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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.

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The usual way to prove a set equality is to demonstrate that anything in the left-hand side is in the right hand side and vice versa. – Ben Millwood Sep 8 '12 at 19:56
up vote 6 down vote accepted

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: $$s = \begin{cases} y & \text{if } 0 < y < 1 \\ 1/y & \text{if } 1 < y \\ 1/2 & \text{otherwise.} \\ \end{cases}$$

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The only thing I'm unclear about is the second step. I understand how to do it, but I'm not sure how it fits into the proof. – emka Sep 8 '12 at 20:25
By definition, $y \in \bigcap_{s \in S}A_s$ if and only if $y \in A_s$ for all $s \in S$. If you find an $s$ that disproves the latter, you've disproved the former. – Ilmari Karonen Sep 8 '12 at 20:27

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 $x \notin A_{\sigma_x}$.

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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 $$\bigcap_{s\in S}A_s = \{1\}$$

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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

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The second sentence (or, what would be the second sentence if you used punctuation) is unclear. – Trevor Wilson Sep 8 '12 at 19:45

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