# Proof Verification: Nested Interval Theorem $\implies$ Completeness of $\mathbb{R}$

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):

1. If there exists $$s'\in S$$, such that $$s' > a_i$$ put $$s_{i+1} = s'$$.
2. 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?

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 $$S$$.
• I would try something like this, that controls the length of the intervals $I_n$: Pick $a_1$ some element in $S$, $b_1$ an upper bound, let $I_1=[a_1,b_1]$. Then let $m_1 = \frac{a_1+b_1}{2}$. If $[m_1,b_1]$ contains no element of $S$, pick $I_2=[a_1,m_1]$. Otherwise, pick $I_2=[m_1,b_1]$. Rename the endpoints as $I_2=[a_2,b_2]$ and proceed recursively. Jan 18, 2021 at 7:25