$x^2+y^2<1, x+y<3$ is open or closed? I'm trying to figure out if 
$$\{x^2+y^2<1, x+y<3|(x,y)\in \mathbb R^2\}$$
is open or closed. 
I tried to imagine this set. It looks, for me, as a 'pizza', or a circular sector, which have two 'straigth' sides (closed) and a circular side (open). So I'm really confused...
I also need to prove if this set is closed or open, by using open balls. Could somebody help me finding the radius? Quite complicated in a so irregular figure.
 A: The set is just $$\{x^{2}+y^{2}<1\}$$
So it's open.
A: The intersection of two open sets is open. Or, straightforwardly, since $\{x^2+y^2<1|(x,y)\in\mathbb{R}^2\}$ is contained in $\{x+y<3\}$, the set 
$$
\{x^2+y^2<1, x+y<3|(x,y)\in\mathbb{R}^2\}=\{x^2+y^2<1|(x,y)\in\mathbb{R}^2\}
$$
is open. 
A: Since this problem is a bit trivial (one condition defining the set implies the other), let us consider a more general scenario where the second condition is $ax+by < c$ for some $a,b,c \in \mathbb{R}.$
It is easy to see that the set $X := \{ x^2 + y^2 < 1 \}$ is open using the open ball definition, since it's already an open ball of radius $1.$ To see that $Y := \{ ax+by < c \}$ is open, take any point $(x_0,y_0) \in Y;$ since the distance between $(x_0,y_0)$ and the line $ax+by = c$ is $d := |ax_0 + by_0 -c|/\sqrt{a^2+b^2} > 0,$ the open ball of radius $d$ centered at $(x_0,y_0)$ is entirely contained in $Y,$ which shows that $Y$ is open.
Finally, it is easy to see that the intersection of two open sets is open: for any point in the intersection, consider the two open balls given by the definition of being open; then the smallest of these two balls is contained in the intersection.
Hence $X \cap Y = \{ x^2 + y^2 < 1, ax + by < c \}$ is open.
