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.
Best Answer
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 $k$.