$a_{n}=\sup \left\{x_{k}: k \geq n\right\} .$ If $a_{i}j \in \mathbb{N},$ then $a_{j} \in\left\{x_{n}: n \in \mathbb{N}\right\}$ Let $\left\{x_{n}\right\}$ be a sequence, and define $a_{n}=\sup \left\{x_{k}: k \geq n\right\} .$ If $a_{i}<a_{j}$ for some $i>j \in \mathbb{N},$ then $a_{j} \in\left\{x_{n}: n \in \mathbb{N}\right\}$
I am just a few chapters into an intro to real analysis course. This is a prove or disprove. 
Honestly, I am pretty lost on this one even though it seems like it should be simple. $a_n$ is the supremum of a tail of the sequence. However, I know a supremum doesn't have to be in the sequence itself. So I believe this is asking "under these conditions, is $a_j$ actually in the original set that you are taking the supremum of?" But... I have no idea. I feel like we are given very little information. If it's false, I am not sure how I would disprove it. And if it is true, I don't know how to prove it. Any help is greatly appreciated!
 A: You have the $\left\{a_{n}\right\}$ sequence defined in terms of the $\left\{x_{n}\right\}$ sequence as
$$a_{n}=\sup \left\{x_{k}: k \geq n\right\} \tag{1}\label{eq1A}$$
You're asked to either prove or disprove that if $a_{i}<a_{j}$ for some $i>j \in \mathbb{N}$, then $a_{j} \in\left\{x_{n}: n \in \mathbb{N}\right\}$.
Note that you have for any $m \gt n$ that
$$a_{n} = \max(\sup \left\{x_{k}: n \le k \lt m\right\}, \sup \left\{x_{k}: k \geq m \right\}) \tag{2}\label{eq2A}$$
This is relatively easy to verify since, if any value among those with indices $n \le k \lt m$ is larger than or equal to all of the ones for $k \ge m$, then the supremum (actually, maximum since it's a non-empty, finite set) of this first subset would be at least as large as that of the other values and, thus, would be equal to the supremum overall. Otherwise, the supremum of the first finite set would be less than the supremum of the second set, and thus the supremum of the entire set would be equal to the supremum of the second set, i.e., for all $k \ge m$.
Using this, for $a_i \lt a_j$ for some $i \gt j$  means
$$\begin{equation}\begin{aligned}
a_{j} & = \max(\sup \left\{x_{k}: j \le k \lt i\right\}, \sup \left\{x_{k}: k \geq i \right\}) \\
& = \max(\sup \left\{x_{k}: j \le k \lt i\right\}, a_{i}) \\
& = \sup \left\{x_{k}: j \le k \lt i\right\}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since the number of elements in the non-empty set $\left\{x_{k}: j \le k \lt i\right\}$ is finite, in particular it has $i - j$ elements, the supremum is the maximum of the elements and, thus, is equal to one of the elements of the set. In other words, you then have that
$$a_{j} \in\left\{x_{n}: n \in \mathbb{N}\right\} \tag{4}\label{eq4A}$$
This shows the requested statement is always true.
