Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: $$s = \begin{cases} y & \text{if } 0 < y < 1 \\ 1/y & \text{if } 1 < y \\ 1/2 & \text{otherwise.} \\ \end{cases}$$

share|cite|improve this answer
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 $x \notin A_{\sigma_x}$.

share|cite|improve this answer

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

share|cite|improve this answer

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.