A convergent sequence has either a maximum or a minimum or both. At odd times when I am not writing my Ph.d thesis I am solving some problems from Polya-Szego book which seems to me really interesting. Here is the one of the problems (Problem 106 from Volume 1) which I've solved and would be grateful if you can take a look.

A convergent sequence has either a maximum or a minimum or both.

Suppose $\{a_n\}$ be a convergent sequence but there is no $\min \limits_{n\in \mathbb{N}} a_n$ and there is no $\max \limits_{n\in \mathbb{N}}a_n$.
It is easy to show the following fact: If the sequence $\{a_n\}$ does not have a minimum then $\forall n\in \mathbb{N}$ $\exists k>n$ such that $a_k<a_n$.
In the same way you can show that if the sequence $\{a_n\}$ does not have a maximum then $\forall n\in \mathbb{N}$ $\exists m>n$ such that $a_m>a_n$.
Then using it we can construct two increasing subsequences $n_1<n_2<n_3<\dots$ and $m_1<m_2<m_3<\dots$ such that $n_1=m_1=1$ and $a_{n_1}>a_{n_2}>a_{n_3}>\dots $ and $a_{m_1}<a_{m_2}<a_{m_3}<\dots$.
Since $\{a_n\}$ is convergent then $a_{n_k}\to a$ and $a_{m_k}\to a$ as $k\to \infty$.
Then $a=\inf \limits_{k\in \mathbb{N}}a_{n_k}$ and $a=\sup \limits_{k\in \mathbb{N}}a_{m_k}$.
Hence $a_{n_1}>a$ and $a_{m_1}<a$ but this is a contradiction because $n_1=m_1=1$.
Please let me know is everything valid in this short proof?
 A: Your proof is correct! Here is an alternative proof that I like.
If $a_n$ is a constant sequence, then it trivially has a maximum and minimum. Let $\{a_n\}$ be a non-constant convergent sequence with limit $a$. Let $m$ be such that $a_m \neq a$. By the definition of convergence, there exists a $k \in \Bbb N$ such that $n > k \implies |a_n - a| < |a_m - a|$.
Suppose that $a_m > a$. We see that for all $n > k$, $a_n < a + (a_m - a) = a_m$. It follows that
$$
\sup(\{a_n\}_{n \in \Bbb N}) = \sup \{a_1,\dots,a_k\}.
$$
Thus, the supremum of the sequence is equal to $a_n$ for some $n$ between $1$ and $k$, which is to say that $\{a_n\}$ attains a maximum.
Similarly, if $a_m < a$, we can conclude that $\{a_m\}$ attains a minimum. The conclusion follows.
A: Consider a  sequence $(a_{n})$ converging to $a$ which has no maximum and minimum. Suppose that the sets $A=\{a_{n}| a_{n}\lt{d}\}$ and $B=\{a_{n}| a_{n}\gt{d}\}$ have infinite elements for some real number $d\neq{a}$ which is not a limit point of both $A$ ,$B$(if $A$ (or $B$ ) is finite then the minimum of $A$ is the minimum of the sequence $(a_{n})$ and the maximum of $B$ is the maximum of the sequence $(a_{n})$). Note that the sequence $(a_{n})$ is bounded as a convergence sequence. Then the sets $A$ and $B$ have at least a limit point and we can obtain two subsequences of $A$ and $B$ with different limits $x$ and $y$ satisfy $x \lt{d}\lt{y}$. This contradicts the convergence of the sequence
