously prove this method gives a bijection, or salvage it if it doesn't?
 A: You could first define a mapping
$$
\Gamma : (0,1) \times (0,1) \longrightarrow (0,1) \\
 \left(\sum_{i \geq 0}a_i10^{-i} , \sum_{i \geq 0}b_i10^{-i}\right) \mapsto \sum_{i \geq 0}a_i10^{2i} + \sum_{i \geq 0}b_i10^{2i-1}
$$
that maps $(0.a_1a_2\dots \ , \ 0.b_1b_2\dots)$ to $0.a_1b_1a_2b_2\dots \ $ and to have uniqueness in this representation, we won't allow an end of consecutive nines (e.g. instead of $0.34999\dots$ we will write $0.35$). You can check that this application has the following inverse,
$$
\Lambda : (0,1) \longrightarrow (0,1) \times (0,1) \\
\sum_{i \geq 0}a_i10^i \mapsto \left(\sum_{i \geq 0}a_{2i}10^i,\sum_{i \geq 0}a_{2i-1}10^i\right)
$$
thus proving that $(0,1) \simeq (0,1)^2$. On the other hand, one can biject $(0,1)$ to $\mathbb{R}$ via $\tan(t)$, for example by considering $\tan(\frac{\pi}{2}(2t-1))$. In particular, we have that 
$$
\mathbb{R}^2 \simeq (0,1)^2 \simeq (0,1) \simeq \mathbb{R}
$$ 
We can proceed to show $\mathbb{R}^n \simeq \mathbb{R}$ by induction on $n \in \mathbb{N}$. If $n = 1$, this is evident. If the statement is true for $n$, 
$$
\mathbb{R}^{n+1} \simeq \mathbb{R}^n \times \mathbb{R} \stackrel{\text{I.H.}}\simeq \mathbb{R} \times  \mathbb{R} \simeq \mathbb{R}^2 \simeq \mathbb{R} 
$$
which completes the proof.
