My proof is similar in spirit to this one, but I'm not able to figure out if it's correct or not.
Consider $S\subset \mathbb{R}$ that is bounded above. Consider an upper bound of $S$, say $u$. If $u\in S$, we are done, so let $u\notin S$. We will construct $I_n = [a_n,b_n]$ to apply the nested-interval theorem. Define $b_i = u$, for all $i$.
For $a_1$, pick some $s\in S$. Now define $a_n$ recursively (is it okay to make recursive definitions like this one):
- If there exists $s'\in S$, such that $s' > a_i$ put $s_{i+1} = s'$.
- Otherwise, $s_{i+1} = s_i$.
We can see that $$I_1\supseteq I_2 \supseteq …$$
By the nested interval theorem $$\bigcap_{n\ge 1} I_n = [a,b] \neq \varnothing$$
We can now show that $a$ is the least upper bound of $S$.
Is this proof alright?
Best Answer
I think your definition of $a_n$ is a problem. Suppose $S = [0,2]$. This is clearly bounded. Pick $u=3$ and the we can pick $a_n=1-\frac{1}{n}$. We then get a nested series of closed intervals but their intersection is $[1,3]$. $1$ is clearly not an upper bound for $S$.