Prove that a convergent real sequence always has a smallest or a largest term My attempt: 
Suppose not, i.e., suppose there exists a convergent sequence $(a_n)$ that does not have a smallest or largest term.
$\implies (a_n)$ is bounded sequence.
$\implies A=\{a_n:n\in\mathbb{Z}^+\}$ is a bounded subset of $\mathbb{R}$.
$\implies \sup{A},\inf{A}$ exists in $\mathbb{R}$.
Now, we need to show that at least one of $\sup{A},\inf{A}\in A$.
Equivalently, we need to show that the following case is impposible:
"Both $\sup{A}\notin A$ and $\inf{A}\notin A$".  
I don't know how to proceed or if I am working it out correctly.
 A: $A:=${$a_n| n \in \mathbb{Z^+}$}
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.
A: 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: Case $\bf{1}$:
If there is a $k$ so that $a_k\lt\lim\limits_{n\to\infty}a_n$, then $\inf\limits_{n\ge0}a_n\lt\lim\limits_{n\to\infty}a_n$. Since the limit exists, there is an $n_0$ so that
$$
n\ge n_0\implies a_n\ge L=\frac12\left(\inf_{n\ge0}a_n+\lim_{n\to\infty}a_n\right)\tag1
$$
Thus, there are only a finite number of terms where $a_n\lt L$. Since an infimum over a compact set is attained, there must be an $n_1$ so that $a_{n_1}=\inf\limits_{n\ge0}a_n$.
That is, the infimum is attained.
Case $\bf{2}$:
If there is a $k$ so that $a_k\gt\lim\limits_{n\to\infty}a_n$, then $\sup\limits_{n\ge0}a_n\gt\lim\limits_{n\to\infty}a_n$. Since the limit exists, there is an $n_0$ so that
$$
n\ge n_0\implies a_n\le L=\frac12\left(\sup_{n\ge0}a_n+\lim_{n\to\infty}a_n\right)\tag2
$$
Thus, there are only a finite number of terms where $a_n\gt L$. Since a supremum over a compact set is attained,  there must be an $n_1$ so that $a_{n_1}=\sup\limits_{n\ge0}a_n$.
That is, the supremum is attained.
Case $\bf{3}$:
If neither Case $1$ nor Case $2$ hold, then for all $k$, $a_k=\lim\limits_{n\to\infty}a_n$.
That is, both supremum and infimum are attained.
A: The sequence is convergent, hence bounded; if $q=\sup\{a_n:n\ge0\}$ and $p=\inf\{a_n:n\ge0\}$, then we know that $l=\lim_{n\to\infty}a_n\in[p,q]$. The case $p=q$ is obvious, so we can assume $p<q$.
Suppose $l>p$. Then, for $n>N$, $a_n>(l+p)/2>p$ and so $p=\inf\{a_n:0\le n\le N\}$ is actually a minimum.
Similarly, if $l<q$.
