Show that $\{(x,y) \in \mathbb{R}^2|x^2+y^2=1 \text{ and } x>0 \}$ is Borel

Question

I have the following set $\{(x,y) \in \mathbb{R}^2|x^2+y^2=1 \text{ and } x>0 \}$ which I want to show is Borel

It seems obvious that is it Borel because all are single points which can be expressed as $\bigcap_{n \in \mathbb{N}} (x-1/n,x+1/n) \times (y-1/n,y+1/n)$ But this describes any point in $\mathbb{R}^2$.

My problem is to describe the set as it passes through the right-half of the unit circle

Any hint would be appreciated

Answer 1

The set in the post is the intersection of the closed set $\{(x,y):x^2+y^2=1\}$ and the open set $\{(x,y):x>0\}$

So it is Borel as an intersection of Borel sets.

Answer 2

Let $f(x,y)=x^2+y^2$ defined on $(0,\infty )\times \mathbb R$. It's a continuous function. Then, your set is $f^{-1}(\{1\})$ which is closed in $(0,\infty )\times \mathbb R$ because $f$ is continuous, so obviously a Borel set in $(0,\infty )\times \mathbb R$ and thus in $\mathbb R^2$.