How does the existence of $n_0$ follow in this partial proof of the Monotone Convergence Theorem for sequences?

Question

The following is a partial proof of the Montone Convergence Theorem for sequences transcribed from one of Alfonso Gracia-Saz' Calculus videos.

More specifically, what it's proving is that if a sequence is increasing and bounded above, then it is convergent.

1. Let $\big\{ a_n\big\}_{n = 0}^\infty $ be an increasing sequence that's bounded above.

2. Let's consider the set $\mathcal{A} = \{ a_n \mid n \in \mathbb{N} \} $. It is non-empty, and bounded above, so by the Least Upper Bound Principle, it must have a supremum.

3. We will take $L = \sup \mathcal{A}$. We will prove that $ L = \displaystyle \lim_{n \to \infty} a_n$.

4. Let $\epsilon > 0$.

5. By the definition of supremum, $ \exists n_0 \in \mathbb{N}, L - \epsilon < a_{n_0}$. We take that value of $n_0$.

6. Let $n \in \mathbb{N}$. Assume $n \geq n_0$. We want to show that $L - \epsilon < a_n < L + \epsilon$.

7. We know $L - \epsilon < a_{n_0}$.

8. Because the sequence is increasing, $a_{n_0} \leq a_n$.

9. By the definition of supremum, $a_n \leq L$.

10. Thus, $$ L - \epsilon < a_{n_0} \leq a_n \leq L < L + \epsilon $$ $ \blacksquare$

My question is regarding line (5) above. I don't understand how the existence of $n_0$ follows from the previous lines and the definition of supremum. What if the supremum is not contained within the sequence (it's not the maximum after all), and $\epsilon$ is a very, very small real number? Wouldn't that make it possible for such a natural number $n_0$ not to exist?

I'm sure I'm missing something, but I can't seem to figure it out on my own.

Answer 1

You can prove 5) by contradiction. If the statement in 5) is false, then $a_n \leq L-\epsilon$ for all $n$. This means $L-\epsilon$ is an upper bound for $\mathcal A$. But you cannot have an upper bound smaller than the least upper bound.

Answer 2

In general, you would be right and the correct expression would be: Given $\epsilon>0$ $\exists n_0 \in \mathbb{N}, L - \epsilon \leq a_{n_0} $

For example, for the set $(0,1)\cup\{2\}$ you need the inequality to be non strict

Equally, if $\big\{ a_n\big\}_{n = 0}^\infty$ is just increasing, you're correct in your suspicion, that such $n_0$ might not exist, take for example$a_n=$constant

This is however a minor detail that's often ignored, becuase you the strictness of the inequality in the definition of sequence converging to a point is not relevant and can be changed at will(Still, if it really bother you, change $\epsilon$ to $\frac{\epsilon}{2}$