# Elaborating on Abbot's Nested Interval Property

I'm currently reading Abbot's Understanding Analysis, where in page 19 he introduces the Nested Interval Property and writes "because the intervals are nested, we see that every $b_n$ serves as an upper bound for $A$." I was wondering if this claim could be shown more rigorously, and have provided a proof I attempted. Please let me know if there is any error in understanding, or if the proof is sufficient. Thank you in advance.

Hypothesis: Let $A=\{a_n:n\in\mathbb{N}\}$. For all $n,m\in\mathbb{N}$, where $n>m$, $a_n<b_n$, $a_n<a_m$ and $b_n>b_m$. These are the assumptions I made which are relevant to the proof.

Claim: For all $n\in\mathbb{N}$, $b_n$ is an upper bound to $A$.

Proof: Let $i,j\in\mathbb{N}$. We proceed to show that $a_i<b_j$. If $i>j$, then $b_i\leq b_j$; since $a_i<b_i$, we have $a_i<b_j$. If $i<j$, then $b_j\leq b_i$; since $a_i\leq a_j$ and $a_j<b_j$, we have $a_i<b_j$. If $i=j$, then $a_i=a_j<b_j$. Therefore, $a_i<b_j$.

• Looks fine. ${}$ Commented May 14, 2018 at 18:51

Thus $a_1$ is an upper bound of A.
Since $a_1 \lt b_1$ it follows that $b_1$ and all subsequent