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Theorem 2.38 states the following: If $\{I_n\}$ is a sequence of intervals in $R^1$, such that $I_n$ contains $I_{n+1}$ $(n=1,2,3,…),$ then the infinite intersection of the sets $\{I_n\}$ is not empty.

In the proof, Rudin then defines $\{I_n\} = [a_n, b_n]$, and set $E$ as the set of all $a_n$. Letting $x =$ sup $E$ and arbitrary $m, n \in \mathbf{N}$, the following is true $$a_n \leq a_{m+n} \leq b_{m+n} \leq b_{m}.$$

He then establishes the fact that $x \leq b_m$ for any $m \in \mathbf{N}$. This is the step that I fall short on. Why is this statement true?

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    $\begingroup$ From the above $a_n\le b_m$, so $b_m$ is an upper bound of $E$, and so $\ge$ the least upper bound of $E$. $\endgroup$ Jun 22, 2017 at 6:18
  • $\begingroup$ Oh that was easy $\endgroup$
    – lithium123
    Jun 22, 2017 at 6:20

1 Answer 1

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Suppose there exists $n\in\mathbb N$ such that $x > b_n$. Letting $\varepsilon = x-b_n>0$, there then exists $m\in\mathbb N$ such that $x-a_m < \varepsilon/2$. But then $a_m-b_n = a_m-x+x-b_n>\varepsilon/2$, and so $a_m > b_n$, which is impossible.

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