# bijection between unit interval and unit square

The real numbers in the open unit interval  $I\,=\,(0,\,1)$  can be uniquely represented by their decimal expansions, when these must not end in an infinite string of 9’s.  Correspondingly, the elements of the open unit square $I\!\times\!I$ are represented by the pairs of such decimal expansions.

Let

 $P\;:=\;(0.x_{1}x_{2}x_{3}\ldots,\,0.y_{1}y_{2}y_{3}\ldots)$

be such a pair representing an arbitrary point in $I\!\times\!I$ and let

 $p\;:=\;0.x_{1}y_{1}x_{2}y_{2}x_{3}y_{3}\ldots$

Then it’s apparent that

 $\displaystyle P\mapsto p$ (1)

is an injective mapping from $I\!\times\!I$ to $I$.  Thus

 $|I\!\times\!I|\;\leq\;|I|.$

But since $I\!\times\!I$ contains more than one horizontal open segment equally long as $I$ (and accordingly there is a natural injection from $I$ to $I\!\times\!I$), we must have also

 $|I\!\times\!I|\;\geq\;|I|.$

The conclusion is that

 $|I\!\times\!I|\;=\;|I|,$

i.e. that the sets $I\!\times\!I$ and $I$ have equal cardinalities, and the Schröder$-$Bernstein theorem even garantees a bijection between the sets.

Remark 1.  Georg Cantor utilised continued fractions for constructing such a bijection between the unit interval and the unit square; cf. e.g. http://www.maa.org/pubs/AMM-March11_Cantor.pdfthis MAA article.

Remark 2.  Since the mapping  $g\!:\,I\to\mathbb{R}$  defined by

 $g(x)\;=\;\tan\left(\pi{x}-\frac{\pi}{2}\right)$

is bijective, we can conclude that the sets $\mathbb{R}$ and $\mathbb{R}\!\times\!\mathbb{R}$, i.e. the set of the points of a line and the set of the points of a plane, have the same cardinalities.  This common cardinality is $2^{\aleph_{0}}$.

Title bijection between unit interval and unit square BijectionBetweenUnitIntervalAndUnitSquare 2015-02-03 21:45:39 2015-02-03 21:45:39 pahio (2872) pahio (2872) 18 pahio (2872) Result msc 03E10 JuliusKonig BijectionBetweenClosedAndOpenInterval