# How can you prove that $x_n$ is a constant sequence if and only if $x_n$ is both monotone increasing and monotone decreasing?

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 $x_{n+1} = x_n$.

• Recall that two numbers $a$ and $b$ are equal iff $a\geq b$ and $a\leq b$. Are you still confused? – Juniven Jan 9 '17 at 0:48
• I know, but is there mathematically rigorous proof to show that is true? Or is it too obvious to do so? – user3000482 Jan 9 '17 at 0:50
• 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. – mathworker21 Jan 9 '17 at 0:52

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 $x_n=x_{n+1}$.

• But this is when the sequence is not monotone increasing and monotone decreasing, that is, $x_{n+1} \gt x_n$ $x_{n+1} \lt x_n.$ – user3000482 Jan 9 '17 at 0:57
• 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. – Tsemo Aristide Jan 9 '17 at 1:00
• So, I guess the problem is not valid? – user3000482 Jan 9 '17 at 1:02
• 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. – Tsemo Aristide Jan 9 '17 at 1:03
• I see it now, thank you! – user3000482 Jan 9 '17 at 1:04

Hint: Logically, $x_{n+1}\geq x_n$ means $x_{n+1}>x_n$ or $x_{n+1}=x_n$.

• Actually, that's what it means by definition I think. – Simply Beautiful Art Jan 9 '17 at 0:44
• @SimpleArt Sure. I want to emphasizes the fact we have $A\vee B$. – Jacky Chong Jan 9 '17 at 0:46
• 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? – user3000482 Jan 9 '17 at 1:06

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.

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 $$a or $$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 $$a or $$a=b$$.

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 $n$.