# Can you explain the intuition of Tom Apostol's proof of the Additive Property of sets?

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.

• Do you know what "adding the inequalities" means? Most of these steps are just very simple algebra. Jun 10, 2018 at 0:46
• @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.
– user537153
Jun 10, 2018 at 0:47

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 $\sup A+\sup B=\sup C$.

• Thank you. As one quick question, how do you know that $a + b \le \sup C$?
– user537153
Jun 10, 2018 at 1:50
• $a+b$ is an element of $C$. Jun 10, 2018 at 1:51
• 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.
– user537153
Jun 10, 2018 at 1:57
• 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 $\sup A+\sup B$. Jun 10, 2018 at 2:01
• @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. Jun 10, 2018 at 2:57