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.

  • $\begingroup$ Well, $a_n \leq \sup_{n} a_n$, so just taking $\sup_n a_n = b$ in the first part gives the second part. $\endgroup$ Commented Dec 22, 2020 at 14:59
  • $\begingroup$ 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 $a_n=-1/n$. $\endgroup$
    – Masacroso
    Commented Dec 22, 2020 at 15:01

1 Answer 1


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.


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