Prove that the set, upper half unit circle, $S=\{(x,y) \in \mathbb{R}\times \mathbb{R} : y=\sqrt{1-x^2}, y>0\}$ has cardinality $\mathfrak{c}$.

Question

Please check my proof: Prove that the set, upper half unit circle, $S=\{(x,y) \in \mathbb{R}\times\mathbb{R} : y=\sqrt{1-x^2}, y>0\}$ has cardinality of $\mathfrak{c}$.

Edited with my updated work.

Define S by ${(x,y) \in \mathbb{R}\times\mathbb{R} : x \in (-1, 1), y^2+x^2=1}$ Then $f: (-1, 1) \to S$ given by $f(x)=(x,\sqrt{1-x^2}, y>0)$. Then $(1-x)(1+x)>0$. So $(1-x)>0$ and $(1+x)>0$ or $(1-x)<0$ and $(1+x)<0$. Thus, $-1<x<1$ and $x\in (-1, 1)$.

Then let $x, y \in (-1, 1)$ and $f(x)=f(y)$ which implies $(x, \sqrt{1-x^2})=(y, \sqrt{1-y^2})$. Thus $x=y$. Therefore, $f$ is one to one.

Then let $(x, y)\in S$. So $y=\sqrt{1-y^2}$ and $x\in (-1, 1)$. Then $f(x)=(x, \sqrt{1-x^2}=(x, y)$. Thus $f$ is onto.

Therefore $S$ is equivalent to $(-1, 1)$ and by the Theorem, every open interval is uncountable and has cardinality $c$.

Answer 1

Your $f:\mathbb R\to S$ is not a function as it is not well-defined.

Hint to get the bijection required:

You have upper half of a circle, drop a perpendicular from a given point on the circle to $x-$ axis. (Draw it! think of it as projecting the half circle to $x-$ axis) It should give an idea that your $S\sim (0,1)$. Can you write the bijection now? I'm adding it here as a spoiler.

Define $f:S\to \mathbb (0,1)$ by $f(x,\sqrt { 1-x^2})=x$ and now verify that it is a bijection from $S$ onto $(0,1)$

And $(0,1)\sim \mathbb R$ as you have a bijection $h:(0,1)\to \mathbb R$ defined by $h(x)=\tan (\pi(x-\frac 12))$

Now using the transitivity of similarity of sets, can you take it from here?