In his book Calculus Vol. 1, Apostol writes the following proof:

Theorem I.32. Let $h$ be a given positive integer and let $S$ be a set of real numbers.

(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > \sup S - h$$

Theorem I.33 Additive Property: Given nonempty subsets $A$ and $B$ of $\mathbb R$, let $C$ denote the set$$C = \left\{a + b \mid a \in A, b \in B\right\}.$$

(a) If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and $$\sup C = \sup A + \sup B.$$

Proof. Assume each of $A$ and $B$ has a supremum. If $c \in C$, then $c = a+b$, where $a \in A$ and $b \in B$. Therefore, $c \le \sup A +\sup B$; so $\sup A + \sup B$ is an upper bound for $C$. This shows that $C$ has a supremum and that $$\sup C \le \sup A + \sup B.$$

Now let $n$ be any positive integer. By Theorem I.32 (with $h = \frac{1}{n}$) there is an $a$ in $A$ and a $b$ in $B$ such that $$a > \sup A - \frac1n, \quad b > \sup B - \frac1n.$$

Adding those inequalities, we obtain $$a + b > \sup A + \sup B - \frac2n, \quad or \quad \sup A + \sup B < a + b + \frac2n \le \sup C + \frac2n,$$

since $a + b \le \sup C$. Therefore we have shown that $$\sup C \le \sup A + \sup B < \sup C + \frac2n$$

for every integer $n \ge 1$. By Theorem I.31, we must have $\sup C = \sup A + \sup B.$ This proves (a).

Theorem I.31 is Theorem 1.2 on the question I asked here:

Is my proof of "Theorem 1.2" correct?

My problem is, I don't understand the steps from adding the inequalities and onwards. Any insight you could provide would be most helpful.

  • $\begingroup$ Do you know what "adding the inequalities" means? Most of these steps are just very simple algebra. $\endgroup$ Jun 10, 2018 at 0:46
  • $\begingroup$ @EricWofsey Yes. I understand how they are added, I just get stuck when amending the $\sup C$, and figuring out whether the sign should be inclusive or exclusive. $\endgroup$
    – user537153
    Jun 10, 2018 at 0:47

1 Answer 1


After adding the inequalities we have $$a+b> \sup A + \sup B - \frac2n.$$

Adding $\frac{2}{n}$ to both sides and flipping the inequality around gives $$\sup A + \sup B < a + b + \frac2n.$$ But $a+b\leq\sup C$, so $$a+b+\frac{2}{n}\leq \sup C+\frac{2}{n}.$$ Combining the last two inequalities we conclude that $$\sup A + \sup B < \sup C+\frac{2}{n}.$$ Since it was shown earlier in the proof that $\sup C \le \sup A + \sup B$, we thus have $$\sup C \le \sup A + \sup B < \sup C + \frac2n.$$ Moreover, $n$ here can be any positive integer. So, Theorem I.31 with $y=2$ tells us that $\sup A+\sup B=\sup C$.

  • $\begingroup$ Thank you. As one quick question, how do you know that $a + b \le \sup C$? $\endgroup$
    – user537153
    Jun 10, 2018 at 1:50
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    $\begingroup$ $a+b$ is an element of $C$. $\endgroup$ Jun 10, 2018 at 1:51
  • $\begingroup$ I can't believe I never thought of that. It all makes sense now. Because your answer has been so helpful, do you have any advice on the intuitive understanding of proofs? I ask because I understand why Apostol's proof is true, but I know that, from scratch, I would not have been able to formulate it myself. I can prove the case for the infimum, but only using the supremum proof as a blueprint. $\endgroup$
    – user537153
    Jun 10, 2018 at 1:57
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    $\begingroup$ When you read proofs like this, I would recommend thinking about what they mean and trying not to get lost in the algebra. For instance, in this proof, $\frac{1}{n}$ is not actually important--it's just some small number (that can be arbitrarily small). So, the idea is, you pick $a$ and $b$ that are really close to $\sup A$ and $\sup B$. Then, $a+b$ will be really close to $\sup A+\sup B$. But $a+b\in C$, so this shows $\sup C$ can't be very much smaller than $\sup A+\sup B$. $\endgroup$ Jun 10, 2018 at 2:01
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    $\begingroup$ @JamieCorkhill I upvoted because I think your question is excellent. I really like Apostol's book. I recommend that you go slow and careful: each time that you are not sure of the validity of one of Apostol's proofs, do just what you did: post it here, showing the work that you did trying to verify the proof. $\endgroup$ Jun 10, 2018 at 2:57

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