Find $\bigcup_{s \in \mathbb R} \bigcap_{t \geq s} A_t$ 
$A_t = (\cos t - 2, \cos t + 2) $ for $t \in \mathbb R$. Find $\bigcup_{s \in \mathbb R} \bigcap_{t \geq s} A_t$.

Please, help. I have no idea how to correctly solve a task like this, and I can't even find a $\bigcup_{s \in \mathbb R} \bigcap_{t \geq s}$. I understand how to solve much simpler examples, e.g. finding just a $\bigcup$ or $\bigcap$. But what should I do in this case? How to start? 
 A: Just start out with some examples. Suppose that $s = 0$. Then what is:
$$
\bigcap_{t \geq 0} (\cos t - 2, \cos t + 2)
$$
Well let's think about some examples.


*

*If $t = 0$, then we have: $(-1, 3)$.

*If $t = \pi/2$, then we have: $(-2, 2)$.

*If $t = \pi$, then we have: $(-3, 1)$.


Draw out these intervals. Where do they all overlap as we keep varying our choice of $t$? It should be easy to see that:
$$
\bigcap_{t \geq 0} (\cos t - 2, \cos t + 2) = (-1, 1)
$$
Now repeat this procedure for a different choice of $s$.

EDIT: Here's a full proof.

Claim: $\bigcup_{s \in \mathbb R} \bigcap_{t \geq s} A_t = (-1, 1)$



*

*$\boxed{\subseteq}:$ Choose any $x \in\bigcup_{s \in \mathbb R} \bigcap_{t \geq s} A_t$. Then there is some $s_0 \in \mathbb R$ such that $x \in \bigcap_{t \geq s_0} A_t$. Now consider $t_{\textsf{even}} = 2|\lceil s_0 \rceil| \pi$ and $t_{\textsf{odd}} = (2|\lceil s_0 \rceil| + 1) \pi$, which are both certainly greater than or equal to $s_0$. Observe that:
$$
A_{t_\textsf{even}} = (-1, 3) \\
A_{t_\textsf{odd}} = (-3, 1)
$$
Thus, we conclude that:
$$
x \in \bigcap_{t \geq s_0} A_t \subseteq (A_{t_\textsf{even}} \cap A_{t_\textsf{odd}}) = (-1, 3) \cap (-3, 1) = (-1, 1)
$$
as desired.

*$\boxed{\supseteq}:$ Choose any $x \in (-1, 1)$. To prove our claim, it is enough to show that $x \in \bigcap_{t \geq 0} A_t$, since this is a subset of $\bigcup_{s \in \mathbb R} \bigcap_{t \geq s} A_t$. To this end, choose any $t \geq 0$. We need to show that $x \in A_t$. Indeed, notice that:
\begin{align*}
x
&> -1 &\text{since } x \in (-1, 1) \\
&= 1 - 2 \\
&\geq \cos t - 2 &\text{since } \cos t \in [-1, 1] 
\end{align*}
Likewise:
\begin{align*}
x
&< 1 &\text{since } x \in (-1, 1) \\
&= -1 + 2 \\
&\leq \cos t + 2 &\text{since } \cos t \in [-1, 1] 
\end{align*}
Thus, we conclude that $x \in (\cos t - 2, \cos t + 2) = A_t$, as desired.
