Expressing the union and intersection of sets Let $A=\left\{(x,y)\in \mathbb{R^2}: x+2y\leq 1\right\}$ and $B=\left\{(x,y)\in \mathbb{R^2}: x^2+y^2\leq 1\right\}$
Then we have $$1) C = A\cup B$$ $$2) D = A\cap B$$
How can I express $C$ and $D$ in the same way as $A$ and $B$?(it is possible?)
I can not formalize this, and I need to be able to see if $C$ and $D$ are convex sets
 A: Think of union as "or" and intersection as "and." That is, $A \cup B$ is the set of points $(x,y)$ so that one equation is satisfied OR the other is. On the other hand, $A \cap B$ is the set of points $(x,y)$ so that one equation AND the other are satisfied.
$$
C = A \cup B = \{ (x,y) \in \mathbb{R} : x + 2y \le 1 \text{ or } x^2 + y^2 \le 1 \} \\
D = A \cap B = \{(x,y) \in \mathbb{R}: x+2y \le 1 \text{ and } x^2 + y^2 \le 1\}
$$
Graphically, if you draw both $A$ and $B$ as shaded subsets of the plane (in one graph), $C$ is the part of the plane that is shaded. $D$ is the part of the plane where the shaded regions overlap. After drawing it like this, it should be immediately obvious which ones are convex.
EDIT: To prove that a set $S$ isn't convex, it's sufficient to give two points $p,q$ so that the line segment from $p$ to $q$ is not contained in $S$. Look at your drawing, and visually estimate an example of such a pair $p,q$. Then do some algebra to find a point on the line segment between them that doesn't satisfy the inequalities defining the set.
