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Let $(x_n),(y_n)$ be sequences as in the Nested Interval Property.

Show that for all $n,m \in N$, $x_n \leq y_m$

We know from Nested Interval Property that

  1. $x_1 \leq x_2 \leq x_3 \leq \cdots$

  2. $y_1 \geq y_2 \geq y_3 \leq\cdots$

  3. $x_n \leq y_n$ for all $n$

  4. $\lim _{n\to\infty} (y_n-x_n) =0$

  5. Squeeze Theorem

How can I use them to prove the question please?

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  • 1
    $\begingroup$ If $m < n$ then $x_m \leq x_n \leq y_n \leq y_m$. If $m > n$ then $x_n \leq x_m \leq y_m \leq y_n$. In both cases, $x_n \leq y_m$. $\endgroup$ – Bungo Mar 28 '17 at 21:17
  • $\begingroup$ how about if m=n ? $\endgroup$ – qwer tyui Mar 28 '17 at 21:40
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    $\begingroup$ If $m=n$ then simply observe that $x_n \leq y_n = y_m$. $\endgroup$ – Bungo Mar 28 '17 at 21:42
  • $\begingroup$ I think before I had to use sequence definition to approach it but you give me easy way $\endgroup$ – qwer tyui Mar 28 '17 at 21:50

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