# Prove that the set $\mathbb{R^{n}}$ has the the same size as $\mathbb{R}$.

I would very much appreciate if anyone would be so kind as to review whether or not the following attempt is successful or not.

Theorem: The set $\mathbb{R^{n}}$ has the the same size as $\mathbb{R}$.

Proof

We will use mathematical induction. The base case $n=1$ is clearly true, and so we assume that the proposition holds for all natural numbers up to $n-1$. Our goal is to show that the theorem holds for $n$ as well. It suffices to prove that the set of $(x_{1},...,x_{n})$, $0<x_{1},...,x_{n}\leq 1$ can be mapped bijectively onto $(0,1]$. Now, consider $(x_{1},...,x_{n})$ and write these numbers in their unique non-terminating decimal expansion as in this example:$$x_{1}=0.42\mathbf{0}1\mathbf{221} 07128...$$ $$x_{2}=0.12\mathbf{3}0\mathbf{321}36719...$$ $$\vdots$$ $$x_{n}=0.07\mathbf{1}2\mathbf{075}46621...$$

Note that we separate the digits of $x_{1},...,x_{n}$ into groups by always going to the next nonzero digit, inclusive. We then proceed to associate to $(x_{1},...,x_{n})$ the number $\alpha \in (0,1]$ by writing down the first $x_{1}$-group, after that the $x_{2}$-group etc up to the first $x_{n}$-group, before proceeding to the second $x_{1}$-group. Thus, our example gives: $$\alpha=0.4212...0703...110....2...$$

As none of $(x_{1},...,x_{n})$ exhibits only zeroes from a certain point on, we have that $\alpha$ is also a non-terminating decimal expansion. Conversely, we see from the expansion of $\alpha$ that it is possible to extract each of $(x_{1},...,x_{n})$, and so the map is bijective. $\square$

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Your interleaving scheme isn't exactly clear cut. In particular, I'm not sure what you mean by "always going to the next nonzero digit, inclusive." –  Cameron Buie Apr 7 '13 at 15:54
Not clear to me why you are using induction or where it is used in your proof. Why not just set up explicitly an injection from $\Bbb R^n$ into $\Bbb R$ using the techniques you're already using? (Mapping $\Bbb R$ into $\Bbb R^n$ also necessary to show same size, this is trivial). Edit: @DominicMichaelis 's comment shows the theorem to be using. –  muzzlator Apr 7 '13 at 15:55
well i don't see the bijection, but on the other hand i would surely use cantor schröder bernstein for this one –  Dominic Michaelis Apr 7 '13 at 15:56
The reason as to why I use induction, rather than Cantor-Bernstein-Schroeder Theorem, is that I prefer to have different types of proofs for theorems that simply are wondrous. As for the interleaving scheme, I completely agree with Cameron Buie that it is far from satisfactory. What I tried to convey was that {42, 12,...,07} form a group and so does {0,3,...,1} etc. The point then is that this can then be used as a foundation for construction a bijective mapping. –  user71429 Apr 7 '13 at 16:23

I don't think this checks out perfectly because this cannot deal with rational numbers. In particular, there is nothing that maps to $\alpha = 0.90909090\dots$. Your corresponding $x_1$ and $x_2$ would be $0.9999\dots$ and $0.0000\dots$, which are not in $( 0,1 )$.
However, you can proceed much more simply by noting that if $\mathbb R \sim \mathbb R^2$, we can conclude the statement by induction.
There is a theorem in set theory that for any two infinite cardinals, $|\aleph_\alpha \times \aleph_\alpha| = \aleph_\alpha$. Using this, we could conclude the theorem directly.
Another strategy: if you know that $2^{\aleph_0} = |\mathbb R|$, you can conclude this statement by taking two disjoint countably infinite sets $A$ and $B$. Since $|A \cup B| = |A| = |B| = \aleph_0$,
$$|\mathbb R| = 2^{|A \cup B|} = |2^A \times 2^B| = |\mathbb R \times \mathbb R|$$