elementary-set-theory
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.
Let's do a fundamentally different approach. Consider projecting from the open bottom half of the circle $x^2 + (y-1)^2 = 1$, as I drafted below.
In this, by projecting from $(0,1)$, we get a 1-1 correspondence from the angle $(0, -\pi) \to \mathbb{R}$. There is a clear 1-1 correspondence from $(0,1)$ and $(0, -\pi)$.
Thus we have our bijection from $(0,1)$ to $\mathbb{R}$. For $[0,1]$, we could use big theorems, or we could just modify this one. So let's "make space" by saying that whatever was associated to $0$ is now associated to $2$, what was associated to $1$ is now associated to $3$, $2$ now goes to $4$, and so on, effectively freeing up the real numbers $0,1$ by displacing the natural numbers. Send $0$ to $0$, and $1$ to $1$, and now we have a bijection $[0,1] \to \mathbb{R}$.
Nice, direct, and constructive.
Your first function is a bijection and proves that any two finite intervals have the same cardinality. Your second fails because $g(-\frac 12) \not \in (0,1)$ but you have the right idea. There are many bijections between $\Bbb R$ and some finite interval. One of the simplest is $\arctan(x)\to (\frac {-\pi}2,\frac \pi 2)$ This solves the open intervals. For closed intervals, you need to "swallow" the endpoints somehow.
Best Answer
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.