Problem: Prove that a monotonically increasing sequence that is bounded above always has a least upper bound.
This theorem is present in calculus and real analysis books along with proofs for it. I want to know if my proof is convincing enough.
Solution: let's start with the definitions
If $ \{a_n\} $ is monotonically increasing, then $ a_{n+1} \geq a_n $ for all $n$.
If $ \{a_n\} $ is bounded above then there exists $M$ such that $a_n \leq M$ for all $n$.
If $ M $ is an upper bound, but no number less than $ M $ is an upper bound for $ \{a_n\} $, then $ M $ is called the least upper bound for $ \{a_n\}$.
It has already been established in the previous exercise that if a monotonically increasing sequence $ \{a_n\} $ is bounded above, then the sequence converges to some value. Let's call this value of convergence $\ell$. Then
$$ lub\{a_n\} = max\{a_n\} = \ell = \lim_{n \to \infty} a_n $$
For if $ \ell $ wasn't the least upper bound of $ \{a_n\} $ then $\{a_n\}$ wouldn't converge to it.
Best Answer
It's not quite convincing enough for me. You didn't explicitly show why it was true. Let $L$ be the limit of the sequence $\{a_n\}$ We'll use the monotone convergence theorem as you said it was already proven to you. Clearly $L \geq a_n$ for all $n$. And so $L$ is an upper bound.
Assume there is some upper bound which is less than $L$, call it $k$. Let $L - k = p$. From the definition of the limit of a sequence, there exists an $N$ such that for all $n > N$, $|a_n - L| < \varepsilon$ for all $\varepsilon > 0.$ Now take $\varepsilon = p$. This shows that there is a term which is greater than $k$, contradicting the choice of $k$ as an upper bound.
Now we have shown that $L$ is a upper bound, and any value smaller than $L$ is not an upper bound, finishing the proof.