Show that $A=\{(x,y) \in \mathbb R^2 \mid x^2+y^2 \leq 1 , \, x<y\}$ is Borel In my homework I want to show that $A=\\{(x,y) \in \mathbb R^2\mid x^2+y^2 \leq 1 , \, x<y\\}$ is Borel and determine if A is open or closed My idea was to say that: $A$ is the intersection or all $\\{(x,y) \in \mathbb R^2\mid x^2+y^2 \leq 1 $ but I don't know how to handle the x Also it seems closed on the half-circle but open on the $x<y$ but it doesn't make sense that it is both open and closed. Any help would be appreciated

$A_1=\\{(x,y)\in\mathbb{R}^2:x^2+y^2\le 1\\}$ is a Borel set ($\because$ it is closed), $A_2=\\{(x,y)\in\mathbb{R}^2:x-y<0\\}$ is a Borel set ($\because$ it is open). Then $A=A_1\cap A_2$ is also a Borel set.