# Rudin's Principles of Mathematical Analysis Theorem $1.20 (a)$

$$1.20 (a)$$: If $$x,y \in \mathbb{R}$$ and $$x \gt 0$$, then there is a positive integer n such that $$nx \gt y$$.

The proof in the book:

Let $$A$$ be the set of all $$nx$$, where $$n$$ runs through the positive integers. If (a) were false, then $$y$$ would be an upper bound of $$A$$. But then $$A$$ has a least upper bound in $$R$$. Put $$\alpha = \sup A$$. Since $$x \gt 0$$, $$\alpha -x \lt \alpha$$, and $$\alpha -x$$ is not an upper bound of $$A$$. Hence $$\alpha -x \lt mx$$ for some positive integer $$m$$. But then $$\alpha \lt (m+1)x \in A$$, which is impossible, since $$\alpha$$ is an upper bound of $$A$$. $$\blacksquare$$

My question: for $$A$$ to have an upper bound in $$R$$ it also has to be non-empty. The author doesn't say that anywhere in the proof. Why? The proof must be incomplete, right?

• Well $x\in A$ so it is non-empty. Commented Jul 5, 2019 at 17:03
• Should I delete this question? Commented Jul 5, 2019 at 17:05
• @RUBENGONÇALOMOROUÇO No need to delete the question, it is a legitimate query. Your question could help others who also have the same question and find this post. Do consider upvoting and accepting an answer if you find it helpful, though.
– user279515
Commented Jul 5, 2019 at 17:11
• In general, Rudin doesn't spell out every detail in his proofs, expecting the reader to fill in anything that's missing. This is a mild example; he leaves more details to the reader in later chapters. This is not out of laziness on Rudin's part; he wants the reader to be actively engaged in order to learn more effectively.
– user169852
Commented Jul 5, 2019 at 18:02

It is clear from the context that $$A$$ is non empty, since $$A=\{nx: n \in \Bbb N\}$$, so atleast your $$x=1 \cdot x \in A$$