Prove that for a convergent sequence with no max, sup(an)=L Let ${a_n}$ be a convergent sequence such that $\lim_{n \to \infty}=L$. Suppose $a_n$ has no maximum. Prove that $\sup{(a_n)}=L$.
In a similar way, Let ${a_n}$ be a convergent sequence such that $\lim_{n \to \infty}=L$. Suppose $a_n$ has no minimum. Prove that $\inf{(a_n)}=L$.
I'm having some trouble formalizing this idea... i did try:
$a_n$ converges to $L \Rightarrow$ let $\varepsilon \gt 0.$ for some $n \geq N\in\Bbb N, \ |a_n-L|\lt \varepsilon.$ From the completeness axiom, any non-empty bounded sequence has a $\sup \Rightarrow$ let $M=sup(a_n)$.
For any $\varepsilon \gt 0\ $ :   $\ M- \varepsilon \leq a_n \lt M$
Now im pretty confused. how should i link M to L, the limit? Is this even the right direction?
 A: Clearly $M\geq L$, as the definition of limit implies that no upper bound can be less than $L$. 
If $k$ is such that $a_k>L$, then there are only finitely many terms of the sequence that are greater than or equal to $a_k$, and one of them is going to be a maximum. And if no such $k$ exists, but there is an $\ell$ such that $a_\ell = L$, then that is a maximum.
By the above paragraph, assuming that there is no maximum implies that $a_n<L$ for all $n$, which again means that $L$ is an upper bound for the sequence. And since $M$ is the least upper bound, we have $M\leq L$.
A: Let $M=\sup_na_n$
The supremum exists since $a_n$ is bounded.
By the know property of supremum we have that,for $\epsilon=\frac{1}{n}$ you can find a subsequence of distinct(Why??) points(since $a_n$ has no maximum) $a_{m_n}$ of $a_n$ such that $M-\frac{1}{n}<a_{m_n}\leq M$
So $a_{m_n} \to M$ and also $a_{m_n} \to L$(since every subsequence of a convergent sequence has the same limit as the original sequence)and by uniqueness of limit we have that $L=M$
A: 
1) $L\leq \sup_{n\in\mathbb N}a_n$.

Indeed, if $L>\sup_{n\in\mathbb N}a_n$, then, if $0<\varepsilon<L-\sup_{n\in \mathbb N}a_n$, by definition of the convergence, there is $N$ s.t. $$\sup_{n\in \mathbb N}a_n<L-\varepsilon <a_N,$$ which is not possible.

2) $L=\sup_{n\in \mathbb N}a_n$.

Suppose $L<\sup_{n\in\mathbb N}a_n$. In particular, if $0<\varepsilon <\sup_{n\in \mathbb N}a_n-L$, by definition of the convergence, there is $N$ s.t. $$a_k\leq L+\varepsilon<\sup_{n\in \mathbb N}a_n,$$
for all $k\geq N$. Therefore $$\sup_{n\in\mathbb N}a_n=\sup_{n=1,...,N}a_n,$$
and thus, there is $k\in\{1,...,N\}$ s.t. $$a_k=\sup_{n\in\mathbb N}a_n.$$
Contradiction with the fact that $(a_n)$ has no maximum.  
