Prove that a convergent real sequence always has a smallest or a largest term

Let $c=\sup A$ and $d=\inf A$. If these values are not attained by the sequence then there is a subsequence $a_{n_k}$ strictly increasing to $c$ and a subsequence $a_{m_k}$ strictly decreasing to $d$. But the sequence is convergent so we must have $c=d$. But then $a_n$ is independent of $n$ contradicting the assumption that sup and inf are not reached.

$A:=${$a_n| n \in \mathbb{Z^+}$}

Since $a_n$ is convergent, it is bounded,

$S:= \sup A$, $I:=\inf A$ exist.

Assume $S, I \not \in A$.

1)$ \sup A \not = \inf A$ ;

There exists a subsequence $a_{n_k}$ of $a_n$ converging to $S$.

There exist a subsequence $a_{n_l}$ of $a_n$ converging to $I$.

Since $a_n$ is convergent every subsequence converges to the same limit.

A contradiction.

2) $S=I$ , and by assumption $S,I \not \in A$,

we have $I <a_n<S$ , i.e .

$I <S$, a contradiction.