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I am trying to give another proof for the theorem:

If $(a_n)$ is monotone and bounded the $(a_n)$ converges.

I saw the question at Lingnoi401 (https://math.stackexchange.com/users/353503/lingnoi401), Show that a bounded, monotone increasing sequence is a Cauchy sequence, URL (version: 2017-03-03): Show that a bounded, monotone increasing sequence is a Cauchy sequence,

and the answer given by Mark Bennet (https://math.stackexchange.com/users/2906/mark-bennet), Show that a bounded, monotone increasing sequence is a Cauchy sequence, URL (version: 2017-03-03): https://math.stackexchange.com/q/2169936.

I understand why for $s=\sup\{a_n|n \in \mathbb{N}\}$ we have $s-\varepsilon < a_N$ for sufficiently large $N$, and therefore for every $n,m>N$.

However, I have trouble understanding formally how to get from:

$s-\varepsilon <a_N \leq a_n \leq a_m \leq s$

to:

$0 \leq a_m - a_n < \varepsilon$

The left inequality is understood due to $(a_n)$ being monotone (increasing for example in this case). I don't understand how to get to the right inequality from the first chain of inequalities algebraically.

Thank you for the help!

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    $\begingroup$ We have $a_n> s-\varepsilon$, so $-a_n<\varepsilon-s$, along with $a_m\leq s$, it follows $a_m-a_n<s+(\varepsilon-s)=\varepsilon$ $\endgroup$ Commented Nov 5, 2023 at 11:44

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You missed something from the proof that you are trying to understand, namely that $m>n$. Then, since the sequence is increasing, $a_m-a_n\geqslant0$.

On the other hand, both numbers $a_m$ and $a_n$ belong to the interval $(s-\varepsilon,s]$. So, $a_m\leqslant s$ and $a_n>s-\varepsilon$. Therefore$$a_m-a_n<s-(s-\varepsilon)=\varepsilon.$$

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  • $\begingroup$ Thank you! I would just like to make sure I understood correctly: the reason for the strict inequality is that both $a_m,a_n$ are strictly greater than $s-\varepsilon$ Additionally, since they are both elements of the interval $(s-\varepsilon,s]$ we can deduce the fact they are between $s$ and $s-\varepsilon$ and get that last inequality. $\endgroup$ Commented Nov 6, 2023 at 17:17
  • $\begingroup$ No. It's because $a_m\leqslant s$ and $a_n>s-\varepsilon$ (and therefore $-a_n<-(s-\varepsilon)$). $\endgroup$ Commented Nov 6, 2023 at 17:21
  • $\begingroup$ All right, I'll try again: $a_m \leq s$, and $-a_n < -(s-\varepsilon)$ thus $a_m \leq s$ / $(-a_n)$ $\Rightarrow$ $a_m-a_n \leq s-a_n < s-(s-\varepsilon) = \varepsilon$ $\endgroup$ Commented Nov 6, 2023 at 17:27
  • $\begingroup$ Maybe that you meant that$$a_m\leqslant s\wedge-a_n<-(s-\varepsilon)\implies a_m-a_n<s-(s-\varepsilon)=\varepsilon.$$ $\endgroup$ Commented Nov 6, 2023 at 17:33
  • $\begingroup$ Yes, thank you! $\endgroup$ Commented Nov 9, 2023 at 16:52

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