My Solution to 10.7 in Ross Analysis 2nd Edition

Question

This is not the solution they have in the back of the book, but I want to see if my solution it is also correct.

Problem 10.7: Let $S$ be a bounded nonempty subset of $\mathbb{R}$ such that $\sup S$ is not in $S$. Prove that there is a sequence ($s_n$) of points in S such that $\lim s_n = \sup S$.

My Proof:

Let $n\in \mathbb{N}$, such that $\frac{1}{n} <(\sup S - \inf S)$. Then $\sup S-\frac{1}{n} >\inf S$. Also $\sup S-1/n<\sup S$ so we get $\inf S<\sup S-\frac{1}{n}<\sup S$. Define $s_n=\sup S-\frac{1}{n}$, then $s_n$ is an element of $S$, and $\lim s_n=\sup S$.

Answer 1

Your proof is not correct. You chose an $n$ that specifically satisfies $$\inf S < \sup S - \tfrac{1}{n} < \sup S$$ but it is not entirely correct to say that each $s_k = \sup S - \tfrac{1}{k}$ is in $S$. You see?

Answer 2

Just to complement the other answer: suppose $S= [0,1] \setminus \mathbb Q$, so that $S$ is the set of irrationals between $0$ and $1$. None of the elements of the sequence you construct are in $S$.