Infinite sums and intersections of subsets of $\mathbb{R}^2$ In this exercise I am asked to find $\bigcup_a \bigcap_b X_{a,b}$ (as well as other variations of sums and intersections of $X_{a,b}$) where $X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert y \leq ax(x-b)\} a,b \in \mathbb{R}, a,b >0 $. I drew the graph of $y =ax(x-b)$ but it's really hard for me to visualize what $\bigcup_a \bigcap_b X_{a,b}$ is going to look like by sheer looking at the graph. I also plugged the formula into desmos to see how it behaves when $a$ and $b$ are changing, but this didn't help either as the results I got were different from the answers provided by the teacher. Is there any systematic approach to a problem like that? Working on exercises of this type becomes dreadful without any good method as the sets get more complicated - what about this monstrosity for instance:
$X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert ax^2 < y \leq \sqrt[3]{ab^2} a,b \in \mathbb{R}^2 a,b>0\}$?!
 A: I would first try to determine what $S_a = \bigcap_b X_{a,b}$ looks like for a fixed $a$.
Then you can consider how $a$ affects its shape, and try to determine what $\bigcup_a S_a$.
Is that helpful?
A: Denote $X_{a,b}^+ = X_{a,b} \cap \{(x,y) \in \mathbb R^2 \mid x \gt 0\}$ and $X_{a,b}^- = X_{a,b} \cap \{(x,y) \in \mathbb R^2 \mid x \le 0\}$. Obviously $X_{a,b}= X_{a,b}^+ \cup X_{a,b}^-$.
For $a,b >0$ $f_{a,b}(x) = ax(x-b)$ represents a parabola whose minimum is equal to $-\frac{a b^2}{4}$ and attained at $x_{a,b}=b/2$.
Now take a fix $a > 0$ and $0 \lt b_1 \le b_2$. Looking at the sign of $f_{a,b_1}(x) - f_{a,b_2}(x)$, you get that $X_{a,b_2}^+ \subseteq X_{a,b_1}^+$ and $X_{a,b_1}^- \subseteq X_{a,b_2}^-$.
From above inclusions, considering $(x,y)$ with $x>0$ and noticing that $\lim\limits_{b \to \infty} f_{a,b}(x) = -\infty$ we obtain $\bigcap_b X_{a,b}^+ = \emptyset$. While for $x \le 0$ we notice that $\lim\limits_{b \to \infty} f_{a,b}(x) = a x^2$ and therefore $\bigcap_b X_{a,b}^- = \{(x,y) \in \mathbb R^2 \mid x \le 0\  \mathrm{and} \ y \le ax^2\}.$
From this we get
$$\bigcup_a\bigcap_b X_{a,b}^+ = \emptyset,\, \bigcup_a\bigcap_b X_{a,b}^- = \{(x,y) \in \mathbb R^2 \mid x \le 0\} \setminus \{(0,y) \mid y \gt 0\}$$ and finally
$$\bigcup_a\bigcap_b X_{a,b} = \{(x,y) \in \mathbb R^2 \mid x \le 0\} \setminus \{(0,y) \mid y \gt 0\}.$$
Now providing you with global hints to find those results is not so easy... I would say:

*

*First try to find for $a$ fixed what $\bigcap_b X_{a,b}$ is.

*For that take two distinct values $b_1,b_2$ and try to find inclusions between the $X_{a,b_1},X_{a,b_2}$.

*This is what I did when I noticed the inclusions $X_{a,b_2}^+ \subseteq X_{a,b_1}^+$ and $X_{a,b_1}^- \subseteq X_{a,b_2}^-$.

*From such inclusions, it is easier to find $\bigcap_b X_{a,b}$ as a set limit.

