# How to conclude that $a \leq \mathrm{sup}_{n} a_n = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$

This is part of the exercise 5 of chapter 1 in Real Analysis by N. L. Carothers. The whole problem is the following:

Suppose that $$a_n \leq b$$ for all $$n$$, and that a = $$\lim_{n\to\infty}a_n$$ exists. Show that $$a \leq b$$. Conclude that $$a \leq \mathrm{sup}_{n} a_n = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$$.

So far what I have done is that I have used the definition of a limit to conclude that $$\left| a_n - a \right| < \epsilon \leftrightarrow a_n + \epsilon > a > a_n - \epsilon \leftrightarrow b + \epsilon \geq a_n + \epsilon > a > a_n - \epsilon \leftrightarrow a_n - \epsilon < a < b + \epsilon$$. Since $$\epsilon$$ can be made arbitrarily small, we conclude that $$a < b$$.

So I am not sure how to show the missing equality, and how to use the first part of the proof in the $$\mathrm{sup}_n$$ part.

• Well, $a_n \leq \sup_{n} a_n$, so just taking $\sup_n a_n = b$ in the first part gives the second part. Commented Dec 22, 2020 at 14:59
• the last part of your proof is not totally right, from the condition $a<b+\epsilon$ for arbitrary $\epsilon >0$ you deduce that $a\leqslant b$, but you cannot deduce that $a<b$, you can see why choosing $a=b=0$ and $a_n=-1/n$. Commented Dec 22, 2020 at 15:01

Your conclusion that $$a < b$$ is not right, for example $$\forall n\ \left(1 - \frac{1}{n}\right) < 1$$, but $$\lim_\limits{n\to\infty}\ \left(1 - \frac{1}{n}\right) = 1 \not < 1$$
So you can conclude only that $$a\le b$$.
(formally, you show that $$\forall \epsilon > 0\ a < b + \epsilon$$
hence $$a$$ is lower bound of $$\{b + \epsilon\ |\ \epsilon > 0\}$$ and by definition of infinum
$$a \le \inf{\{b + \epsilon\ |\ \epsilon > 0\}} = b$$)
Second part: by definition of supremum $$\forall n\ a_n \le \mathrm{sup} \{a_n : n \in \mathbb{N}\}$$
You can take $$b = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$$, and use first part.