Need help proving that an intersection has a single element. Let $S=\{s : 0 &lt; s &lt; 1 \}$, and $A_s = \{x : s &lt; x &lt; 1/s \}$.

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: $$s = \begin{cases}
y &amp; \text{if } 0 &lt; y &lt; 1 \\ 
1/y &amp; \text{if } 1 &lt; y \\ 
1/2 &amp; \text{otherwise.} \\ 
\end{cases}$$
Any number less than 1 take its inverse it is greater than 1 so everfy set contains 1 intersection of those contains 1
Also if s ->1 then 1 / s ->1 since 1 / s is continuous around 1
The result follows