Set inclusion – proof Let $K_s := \{(x,y) \in R^2:  x^2 + (y-s)^2 \le s \}$ for $s \ge 0$. Prove, that:
$\{(x,y) \in R^2: y \ge x^2 \} \subset \bigcup_{s \in N}K_s \subset \bigcup_{s\in(0,\infty)}K_s=\{(x,y): y\ge x^2 -\frac{1}{4}\}$
I perfectly know what all the sets look like, but I struggle a bit in naming it in the set's theory language. I could put $y =x^2$ into the $x^2 + (y-s)^2 \le s$ and check if there are always 4 roots, but I guess it is not the best way to deal with this problem. And it won't work for succeeding inclusions.
 A: Let's rephrase the three things you have to prove.
Task #1: Prove
$$
\{(x,y) \in R^2\colon y \ge x^2 \} \subset \bigcup_{s \in N}K_s.
$$
In other words, prove
$$
\text{If } (x,y)\in \{(x,y) \in R^2\colon y \ge x^2 \} \text{, then there exists $s\in N$ such that } (x,y)\in K_s.
$$
In other words, prove
$$
\text{If } y \ge x^2 \text{, then there exists $s\in N$ such that } x^2+(y-s)^2 \le s.
$$
Task #2: Prove
$$
\bigcup_{s \in N}K_s \subset \bigcup_{s\in(0,\infty)}K_s.
$$
In other words, prove
$$
\text{If there exists $s\in N$ such that } (x,y)\in K_s \text{, then there exists $s\in(0,\infty)$ such that } (x,y)\in K_s.
$$
Task #3: Prove
$$
\bigcup_{s\in(0,\infty)}K_s=\{(x,y)\colon y\ge x^2 -\tfrac{1}{4}\}.
$$
In other words, prove
$$
\text{There exists $s\in (0,\infty)$ such that } (x,y)\in K_s \text{ if and only if } x\in \{(x,y)\colon y\ge x^2 -\tfrac{1}{4}\}.
$$
In other words, prove
$$
\text{There exists $s\in (0,\infty)$ such that } x^2+(y-s)^2 \le s \text{ if and only if } y\ge x^2 -\tfrac{1}{4}.
$$
Hopefully these rephrasings help you connect the set-theoretic statements with the algebraic understanding you seem to already have.
