Question

Show that $|R\times R|=\left|R\right|$. [Hint: Use the Schroder Bernstein theorem to show that $\left|\right(0,1)\times (0,1\left)\right|=\left|\right(0,1\left)\right|$. To construct an injection from (0, 1) X (0,1) to (0,1), suppose that $(x,y)\in (0,1)\times (0,1)$. Map (x, y) to the number with decimal expansion formed by alternating between the digits in the decimal expansions of x and y, which do not end with an infinite string of 9s.]

Short answer

$|R\times R|=\left|R\right|$

Step-by-step solution

Step: 1

Because there is one-to-one correspondence between R and the open interval (0,1) given by $f\left(x\right)=2ta{n}^{-1}\left(x\right)/\pi$, it suffices to shows that$\left|\right(0,1)\times (0,1\left)\right|=\left|\right(0,1\left)\right|$

Step: 2

By the Schroder-Bernstein theorem it suffices to find injective function

$$\begin{gathered} f:(0,1)\to (0,1)\mathrm{X}(0,1) \\ g:(0,1)X(0,1)\to (0,1) \end{gathered}$$

Let $f\left(x\right)=(x,\frac{1}{2})$.

Step: 3

For g we follow the hint. Suppose$(x,y)\in (0,1)\times (0,1)$ and represent x and y with their decimal expansions$x=0.x_1{x}_2{x}_3..$ and$y=0.y_1{y}_2{y}_3..$ never choosing the expansion of any number that ends in an infinite string of 9s. Let$g(x,y)$ be the decimal expansion obtained by interweaving these two strings, namely,$0.x_1{y}_1{x}_2{y}_2{x}_3{y}_3..$