Answer
(a) closed and bounded
(b) open and bounded
(c) closed and unbounded
(d) open and unbounded
(e) closed and bounded
(f) open and bounded
Work Step by Step
(a) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:{x^2} + {y^2} \leqslant 1} \right\}$
We can choose $M=2$ such that $D$ is contained in a disk of radius $M$ centered at the origin. Therefore, $D$ is bounded. Notice that the boundary points of $D$ are on the circle of radius $1$. Since $D$ contains all its boundary points, it is closed.
Hence, $D$ is closed and bounded.
(b) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:{x^2} + {y^2} < 1} \right\}$
We can choose $M=1$ such that $D$ is contained in a disk of radius $1$ centered at the origin. Therefore, $D$ is bounded. Notice that every point of $D$ is an interior point, therefore it is open.
(c) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:x \geqslant 0} \right\}$
Since $x \ge 0$, we can always choose a number $x$ such that $D$ contains point arbitrarily far from the origin. Therefore, $D$ is unbounded.
Notice that the boundary points of $D$ is $\left( {0,y} \right)$ for $y \in \mathbb{R}$. Since $D$ contains all its boundary points, it is closed.
(d) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:x > 0,y > 0} \right\}$
Since $x>0$ and $y>0$, we can always choose a number $x$ and $y$ such that $D$ contains point arbitrarily far from the origin. Therefore, $D$ is unbounded.
Notice that every point of $D$ is an interior point, therefore it is open.
(e) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:1 \leqslant x \leqslant 4,{\ \ }5 \leqslant y \leqslant 10} \right\}$
We can choose $M=12$ such that $D$ is contained in a disk of radius $M$ centered at the origin. Therefore, $D$ is bounded.
Notice that the boundary points of $D$ are on the rectangular with sides $1 \le x \le 4$ and $5 \le y \le 10$. Since $D$ contains all its boundary points, it is closed.
(f) We have the domain:
$D = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}:x > 0,{x^2} + {y^2} \leqslant 10} \right\}$
We can choose $M=11$ such that $D$ is contained in a disk of radius $M$ centered at the origin. Therefore, $D$ is bounded. Notice that the boundary points of $D$ are on the half-circle of radius $10$ for $x>0$. However, there are boundary points at $x=0$ not contained in $D$. Since $D$ does not contain all its boundary points, it is not closed.
