# Rudin 2.10 (b) Example

Let $$A$$ be the set of real numbers such that $$0 < x \leq 1$$ . For every $$x \in A$$, let $$E_{x}$$ be the set of real numbers $$y$$ such that $$0< y< x$$. Then

1. $$\bigcap_{x \in A} E_{x}$$ is empty.

The Proof provided in textbook: We note that for every $$y > 0$$, $$y \notin E_{x}$$ if $$x < y$$. Hence $$y \notin \bigcap_{x \epsilon A} E_{x}$$

I didn't understand the proof & also here is my understanding.

Counter Argument-1: We can always find a real number between every $$(0,x)$$ where x is $$x > 0$$. So, It can not be empty.

Counter Argument-2: This is somehow similar to Nested interval property, So, it can not be empty.

Please explain how rudin got this result ?

• He is saying that for any $y$, you can find an $x$ in the interval such that $x\leq y$ and so $y$ is not in $E_x$ for that $x$ and so it cannot be in the intersection of all the $E_x$. Since $y$ was arbitrary, the intersection must be empty. – zbrads2 Apr 26 at 1:56
• In addition to Martin's answer, the Nested Interval Property requires closed intervals. – Lucas Corrêa Apr 26 at 1:58

## 2 Answers

Note that $$\bigcap_{x\in A}E_x\subset(0,\infty)$$. If $$y\in \bigcap_{x\in A}E_x\subset(0,\infty)$$, this means that $$y for all $$x>0$$. As Rudin says, this is a contradiction because given $$y>0$$, you can always find $$x$$ with $$0 (for instance, $$x=y/2$$).

Regarding your comment, the nested interval property is about compact sets. These are open and not closed.

Note that $$x\notin E_x$$ for every $$x$$