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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 ?

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    $\begingroup$ 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. $\endgroup$ – zbrads2 Apr 26 at 1:56
  • $\begingroup$ In addition to Martin's answer, the Nested Interval Property requires closed intervals. $\endgroup$ – Lucas Corrêa Apr 26 at 1:58
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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 $y<x$ for all $x>0$. As Rudin says, this is a contradiction because given $y>0$, you can always find $x$ with $0<x<y$ (for instance, $x=y/2$).

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

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

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