Proof regarding families of sets and Cartesian products I have to prove the following and I believe that the choose an element method is likely the best. I gave an attempt but it was lackluster, I don’t know that I fully understand families of sets. Any help at all in starting would be greatly appreciated.
\begin{equation*}
    \bigcup_{t\in (0,1)}\bigl( (t,1) \times (0,t^2) \bigr)
= \{\, (x,y) \in \mathbb{R}^2 \mid 0<x<1 \text{ and } 0<y<x^2 \,\}.
\end{equation*}
 A: Denote:
$$\begin{cases}
A & = \bigcup_{t\in (0,1)}\bigl( (t,1) \times (0,t^2) \bigr)\\
B &= \{\, (x,y) \in \mathbb{R}^2 \mid 0<x<1 \text{ and } 0<y<x^2 \,\}
\end{cases}$$
You have to prove that the two sets are equal. A usual way to do so is to prove that $A \subseteq B$ and $B \subseteq A$. Also notice that $A,B$ are subsets of $\mathbb R^2$. Notice that $A$ is a union of open rectangles.
Proof that $A \subseteq B$
Take $(x,y) \in A$. By definition, it exists $t \in (0,1)$ such that $(x,y) \in (t,1) \times (0,t^2)$. Note that the notations are unfortunately confusing... $(x,y)$ is here an ordered pair while $(t,1)$ is an interval. That's our poor life of mathematicians!
Anyhow, that means that $0 \lt t \lt x \lt 1$ and $0 \lt y \lt t^2 \lt x^2$. So $(x,y) \in B$: we've proven first inclusion. Let's prove the reverse one.
Proof that $B \subseteq A$
So take $(x,y) \in B$. We have to prove that $(x,y) \in A$, i.e. find a rectangle $(t,1) \times(0,t^2)$ in which $(x,y)$ lie. You'll verify that
$$(x,y) \in (t,1) \times (0,t^2)$$ where $ t = \frac{x + \sqrt y}{2}$. A drawing will help to understand why... and computations validate it.
We're done!
