Show that the following set has the same cardinality as $\mathbb R$ using CSB We have to show that the following set has the same cardinality as $\mathbb R$ using CSB (Cantor–Bernstein–Schroeder theorem).
$\{(x,y)\in \Bbb{R^2}\mid x^2+y^2=1 \}$
I think that these are the two functions:
$f:(x,y)\to \Bbb{R} \\f(x)=x,\\f(y)=y $
$g:\Bbb{R}\to (x,y)\\ g(x)=\cos(x),\\g(y)=\sin(y)$
Is this correct ? 
Thanks.
 A: HINT: There is no continuous bijection between the two sets. Find a bijection from the unit circle to $[0,2\pi)$, and an injection from $\Bbb R$ into $[0,2\pi)$.
Also, when you define a function $g\colon\Bbb R\to\Bbb R^2$ you don't write $g(x)=\cos x$ and $g(y)=\sin y$. You should write $g(x)=(\cos x,\sin x)$ instead. Similarly when defining $f\colon\Bbb R^2\to\Bbb R$, you should define $f(x,y)=z$ rather than writing $f(x)=x$ and $f(y)=y$.
Both functions that you have defined are meaningless expressions.
A: Finding an injective function $g\colon\mathbb{R}\to X$, where $X$ is your set is easy: just remember the formulas for $\cos x$ and $\sin x$ in terms of $\tan(x/2)$ 

 Set$$g(t)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$$

For an injective map $f\colon X\to \mathbb{R}$, observe that any point on the unit circle determines an angle.
Alternatively, observe that only two points of $X$ have the same $x$-coordinate

 $$f(x,y)=\begin{cases}x & \text{if $y>0$}\\x+100 & \text{if $y<0$}\\1 &\text{if $x=1$ and $y=0$}\\-1&\text{if $x=-1$ and $y=0$}\end{cases}$$

