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.
 A: 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: $$s = \begin{cases}
y & \text{if } 0 < y < 1 \\ 
1/y & \text{if } 1 < y \\ 
1/2 & \text{otherwise.} \\ 
\end{cases}$$
A: Consider $s_1,s_2 \in S.$ If $s_1 < s_2$ then $A_{s_2} \subsetneq A_{s_1}$. It follows that
$$ \bigcap_{s \in S} A_s = \lim_{s \to 1} A_s = \{1\} \, . $$ 
To understand why the limit is as it is, note that $1 \in A_s$ for all $s \in S$. Furthermore, for all $x > 0$ and different from 1, there exists $\sigma_x \in S$ such that $x \notin A_{\sigma_x}$. 
For an explicit construction, consider the two cases: $0 < x < 1$ and $x > 1$. If $0 < x < 1$ then $\sigma_x = \frac{1}{2}(1 - x)$ satisfies $x \notin A_{\sigma_x}$. If $x > 1$ then $\sigma_x = \frac{2}{x}$ satisfies $x \notin A_{\sigma_x}$.
A: Try starting the proof by showing that $1$ is in $A_s$ for every $s\in S$. This proves that
$$1 \in \bigcap_{s\in S}A_s$$
rather,
$$\{1\} \subseteq \bigcap_{s\in S}A_s$$
Then, suppose to the contrary that there was some $x\ne 1$ such that $x\in A_s$ for all $s\in S$ (emphasis on the "for all" because any fewer would not suffice). Arriving at a contradiction would show that
$$\bigcap_{s\in S}A_s \subseteq \{1\}$$
Put parts A and B together to get
$$\bigcap_{s\in S}A_s = \{1\}$$
A: 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
