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I know this statement makes sense if the sequence were simply both increasing and decreasing without being monotone, that is, $x_{n+1} \geq x_n$ and $x_{n+1} \leq x_n.$

Actually, I am now confused how we know $x_{n+1} \geq x_n$ and $x_{n+1} \leq x_n$ implies that $x_{n+1} = x_n$.

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  • $\begingroup$ Recall that two numbers $a$ and $b$ are equal iff $a\geq b$ and $a\leq b$. Are you still confused? $\endgroup$ – Juniven Jan 9 '17 at 0:48
  • $\begingroup$ I know, but is there mathematically rigorous proof to show that is true? Or is it too obvious to do so? $\endgroup$ – user3000482 Jan 9 '17 at 0:50
  • $\begingroup$ Your comment indicates you don't understand how mathematics works. It's not the case we can't prove things rigorously because they are "too obvious". You should look at the definition of $\le$ and $\ge$ on the reals or whatever space the $x_n$'s lie in. $\endgroup$ – mathworker21 Jan 9 '17 at 0:52
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You can write $x_{n+1}\geq x_n$ and $x_{n+1}\leq x_n$ like so:

$x_n\leq x_{n+1}\leq x_n$ so $x_n=x_{n+1}$.

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  • $\begingroup$ But this is when the sequence is not monotone increasing and monotone decreasing, that is, $x_{n+1} \gt x_n$ $x_{n+1} \lt x_n.$ $\endgroup$ – user3000482 Jan 9 '17 at 0:57
  • $\begingroup$ Strictly increasing and strictly decreasing is impossible: If you have $x_{n+1}>x_n$ and $x_{n+1}<x_n$, then $x_n<x_{n+1}<x_n$ implies $x_n<x_n$ impossible. $\endgroup$ – Tsemo Aristide Jan 9 '17 at 1:00
  • $\begingroup$ So, I guess the problem is not valid? $\endgroup$ – user3000482 Jan 9 '17 at 1:02
  • $\begingroup$ the problem is valid if it is increasing and decreasing as you write in the question which is different from being strictly increasing and strictly decreasing. $\endgroup$ – Tsemo Aristide Jan 9 '17 at 1:03
  • $\begingroup$ I see it now, thank you! $\endgroup$ – user3000482 Jan 9 '17 at 1:04
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Hint: Logically, $x_{n+1}\geq x_n$ means $x_{n+1}>x_n$ or $x_{n+1}=x_n$.

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    $\begingroup$ Actually, that's what it means by definition I think. $\endgroup$ – Simply Beautiful Art Jan 9 '17 at 0:44
  • $\begingroup$ @SimpleArt Sure. I want to emphasizes the fact we have $A\vee B$. $\endgroup$ – Jacky Chong Jan 9 '17 at 0:46
  • $\begingroup$ Okay, I understand the second part now, that if a sequence is both monotone increasing and decreasing, then it must be a constant sequence. But how do I prove the first part, that is, $x_n = c \rightarrow x_n$ is both monotone increasing and monotone decreasing? $\endgroup$ – user3000482 Jan 9 '17 at 1:06
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I believe it's called an antisymmetric relation - if for two two real numbers x and y, both inequalities x≤y and y≤x hold then x=y.

In your case, if xn is both monotone increasing and decreasing, from their definitions, xn ≤ x(n+1) and x(n+1) ≤ x both must hold - hence xn = x(n+1). There's no other way the two inequalities would hold, hence xn must be constant.

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You question can be essentially answered by the following fact.

Suppose $a,b\in\mathbb{R}$. Then the following are equivalent:

  • (1) $a\leq b$ and $a\geq b$;
  • (2) $a=b$.

To see how (1) implies (2), suppose (2) is not true. Then one must have $a<b$ or $a>b$. What contradiction can you get?

To see how (2) implies (1), all you need to do is recalling that $a\leq b$ means $a<b$ or $a=b$.

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You can argue like this:

If there is an $n$ such that $x_n \ne x_{n+1}$, then either, by tricotomy, $x_n > x_{n+1}$ or $x_n < x_{n+1}$.

But $x_n > x_{n+1}$ contradicts the assumption that $x_n \le x_{n+1}$.

Similarly, $x_n < x_{n+1}$ contradicts the assumption that $x_n \ge x_{n+1}$.

Therefore, $x_n = x_{n+1}$ for all $n$.

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