Need help proving that an intersection has a single element. Let $S=\{s : 0 < s < 1 \}$, and $A_s = \{x : s < x < 1/s \}$.
Claim I want to prove: 
$$\bigcap_{s \in S} A_s = \{1\} \, . $$
I'm not sure how to demonstrate this rigorously. However, I do understand that if we pick an $s$ very close to $0$ we will get a very wide interval.  If  we pick numbers close to $1$ we get very narrow intervals.

First. show that $1 \in A_s$ for all $s \in S$.
Next, show that for any $y \ne 1$, you can find an $s \in S$ such that $y \notin A_s$.

For example, the following choice will work: $$s = \begin{cases}
y & \text{if } 0 < y < 1 \\ 
1/y & \text{if } 1 < y \\ 
1/2 & \text{otherwise.} \\ 
\end{cases}$$
