Prove a convergent sequence has either a minimum, a maximum or both. Let $a_n$ be a convergent sequence. Prove $a_n$ has a minimum, a maximum or both. 
I am being prepared for a final exam, which is why it is important to me to know that $I$ am correct in $my$ attempt. Of course if I am completely wrong, hints or solution are welcome. Thanks. 
$Attempt$: $a_n$ converges to a limit $L\in \Bbb{R}$ as $n\to \infty$. Therefore, for $\epsilon=1$ we get, for large enough $N$ that $\forall n\ge N,$ $|a_n-L|<1$ $\Rightarrow$ $L-1 \le a_n\le L+1$, in particular $a_n\le L+1$. Therefore, for $M=max(a_1,a_2,...,a_N,L+1)$, $a_n\le M$ necessarily. i.e, $a_n$ is upper bounded. The same can be shown with lower bound. If $a_n$ is constant, we are done. Otherwise, the lower and the upper bounds are different. Suppose $a_n$ has no minimum nor maximum, then both lower and upper bound are accumulation point, a contradiction. Therefore $a_n$ has a maximum or a minimum in that case. 
 A: We use the notation of the OP. The result is clear if all the $a_k$ are equal to the limit $L$. Thus we can assume that for some $k_0$ we have $a_{k_0}\ne L$. For simplicity let $a_{k_0}=c$. By symmetry we can suppose that $c\gt L$.
By the definition of limit, there is an index $N$ such that $|a_n-L|\lt c-L$ if $n\gt N$. In particular,  $a_n\lt c$ for $n\gt N$.
The (multi)set $\{a_1,a_2,\dots,a_N\}$ has a maximum $b$, and $b\ge a_k$ for all $k$. 
A: Let $(a_n)_n$ a real-valued convergent sequence. As you proved, this sequence is bounded, so that $m = \inf\{a_n\;|\;n\in\mathbf{N}\}$ and $M = \sup\{a_n\;|\;n\in\mathbf{N}\}$ do exist in $\mathbf{R}$.
We want to show that there exist an $n\in\mathbf{N}$ such that either $a_n = m$ or $a_n = M$. If $m=M$ everything is clear, so we may suppose $m<M$.
Suppose the contrary : $\forall n\in\mathbf{N}, a_n \not= m$ and $a_n \not= M$, i.e. $\forall n\in\mathbf{N}, a_n > m$ and $a_n < M$, by definition of $m$ and $M$. Recall what a sup (resp. inf) $m$ (resp. $M$) of a non empty-set $X$ bounded from above (resp. bounded from below) is. It is the unique $m$ (resp $M$) such that $\forall \varepsilon > 0$, there exist an $x\in X$ such that $M-\varepsilon \leq x$ (resp. $x \leq m+\varepsilon$.) Let apply this with $X = \{a_n\;|\;n\in\mathbf{N}\}$. For $n = 0$, there is a $k_0$ (resp $l_0$) such that $M - 2^{-0} \leq u_{k_0} < M$ (resp. $m + 2^{-0} \geq l_{k_0} > m$.) There is now a $k_1$ (resp $l_1$) strictly bigger that $k_0$ (resp $l_0$) such that $\max(M - 2^{-1} , u_{k_0}) < u_{k_1} < M$ (resp. $\min(m + 2^{-1} , u_{l_0}) > u_{l_1} > m$.) (As we know by previous inequalities that the most right hand side term is $< M$ (resp. ($>m$.)
Iterating this process, we are able to construct a strictly increasing sequence $(k_n)_n$ (resp< $(l_n)_n$) such that the sequence $(a_{k_n})_n$ (resp. $(a_{l_n})_n$) is strictly incresasing (resp. decreasing) and such that $\forall n \in\mathbf{N}, a_{k_n} \in ]M-1/2^n, M[$ (resp. $a_{l_n} \in ]m+1/2^n,m[$. Therefore $(a_{k_n})_n$ converges to $m$ and $(a_{l_n})_n$ to $M$, showing that $m=M=\lambda$, where $\lambda$ is the limit of $(a_n)_n$, as every subsequence of a convergent sequence converges to the same limit than the one of the convergent sequence. This is a contradiction, as we supposed that $m<M$. This shows that in any case, there exist an $n\in\mathbf{N}$ such that either $a_n = m$ or $a_n = M$.
As you may remark, you don't really need our two subsequences two be strictly monotonous, so that knowing that $m$ and $M$ are accumulation points of $X$ is not needed. The only thing that is needed is that $m$ and $M$ are adherent to $X$.
