Statement
An increasing sequence converges (that is it has finite limit) if and only if has supremum and its limit is the supremum.
So I ask if the statement is true and so to prove it. On the contrary if the statement is not true I ask to give a counterexample. So could someone, help me?
Lemma
An increasing sequence converges (that is it has finite limit) if and only if has supremum and its limit is the supremum.
Proof. So if $(a_n)_{n\in\Bbb N}$ is an increasing sequence that converges (that is it has finite limit) then we prove that limit $l$ is an upper bound. So if $l$ was not an upper bound then there must exist $n_l\in\Bbb N$ such that $l<a_{n_l}$ and so for $\epsilon:=(a_{n_l}-l)$ by the convergence to $l$ of the sequence $(a_n)_{n\in\Bbb N}$ there exist $n_0\in\Bbb N$ such that $a_n<\epsilon +l=a_l$ for any $n\ge n_0$ and so by the increase of the sequence it would be $n_0<n_l$ so that finally it would be $a_{n_l}<\epsilon+l=a_{n_l}$ that obviously is impossible. So the set of the upper bound of the sequence is not empty and so by well-know property of sets of real numbers we conclude that the sequence is bounded above, that is has a supremum. Now we prove that $l$ is the minimum of the upper bound. So if $l$ is the limit of the sequence $(a_n)_{n\in\Bbb N}$ then for any $\epsilon>0$ there must exist $n_0\in\Bbb N$ such that $l-\epsilon<a_n$ for any $n\ge n_0$ so that by the arbitrariness of $\epsilon>0$ we conclude that $l$ is the minimum of the upper bound, that is it is the supremum.