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"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?


marked as duplicate by Martin R, metamorphy, Cesareo, YuiTo Cheng, cmk Jul 19 at 14:35

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    $\begingroup$ I don't know how Cantor did. I would do it by writing all real numbers and distributing every $n^{th}$ decimal to $n$ distinct reals. $\endgroup$ – Yves Daoust Jul 19 at 6:35
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    $\begingroup$ I am pretty sure, Cantor would have used continued fractions to identify with $\Bbb N^{\Bbb N}$ $\endgroup$ – Hagen von Eitzen Jul 19 at 6:47
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    $\begingroup$ @HagenvonEitzen: It seems so. From one answer to the possible duplicate target: “Cantor originally tried interleaving the digits himself, but Dedekind pointed out the problem of nonunique decimal representations. Cantor then switched to an argument like the one Robert Israel gave in his answer, based on continued fraction representations of irrational numbers.” $\endgroup$ – Martin R Jul 19 at 9:04
  • $\begingroup$ @MartinR Indeed, I linked to the original paper in my answer. You could check it out in the original or read Dauben's description in his very nice book on Cantor. $\endgroup$ – Henno Brandsma Jul 19 at 13:22

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 $[0,1] \times [0,1]\times \cdots \times [0,1]$

  • $\begingroup$ I think you are throwing away too much if you avoid all expansions with infinitely many ones. You only want to throw away the expansions that are constant $1$ after some finite point. $\endgroup$ – quarague Jul 19 at 7:03
  • $\begingroup$ the map isn't injective $\endgroup$ – mercio Jul 19 at 7:08
  • $\begingroup$ @quarague I am avoiding expansions such that $a_k=1$ for all $k$ exceeding some integer. I am not avoiding expansions with in finitely many $1$'s. $\endgroup$ – Kabo Murphy Jul 19 at 7:12
  • $\begingroup$ You wrote " $a_k=1$ for all but finitely many $k$", this includes for example the alternating sequence $a_{2k}=1$ and $a_{2k+1}=0$ which you don't want to exclude. Write what you did in the comment and it is fine. $\endgroup$ – quarague Jul 19 at 7:19
  • $\begingroup$ @quarague I am sorry, you seem to have a problem with English. $a_k=1$ for all but finitely many $k$ means there is a finite set of integers $F$ such that $a_k=1$ for $k \notin F$. If $N$ is the largest integer in $F$ then $a_k=1$ for all $k >N$. So your example with $a_{2k+1}=0$ does not satisfy this. $\endgroup$ – Kabo Murphy Jul 19 at 7:22

The paper that Cantor published his result in is Ein Betrag zur Mannichfaltigkeitslehre from 1878. (the link is to the original German version). In this he reduces to the irrationals in $[0,1]$, which have a unique representation in terms of continued fractions of positive integers, and where he can easily associate $n$ irrationals to a single new one in a bijective way (mixing the sequences of integers of the continued fractions in the obvious way).

He thus showed (as the irrationals in $[0,1]$ have the same size as the set of all irrationals) that $\|\mathbb{P}|^n = |\mathbb{P}|$, where (modern notation) $\mathbb{P}$ is the set of irrationals. This is what the paper starts with (theorem C), if we formulate it in a more modern way. Then he shows in the remainder that $\mathbb{P}$ and $\mathbb{R}$ have the same size.

So he did not give a direct bijection between $I^n$ and $I$, but did it in a more roundabout, but still valid, way.


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