bijection between unit interval and unit square

The real numbers in the open unit interval  $I=(0,\mathrm{\hspace{0.17em}1})$  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

$$P:=(0.x_1{x}_2{x}_3\mathrm{\dots },\mathrm{\hspace{0.17em}0}.y_1{y}_2{y}_3\mathrm{\dots })$$

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

$$p:=\mathrm{\hspace{0.33em}0}.x_1{y}_1{x}_2{y}_2{x}_3{y}_3\mathrm{\dots }$$

Then it’s apparent that

$P\mapsto p$(1)

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

$$|I\times I|\le |I|.$$

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

$$|I\times I|\ge |I|.$$

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  $g:I\to ℝ$  defined by

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

is bijective, we can conclude that the sets $ℝ$ and $ℝ\times ℝ$, 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 $2^{{\mathrm{\aleph }}_0}$.