elementary set theory - How did Cantor demonstrate a bijection from $I=[0,1]$ to $I^n$?

"I See It, but I Don't Believe It."

Georg Cantor showed that sets of different dimensions can have the same cardinality; in particular, he demonstrated that there is a bijection between the interval $I= [0,1]$ and the $n$-fold product $I^{n} = I \times I \times \cdots \times I$.

Does anyone know specifically how this was done?

Answer

I am not sure if Cantor did it this way, but this argument works: any number $x$ in $[0,1]$ has an expansion to base $2$: $x=\sum \frac {a_k} {2^{k}}$ where $a_k =0$ or $a_k=1$ for each $k$. This expansion is not unique but it can be made unique by avoiding expansions with $a_k=1$ for all but finitely many $k$ (except when $x=1$). Now let $r \in \{0,1,2,...,n-1\}$ and form a sequence $(b_k^{(r)})$ using the coefficients $a_k$ with $k=r\, \pmod{n}$. Let $x_r$ be the number whose expansion to base $2$ has the coefficient sequence $(b_k^{(r)})$. Then the map $x \to (x_1,x_2,...,x_n)$ is a bijection.

A correction: it has been pointed out that $x=1$ causes problem in this argument. (See comment by Henno Brandsma). As suggested we can use the proof to show that there is a bijection between $[0,1)$ and $[0,1) \times [0,1)\times \cdots\times [0,1)$ and use the fact that there are bijections between $[0,1)$ and $[0,1]$ as well as between $[0,1) \times [0,1)\times \cdots \times [0,1)$ and $[0,1] \times [0,1]\times \cdots \times [0,1]$