Prove set is closed I need to prove the set 
$A=\left \{ \left ( r\cos t,r\sin t \right ) \mid 0\leq t\leq \theta ,r\geq 0\right \}$ is closed. I have tried to use the theorems for continuous functions, but I always into troubles cause the domain is not bounded. 
 A: hint: a set is closed if its complementary is open.
A: Pick any point $p \in \mathbb{R}^2 \setminus A$. (If such a point does not exist, then $A = \mathbb{R}^2$ is closed.) Let $d > 0$ be the smallest distance between this point and the rays $t = 0$, $t = \theta$. This choice of $d$ ensures that the distance between $p$ and any point in the sector $A$ is at least $d$. Therefore, the open disk of radius $d$ around $p$ lies entirely within $\mathbb{R}^2 \setminus A$, so $\mathbb{R}^2 \setminus A$ is open.
A: Let $\{(x_n,y_n)\}$ be a sequence of points in $A$ such that $\lim_{n\to\infty}(x_n,y_n)=(x,y)$ for some $(x,y)\in\mathbb R^2$. We can assume that $(x,y)\ne(0,0)$ since clearly $0=0\cdot\cos t=0\cdot\sin t$ for any $t$ and so $(0,0)\in A$. Set $\rho = \sqrt{x^2+y^2}$ and $\theta = \mathrm{atan2}(y,x)$, where 
$$\mathrm{atan2}(y,x) = 
\begin{cases}
\arctan\left(\frac yx\right),& x>0\\
\arctan\left(\frac yx\right)+\pi,& x<0,\ y\geqslant0\\
\arctan\left(\frac yx\right),& x<0,\ y<0\\
\frac\pi2,& x=0,\  y>0\\
-\frac\pi2,& x=0,\ y<0.
\end{cases}
$$
Then $(x,y) = (\rho\cos\theta, \rho\sin\theta)\in A$, which implies that $A$ is closed.
